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State a condition which, in complete metric spaces, is
equivalent to the existence of the limit of a function at a
point. This is analogous to the Cauchy criterion for the limit
of a sequence in a complete metric space.

(We're talking about the limit of function f at point x0.)

For every positive e, there is a positive b such that for every x,y satisfying d(x,x0)<b and d(y,x0)<b, it follows that g(f(x),f(y))<e.

page 157 in 1st analysis

For every positive e, there is a positive b such that for every x,y satisfying d(x,x0)<b and d(y,x0)<b, it follows that g(f(x),f(y))<e.

page 157 in 1st analysis

Consider a real-valued function defined on a bounded open
interval. If this function is bounded and continuous, does it
have to be uniformly continuous?

NO. sin(1/x) defined on (0,1)

If it were uniformly continuous, it could be extended to a continuous function on [0,1]. This is impossible because the limit of sin(1/x) as x approaches zero doesn't exist. Hence this function is not uniformly continuous.

If it were uniformly continuous, it could be extended to a continuous function on [0,1]. This is impossible because the limit of sin(1/x) as x approaches zero doesn't exist. Hence this function is not uniformly continuous.

A, B, C are sets.

A \ (B \ C) = ?

A \ (B \ C) = ?

A \ (B \ C) = (A \ B) u (A n C)

Let f be a real-valued function defined on a connected subset
of |R. Suppose that f is differentiable. What is the relation
between the derivative of f and whether f satisfies a Lipschitz
condition? Give the answer without proof.

For every real K, these two conditions are equivalent.

(1) |f'(x)| <= K, for all x

(2) |f(x)-f(y)| <= K|x-y|, for all x,y

(The proof requires Lagrange's mean-value theorem.)

page 156 in golden gate

(1) |f'(x)| <= K, for all x

(2) |f(x)-f(y)| <= K|x-y|, for all x,y

(The proof requires Lagrange's mean-value theorem.)

page 156 in golden gate

Prove that for each natural number n,

e - (1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!) < 1/(n * n!).

e - (1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!) < 1/(n * n!).

page 79 in OLDTIMER

page 176 in the first notebook for analysis

page 176 in the first notebook for analysis

Prove that the number e is irrational.

Hint: use the fact that for each natural number n,

e - (1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!) < 1/(n * n!).

page 175 in the first notebook for analysis

e - (1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!) < 1/(n * n!).

page 175 in the first notebook for analysis

What is the formula for the n-th derivative of the product of
two functions? (Supposing that both the functions have the n-th
derivative.)

the sum from k=0 to k=n

(n) (n-k) (k)

(k)f g

+(k=0 to k=n) Newton(n,k) * f^(n-k) * g^(k)

page 171 in the palace notebook

(n) (n-k) (k)

(k)f g

+(k=0 to k=n) Newton(n,k) * f^(n-k) * g^(k)

page 171 in the palace notebook

What is the formula for the n-th derivative of sin?

sin(n)(x) = sin(x+n*pi/2)

What is the formula for the n-th derivative of cos?

cos(n)(x) = cos(x+n*pi/2)

a,b >= 0 and a+b=1, x<=y

Prove that x <= ax+by <= y.

Prove that x <= ax+by <= y.

very easy

Prove that if x <= c <= y, then c = ax+by, for some a,b>=0,
a+b=1.

a = (y-c)/(y-x)

b = (c-x)/(y-x)

b = (c-x)/(y-x)

Show a function which is defined on the real line and
continuous at only one point.

0 for rationals,

x for irrationals

continuous only at 0

x for irrationals

continuous only at 0

Show a function which is defined on the real line and
differentiable only at one point.

0 for rationals,

x*x for irrationals

differentiable only at 0, because continuous only at 0

x*x for irrationals

differentiable only at 0, because continuous only at 0

Sketch the proof of Rolle's theorem.

First prove that the function is either constant or has a local
extremum.

Later use the fact that if a differentiable function has a local extremum at a point,

then its derivative is equal to zero at that point.

page 154 in Golden Gate

Later use the fact that if a differentiable function has a local extremum at a point,

then its derivative is equal to zero at that point.

page 154 in Golden Gate

State and prove Cauchy's Mean-value Theorem.

You can refer to Rolle's theorem.

You can refer to Rolle's theorem.

page 155 in Golden Gate

Consider a real-valued differentiable function defined on a
bounded closed interval. Does it have to be a Lipschitz function?

NO. g:[0,1]->|R, g(0) = 0, g(x) = x*x*sin(exp(1/x))

This function is differentiable, g'(0) = 0, but g' is not bounded. Hence it is not Lipschitz.

Put x[n]=1/log(n*2*pi). Show that lim(n->oo) g'(x[n]) = -oo.

Thus the derivative is not bounded.

This function is differentiable, g'(0) = 0, but g' is not bounded. Hence it is not Lipschitz.

Put x[n]=1/log(n*2*pi). Show that lim(n->oo) g'(x[n]) = -oo.

Thus the derivative is not bounded.

Let f be a real-valued function defined on an interval on the
real line. Suppose that it is monotonic and that its image is
an interval. Does the function have to be continuous?

YES.

Use the fact that a bounded monotonic function has a limit.

Use the fact that a bounded monotonic function has a limit.

Let f be a function from a metric space, say X, to a complete
metric space. Let A be a nonclosed subset of X. Let f be a
uniformly continuous function on A. Can f be extended to the
closure of A so that the extended function is continuous?

YES. And what's more, the extension is uniformly continuous.

Completeness is essential. In the proof, we use the Cauchy criterion for the existence of the limit of a function at a point.

Completeness is essential. In the proof, we use the Cauchy criterion for the existence of the limit of a function at a point.

Consider a function from one metric space into another. Suppose
it is continuous and uniformly continuous on a dense set. Does
this function have to be uniformly continuous?

YES. The proof can be performed routinely.

(1) f : R -> R

(2) f has n derivatives

(3) the n-th derivative is constantly zero

What can we infer about this function?

(2) f has n derivatives

(3) the n-th derivative is constantly zero

What can we infer about this function?

It's a polynomial of degree less than n.

We can use Taylor's formula for an easy proof, or we can prove it by induction.

We can use Taylor's formula for an easy proof, or we can prove it by induction.

Prove that for each non-closed subset of |R, there exists a
continuous function defined on this set, which is not uniformly
continuous.

If p is the point belonging to the closure of the set but not
to the set itself, put f(x) = log(|x-p|).

page 185 in GOLDEN GATE

page 185 in GOLDEN GATE

Prove that for each non-closed subset of |R, there exists a
bounded continuous function, which has no maximum.

If p is the point belonging to the closure of the set but not
to the set itself, put f(x)=1/(1+|x-p|).

(1) g : (0,oo) -> |R is diffable

(2) lim(x->oo) g(x)+g'(x) = A; A:-|R*

What can we conclude from this?

(2) lim(x->oo) g(x)+g'(x) = A; A:-|R*

What can we conclude from this?

(3) lim(x->oo) g(x) = A

Hint: use the exponential function and L'Hospital's Rule.

page 190 in golden gate

Hint: use the exponential function and L'Hospital's Rule.

page 190 in golden gate

Let f be a real-valued function defined on an interval. We
informally say that f is convex iff for every two points x,y
belonging to the domain of the function the segment joining
points (x,f(x)) and (y,(f(y)) lies above the graph of the
function.

Provide two formalizations of this idea.

Provide two formalizations of this idea.

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<c<y,

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<c<y,

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

f : I -> R, I is an interval on the real line

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<=c<=y or y<=c<=x

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<=c<=y or y<=c<=x

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

Hint:

a = (y-c)/(y-x)

b = (c-x)/(y-x)

a = (y-c)/(y-x)

b = (c-x)/(y-x)

f:I->|R; I is an interval on the real line

(1) /\(x,y:-I) /\(a,b>=0, a+b=1) f(ax+by) <= af(x) + bf(y)

(2) /\(x<c<y :- I) f(c) <= ((f(y)-f(x))/(y-x))*(c-x) + f(x)

Prove that (2) => (1)

(1) /\(x,y:-I) /\(a,b>=0, a+b=1) f(ax+by) <= af(x) + bf(y)

(2) /\(x<c<y :- I) f(c) <= ((f(y)-f(x))/(y-x))*(c-x) + f(x)

Prove that (2) => (1)

page 103 in palace

f : I -> R, I is an interval on the real line

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every x<y, for every x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every x<y, for every x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

Hint:

a,b>=0, a+b=1,

c = ax + by,

x<=c<=y,

a = (y-c)/(y-x)

b = (c-x)/(y-x)

a,b>=0, a+b=1,

c = ax + by,

x<=c<=y,

a = (y-c)/(y-x)

b = (c-x)/(y-x)

(1) J is an open and connected subset of |R

(2) f:J->|R is a convex function

Does f have to be continuous?

(2) f:J->|R is a convex function

Does f have to be continuous?

YES.

page 115 in palace

page 115 in palace

f:A->B; g:B->R; A,B are intervals on the real line.

f,g are convex and g in increasing

Does g(f(x)) have to be convex?

f,g are convex and g in increasing

Does g(f(x)) have to be convex?

YES.

Let I be an interval on the real line. Let f : I -> |R.

(1) /\x,y:-I f(x/2 + y/2) <= f(x)/2 + f(y)/2

(2) f is continuous

Does f have to be convex?

(1) /\x,y:-I f(x/2 + y/2) <= f(x)/2 + f(y)/2

(2) f is continuous

Does f have to be convex?

Yes.

Show that for every natural n, for all natural p,q such that p+q=2^n,

f(p/2^n * x + q/2^n * y) <= p/2^n * f(x) + q/2^n * f(y).

Then show that an appropriate set is dense and then use continuity to conclude the thesis.

Show that for every natural n, for all natural p,q such that p+q=2^n,

f(p/2^n * x + q/2^n * y) <= p/2^n * f(x) + q/2^n * f(y).

Then show that an appropriate set is dense and then use continuity to conclude the thesis.

f:I->R; I is an interval on the real line; f is differentiable

(1) f is convex

(2) f' is increasing

Prove that (1) => (2)

(1) f is convex

(2) f' is increasing

Prove that (1) => (2)

Hint:

x<c<y from I, (f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

x<c<y from I, (f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

f:I->R; I is an interval on the real line; f is differentiable

(1) f is convex

(2) f' is increasing

Prove that (2) => (1)

(1) f is convex

(2) f' is increasing

Prove that (2) => (1)

Use Lagrange's mean-value theorem.

page 119 in palace

page 119 in palace

f, g, F : I -> R; I is an interval

F = max{f,g}

f, g are convex, does F have to be convex?

F = max{f,g}

f, g are convex, does F have to be convex?

Yes.

f, g, F : I -> R; I is an interval

F = min{f,g}

f, g are convex, does F have to be convex?

F = min{f,g}

f, g are convex, does F have to be convex?

No.

f(x)=7

g(x)=x

f(x)=7

g(x)=x

Prove Jensen's inequality for a convex function.

page 122 in palace

Prove that

(a1 * a2 * ... * an)^(1/n) <= (a1 + a2 + ... + an)/n

(a1 * a2 * ... * an)^(1/n) <= (a1 + a2 + ... + an)/n

Use Jensen's inequality for the minus logarithmic function,
which is convex.

Prove that

n / (1/a1 + 1/a2 + ... + 1/an) <= (a1 * a2 * ... * an)^(1/n)

n / (1/a1 + 1/a2 + ... + 1/an) <= (a1 * a2 * ... * an)^(1/n)

Use: (a1 * a2 * ... * an)^(1/n) <= (a1 + a2 + ... + an)/n

(1) f:(0;oo)->|R

(2) f is convex

(3) f is strictly increasing

Prove that lim f(x) = infinity, as x approaches infinity.

(2) f is convex

(3) f is strictly increasing

Prove that lim f(x) = infinity, as x approaches infinity.

Choose 0<a<b. Put k = (f(b) - f(a)) / (b-a). By (3) k>0.

Since f is convex conclude that

for all x > b, kx + (f(b)-kb) <= f(x).

And hence conclude the thesis.

Since f is convex conclude that

for all x > b, kx + (f(b)-kb) <= f(x).

And hence conclude the thesis.

f,g : I -> R; I is an interval on the real line;

f,g are convex; a,b are positive numbers

Prove that af+bg is convex.

f,g are convex; a,b are positive numbers

Prove that af+bg is convex.

page 124 in palace

(1) f : [0,oo) -> |R

(2) f is convex

(3) f(0) = 0

(4) g : (0,oo) -> |R , g(x)=f(x)/x

What can we conclude about function g?

(2) f is convex

(3) f(0) = 0

(4) g : (0,oo) -> |R , g(x)=f(x)/x

What can we conclude about function g?

g is increasing

Hint: If f is convex, then for all x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c).

Use f(0)=0.

page 124 in the palace notebook

Hint: If f is convex, then for all x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c).

Use f(0)=0.

page 124 in the palace notebook

0<x<1

Prove that lim x^n = 0, as n approaches infinity.

Prove that lim x^n = 0, as n approaches infinity.

Put d = 1/x - 1.

Use (1+d)^n >= 1 + n*d.

page 127 in palace

Use (1+d)^n >= 1 + n*d.

page 127 in palace

Let a(n) be a sequence of nonnegative numbers.

Suppose that lim_sup a(n)^(1/n) < 1.

What can we infer about the series a(n)?

Suppose that lim_sup a(n)^(1/n) < 1.

What can we infer about the series a(n)?

It converges.

(this is Cauchy's test)

(this is Cauchy's test)

Let a(n) be a sequence of nonnegative numbers.

Suppose that lim_inf a(n)^(1/n) > 1.

What can we infer about the series a(n)?

Suppose that lim_inf a(n)^(1/n) > 1.

What can we infer about the series a(n)?

It diverges.

page 130 in palace

page 130 in palace

Let a(n) be a sequence of nonnegative numbers.

Suppose that lim_sup a(n)^(1/n) = 1.

What can we infer about the series a(n)?

Suppose that lim_sup a(n)^(1/n) = 1.

What can we infer about the series a(n)?

Nothing as far as convergence goes. It can converge or diverge,
the stipulated condition is inconclusive.

series 1/n diverges

series 1/n^2 converges

series 1/n diverges

series 1/n^2 converges

Let a(n) be a sequence of positive numbers.

Suppose that lim_sup a(n+1)/a(n) < 1.

What can we infer about the series a(n)?

Suppose that lim_sup a(n+1)/a(n) < 1.

What can we infer about the series a(n)?

It converges.

This is the ratio test, or D'Alembert's test.

This is the ratio test, or D'Alembert's test.

Let a(n) be a sequence of positive numbers.

Suppose that lim_inf a(n+1)/a(n) > 1.

What can we infer about the series a(n)?

Suppose that lim_inf a(n+1)/a(n) > 1.

What can we infer about the series a(n)?

It diverges.

This is the ratio test, or D'Alembert's test.

This is the ratio test, or D'Alembert's test.

State and prove Cauchy's Integral Test for convergence of series.

f : [1,oo) -> [0,oo) decreasing

(1) Riemann_integral_from_1_to_x_of_f converges, as x approaches infinity

(2) series f(n) converges

thesis: (1) <=> (2)

page 133 in palace

(1) Riemann_integral_from_1_to_x_of_f converges, as x approaches infinity

(2) series f(n) converges

thesis: (1) <=> (2)

page 133 in palace

Using Cauchy's Integral Test show that the series 1/n diverges.

hint: log(x) tends to infinity, as x approaches infinity

page 133 in golden gate

page 133 in golden gate

Using Cauchy's Integral Test investigate the convergence of the
series 1/n^A, where A is a real number.

converges for A>1

diverges for A<=1

page 135 in palace

diverges for A<=1

page 135 in palace

Let a(n), b(n) be sequences of positive numbers.

(1) series b(n) converges

(2) lim_sup a(n)/b(n) = K, where K is a real number

What can we infer about the series a(n)?

(1) series b(n) converges

(2) lim_sup a(n)/b(n) = K, where K is a real number

What can we infer about the series a(n)?

converges

Let a(n), b(n) be sequences of positive numbers.

(1) series b(n) diverges

(2) lim_inf a(n)/b(n) is greater than zero

What can we infer about the series a(n)?

(1) series b(n) diverges

(2) lim_inf a(n)/b(n) is greater than zero

What can we infer about the series a(n)?

diverges

Let a(n), b(n) be sequences of positive numbers.

(1) series b(n) diverges

(2) lim_inf a(n)/b(n) = K, where K is a real number

What can we infer about the series a(n)?

(1) series b(n) diverges

(2) lim_inf a(n)/b(n) = K, where K is a real number

What can we infer about the series a(n)?

If K>0, then the series a(n) diverges.

If K=0, then we cannot infer anything.

a(n)=1/n^2, b(n)=1/n, a(n) converges

a(n)=1/n, b(n)=1/log(n), a(n) diverges

and in both cases the limit is zero

If K=0, then we cannot infer anything.

a(n)=1/n^2, b(n)=1/n, a(n) converges

a(n)=1/n, b(n)=1/log(n), a(n) diverges

and in both cases the limit is zero

Let a(n) be a sequence of nonnegative numbers.

Suppose lim_sup a(n)^(1/n) > 1.

What can we infer about the series a(n)?

Suppose lim_sup a(n)^(1/n) > 1.

What can we infer about the series a(n)?

diverges

(This is Cauchy's test.)

(This is Cauchy's test.)

(1) the series a(n) is bounded, it is a series of complex numbers

(2) b(n) tends to zero

(3) b(n) is monotonic

What can we infer about the series a(n)b(n)?

(2) b(n) tends to zero

(3) b(n) is monotonic

What can we infer about the series a(n)b(n)?

converges

This is Dirichlet-Abel test.

page 137 in palace

This is Dirichlet-Abel test.

page 137 in palace

Let a(n) be a sequence of complex numbers, b(n) of real numbers.

(1) the series a(n) converges

(2) b(n) is monotonic

(3) b(n) is bounded

What can we infer about the series a(n)b(n)?

(1) the series a(n) converges

(2) b(n) is monotonic

(3) b(n) is bounded

What can we infer about the series a(n)b(n)?

converges

Use Dirichlet-Abel test.

Hint:

(1) the series a(n) is bounded

(2) {b(n)-g} is a monotonic sequence which tends to zero

Use Dirichlet-Abel test.

Hint:

(1) the series a(n) is bounded

(2) {b(n)-g} is a monotonic sequence which tends to zero

[ g(f(x)) ]' = ?

Derive the formula.

Derive the formula.

[ g(f(x)) ]' = g'(f(x)) * f'(x)

Warning: Do not divide by zero!

Warning: Do not divide by zero!

Consider a convex function defined on an open interval (a,b).
Does this function have to have limits at points a and b?

Yes.

Hint: a monotonic function has a limit.

The limits may be improper, though.

Hint: a monotonic function has a limit.

The limits may be improper, though.

Let s be a fixed real number.

Knowing the derivative of the exponential function,

derive the formula (x^s)' = ?

Knowing the derivative of the exponential function,

derive the formula (x^s)' = ?

Use log.

page 190 in 1st analysis

page 190 in 1st analysis

Consider a series of positive numbers.

Suppose that lim_sup a(n+1)/a(n) > 1.

Does the series have to diverge?

Suppose that lim_sup a(n+1)/a(n) > 1.

Does the series have to diverge?

NO.

1/2 + 1/3 + (1/2)^2 + (1/3)^2 + (1/2)^3 + (1/3)^3 + ...

converges

1/2 + 1/3 + (1/2)^2 + (1/3)^2 + (1/2)^3 + (1/3)^3 + ...

converges

Test the series (n!)/(n^n) for convergence.

converges, the ratio test

Test the series (n!)^2 / (n^n) for convergence.

diverges, the ratio test

Test the series 1/ n^(1+1/n) for convergence.

diverges, compare with the series 1/n

Let a[n] be a decreasing sequence of positive numbers.

(1) the series a[n] converges

(2) the series 2^k * a[2^k] converges

Prove that (2)=>(1).

(1) the series a[n] converges

(2) the series 2^k * a[2^k] converges

Prove that (2)=>(1).

page 44 in redcrest

page 58 in the first analysis notebook

page 58 in the first analysis notebook

Test the series 1/ 2^(sqrt(n)) for convergence.

converges, 2^k*a[2^k]

2^(A^k-k) > 2^k, where A>1

2^(A^k-k) > 2^k, where A>1

Test the series 1/ 2^log(n) for convergence.

diverges, 2^k*a(2^k)

Test the series 1/sqrt(n) * tan(1/sqrt(n)) for convergence.

diverges

hint: lim tan(x)/x = 1 (x->0). compare with 1/n

hint: lim tan(x)/x = 1 (x->0). compare with 1/n

Test the series ( 1/2 * arctan(n) )^n for convergence.

converges, Cauchy's test

Test the series (1-1/n)^(n^2) for convergence.

converges, Cauchy's test

b^x = y

Read in English: log(b,y) = x

Read in English: log(b,y) = x

x = the logarithm of y to the base b

Test the series 1/log(n!) for convergence.

diverges

show that the series 1/(n*log(n)) diverges,

and then use: 2^k*a(2^k)

show that the series 1/(n*log(n)) diverges,

and then use: 2^k*a(2^k)

Test the series 1/n * sin(pi/4 * n) for convergence.

converges, Dirichlet-Abel test

Test the series (-1)^n / (n - (-1)^n) for convergence.

The series converges.

Define S[n] = +(k=1 to n) (-1)^k / (k - (-1)^k).

Prove that lim(n->oo) S[2n] = log(2).

Prove that lim(n->oo) S[2n+1] = log(2), too.

See page 83 in OLDTIMER about this log(2).

The number log(2) is not essential here.

Define S[n] = +(k=1 to n) (-1)^k / (k - (-1)^k).

Prove that lim(n->oo) S[2n] = log(2).

Prove that lim(n->oo) S[2n+1] = log(2), too.

See page 83 in OLDTIMER about this log(2).

The number log(2) is not essential here.

The series a(n) converges.

Does the series a(n)*a(n) have to converge?

Does the series a(n)*a(n) have to converge?

NO.

a(n) = (-1)^n / sqrt(n)

a(n) = (-1)^n / sqrt(n)

Let a(n) be a sequence of positive numbers.

Suppose that the series a(n) converges.

Does the series a(n)*a(n) have to converge?

Suppose that the series a(n) converges.

Does the series a(n)*a(n) have to converge?

Yes.

Let a(n) be a sequence of positive numbers.

Suppose that the series a(n) converges.

Does the series n*a(n) / (n+a(n)) have to converge?

Suppose that the series a(n) converges.

Does the series n*a(n) / (n+a(n)) have to converge?

Yes.

n/(n+a(n)) is less than 1

n/(n+a(n)) is less than 1

Let f be a convex function defined on an interval on the real
line. Does it have to be continuous?

NO.

f:[0,1] -> R

f(1) = 1

f(x) = 0

This function is convex but is not continuous.

f:[0,1] -> R

f(1) = 1

f(x) = 0

This function is convex but is not continuous.

What is a contraction?

(X,d) metric space

f : X -> X

f is a contraction iff there exists 0 < K < 1 such that for every x,y belonging to X we have d( f(x), f(y) ) <= K * d(x,y).

f : X -> X

f is a contraction iff there exists 0 < K < 1 such that for every x,y belonging to X we have d( f(x), f(y) ) <= K * d(x,y).

State and prove Banach's Contraction Principle.

In a complete metric space every contraction has a unique fixed
point.

(x is a fixed point <=> f(x) = x)

Also known as Banach's Fixed Point Theorem.

page 56 in golden gate

(x is a fixed point <=> f(x) = x)

Also known as Banach's Fixed Point Theorem.

page 56 in golden gate

Prove that in a complete metric space, a countable intersection
of a collection of open dense sets is dense.

page 133 in golden gate

Baire Category Theorem

Baire Category Theorem

Prove that if a complete metric space is written as a union of
countably many closed sets, then at least one of those closed
sets contains a ball.

Use the fact that in a complete metric space, a countable
intersection of a collection of open dense sets is dense.

page 133 in golden gate

page 133 in golden gate

Define a totally bounded metric space.

For every e>0, the space can be written as a finite union of
sets with diameter less than e.

page 140 in palace

page 140 in palace

What is the name for a metric space satisfying:

For every e>0, the space can be written as a finite union of sets with diameter less than e.

For every e>0, the space can be written as a finite union of sets with diameter less than e.

a totally bounded metric space

Prove that a compact metric space is totally bounded.

Suppose that the space is not totally bounded.

Construct a sequence which has no convergent subsequence.

Construct a sequence which has no convergent subsequence.

Prove that if every sequence contained in a metric space has a
Cauchy subsequence, then this space is totally bounded.

Suppose that the space is not totally bounded.

Construct a sequence which has no Cauchy subsequence.

page 141 in palace

Construct a sequence which has no Cauchy subsequence.

page 141 in palace

Prove that if a metric space is totally bounded, then every
sequence contained in this space has a Cauchy subsequence.

Notice that for a given e>0, every infinite set has an infinite
subset with diameter less than e. Use this fact to construct a
Cauchy subsequence of any sequence.

page 141 in palace

page 141 in palace

What is a finitely bound collection of sets?

Let W be a collection of sets.

W is finitely bound iff its every finite subcollection has nonempty intersection.

W is finitely bound iff every finite intersection of sets in W is nonempty.

W is finitely bound iff its every finite subcollection has nonempty intersection.

W is finitely bound iff every finite intersection of sets in W is nonempty.

What does it mean that a collection of sets has the finite
intersection property?

Let W be a collection of sets.

(Every finite subcollection of W has non-empty intersection.)

W has the finite intersection property iff every finite intersection of sets belonging to W has nonempty intersection.

(1) W is a collection of sets with the finite interec

(Every finite subcollection of W has non-empty intersection.)

W has the finite intersection property iff every finite intersection of sets belonging to W has nonempty intersection.

(1) W is a collection of sets with the finite interec

Prove that in a sequentially compact metric space every
finitely bound collection of closed sets has nonempty
intersection.

page 142 in golden gate

Let W be a finitely bound collection of closed subsets of a sequentially compact metric space X. We know that the space is totally bounded. Hence, for every natural n, X is a finite union of balls of radius 1/n. Suppose that for each of these balls there is a set in W which has empty intersection with this ball. By the nature of W, this finite collection of

Let W be a finitely bound collection of closed subsets of a sequentially compact metric space X. We know that the space is totally bounded. Hence, for every natural n, X is a finite union of balls of radius 1/n. Suppose that for each of these balls there is a set in W which has empty intersection with this ball. By the nature of W, this finite collection of

If there is a point belonging to the intersection of a
collection of sets, we say that this collection has ___________.

it has nonempty intersection

When no point belongs to the intersection of a collection of
sets, we say that the collection has ___________.

it has empty intersection

Contemplate the fact that from every strictly monotonic
function f:R->R we can construct a metric on R which is similar
to the Euclidean metric.

d(x,y) = |f(x)-f(y)|

It's a metric because f is injective.

It's similar to the Euclidean metric because f is continuous and so is its inverse.

It's a metric because f is injective.

It's similar to the Euclidean metric because f is continuous and so is its inverse.

0 <= a <= x

0 <= b <= x

Contemplate the fact that |b-a| <= x.

0 <= b <= x

Contemplate the fact that |b-a| <= x.

_____

Prove that for a,b >= 0, sqrt(a+b) <= sqrt(a) + sqrt(b).

Square it.

Investigate the uniform convergence of f(n,x) = x/(1+nx), on
[0,1].

YES.

hint: 1 + nx >= nx

hint: 1 + nx >= nx

Investigate the uniform convergence of

f[n](x) = sqrt(n) * x * (1-x*x)^n on [0,1].

f[n](x) = sqrt(n) * x * (1-x*x)^n on [0,1].

No uniform convergence on [0,1].

Use: x[n] = 1 / sqrt(n)

If 0<M<1, then uniform convergence on [M,1].

Use: x[n] = 1 / sqrt(n)

If 0<M<1, then uniform convergence on [M,1].

Investigate the uniform convergence of

the series x / ( (1+nx)*(1+(n+1)x) ).

(x>=0)

the series x / ( (1+nx)*(1+(n+1)x) ).

(x>=0)

It does not converge uniformly on [0,oo).

It converges uniformly on [M,oo) for all M>0.

hint: f[n](x) = 1 / (1+nx); f[n](x)-f[n+1](x)

It converges uniformly on [M,oo) for all M>0.

hint: f[n](x) = 1 / (1+nx); f[n](x)-f[n+1](x)

f(n,x) = x^n / (n + x^n), x>=0.

Investigate the uniform convergence of f.

Investigate the uniform convergence of f.

It does not converge uniformly on [0,oo).

It does not converge uniformly on (1,oo).

It converges uniformly on [0,1].

It converges uniformly on [M,oo) for all M>1.

It does not converge uniformly on (1,oo).

It converges uniformly on [0,1].

It converges uniformly on [M,oo) for all M>1.

Given two complex numbers A and B, B != 0, write the equation
of a line passing through points A and A+B.

Im( (z-A)/B ) = 0

see page 188 in the palace notebook

see page 188 in the palace notebook

What are the three equivalent definitions of a line on the
complex plane?

1) Im( (z-a)/b ) = 0, where a,b are complex numbers, b != 0

2) A*Re(z) + B*Im(z) + C = 0, where A,B,C are real numbers, A and B are not both zero

3) |z-A| = |z-B|, where A,B are distinct complex numbers

2) A*Re(z) + B*Im(z) + C = 0, where A,B,C are real numbers, A and B are not both zero

3) |z-A| = |z-B|, where A,B are distinct complex numbers

Prove that for every complex number A,

it is false that lim cos(n*A) = 0.

it is false that lim cos(n*A) = 0.

Use: lim cos(2*n*A) = 0.

page 189 in the palace notebook

page 189 in the palace notebook

Let z be a complex number other than zero. Let n be a natural
number. Describe the set {w:-C : w^n = z}.

|z|^(1/n) * exp( (i*(arg(z) + k*2*pi)) / n )

for some k=0,1,...,n-1

for some k=0,1,...,n-1

Let p be a real number.

What is the radius of convergence of the power series (n^p * z^n)?

What is the radius of convergence of the power series (n^p * z^n)?

R = 1

Find the radius of convergence of the power series

z^(2*n) / (n * 2^n).

z^(2*n) / (n * 2^n).

the square root of 2

Let z be a complex number such that |z| >= 1.

Prove that the series (z^n) diverges.

Prove that the series (z^n) diverges.

(1) |z|>=1

(2) not (|z|^n -> 0)

(3) not (|z^n| -> 0)

(4) not ( z^n -> 0)

(5) series (z^n) diverges

page 189 in OLDTIMER

(2) not (|z|^n -> 0)

(3) not (|z^n| -> 0)

(4) not ( z^n -> 0)

(5) series (z^n) diverges

page 189 in OLDTIMER

(a+b)! / (a! * b!) = ???

the Newton binomial

1) a+b over a

2) a+b over b

1) a+b over a

2) a+b over b

Define sin(z) for complex numbers using a series.

sin(z) = +(n=0 to oo) [ (-1)^n * z^(2n+1) / (2n+1)! ]

page 72 in palace

page 72 in palace

Define cos(z) for complex numbers using a series.

cos(z) = +(n=0 to oo) [ (-1)^n * z^(2n) / (2n)! ]

page 72 in palace

page 72 in palace

Define cos(z) for complex numbers using exp.

cos(z) = ( exp(iz) + exp(-iz) ) / 2

page 69 in palace

page 69 in palace

Define sin(z) for complex numbers using exp.

sin(z) = ( exp(iz) - exp(-iz) ) / 2i

page 69 in palace

page 69 in palace

Given that exp(x), cos(x), sin(x) are defined for real numbers,
define exp(z) for complex numbers.

exp(x + iy) = exp(x) * ( cos(y) + i*sin(y) )

Where does the series log( 1 + x*x / n*log(n)*log(n) ) converge ?

(x is real)

(x is real)

converges for all x

hint:

log(1+x)<=x

2^k*a(2^k)

hint:

log(1+x)<=x

2^k*a(2^k)

Is it true that when a power series with center 0 converges for
z0, then it converges absolutely for |z|<|z0| ?

Yes. This proof is very easy.

Suppose that the limit |a[n]| / |a[n+1]| exists. Show that it
is the radius of convergence of the power series a[n]*z^n.

Use the ratio test for convergence of series.

Prove that 1 / lim_sup |a[n]| ^ (1/n) is the radius of
convergence of the power series a[n]*z^n.

Use the Cauchy test for convergence of series.

Find the radius of convergence of the series

x^n / ( 2^n + 3^n + (-1)^n * (3^n - 2^n) ).

x^n / ( 2^n + 3^n + (-1)^n * (3^n - 2^n) ).

R = 2

Use Cauchy test.

Use Cauchy test.

Find the radius of convergence of the series ( n! / 2^(n*n) ) *
x^n.

R = oo.

the ratio test

the ratio test

Let p[n] be the sequence of consecutive prime numbers. Find the
radius of convergence of the power series x^p[n] / p[n].

R = 1

Notice that the disputed series converges for -1.

Hence the radius must be at least 1.

Notice that if a>1, then the series diverges for a.

Hence the radius cannot be greater than 1.

Conclusion: R=1

Notice that the disputed series converges for -1.

Hence the radius must be at least 1.

Notice that if a>1, then the series diverges for a.

Hence the radius cannot be greater than 1.

Conclusion: R=1

Let p[n] be the sequence of consecutive prime numbers. Find the
radius of convergence of the power series p[n] * x^p[n].

R = 1

way1:

a[n] = { n for prime n; 0 otherwise }

Use the Cauchy test.

way2:

0<x<1 => +(n=1 to oo) p[n]*x^p[n] < +(n=1 to oo) n*x^n < oo

way1:

a[n] = { n for prime n; 0 otherwise }

Use the Cauchy test.

way2:

0<x<1 => +(n=1 to oo) p[n]*x^p[n] < +(n=1 to oo) n*x^n < oo

Where does this series converge?

1/(x^n) * sin(1/(2^n))

1/(x^n) * sin(1/(2^n))

|x| > 1/2

Hint: lim sin(x)/x = 1 as x approaches 0.

Hint: lim sin(x)/x = 1 as x approaches 0.

By referring to the Euclidean two-dimensional space, justify
intuitively why

[A x B = 0] <=> [A,B are perpendicular].

A x B is the scalar product (a,b)x(c,d) = (a*c,b*d)

[A x B = 0] <=> [A,B are perpendicular].

A x B is the scalar product (a,b)x(c,d) = (a*c,b*d)

Hint: |A-B| = |A+B|

Prove that for all real a,b

sqrt(a*a + b*b) <= |a| + |b|

sqrt(a*a + b*b) <= |a| + |b|

______

Prove that for all real a,b

1/sqrt(2) * (|a| + |b|) <= sqrt(a*a + b*b)

1/sqrt(2) * (|a| + |b|) <= sqrt(a*a + b*b)

=

Investigate the limit

lim xy / (x*x + y*y) as (x,y) approaches (0,0).

lim xy / (x*x + y*y) as (x,y) approaches (0,0).

doesn't exist

Investigate the limit

lim x*y*y / (x*x + y*y) as (x,y) approaches (0,0).

lim x*y*y / (x*x + y*y) as (x,y) approaches (0,0).

= 0

hint: |xy/(x^2+y^2)| is bounded

hint: |xy/(x^2+y^2)| is bounded

Investigate the limit

lim exp(-x*x-y*y) as (x,y) approaches (oo,oo)

lim exp(-x*x-y*y) as (x,y) approaches (oo,oo)

= 0

Investigate the convergence of the sequence {n^p * z^n}n where
z is a complex number and p is a real number.

if |z|>1, then diverges

if |z|<1, then converges

if |z|=1, then it depends on p, see page 15 in OLDTIMER for inspiration

if |z|<1, then converges

if |z|=1, then it depends on p, see page 15 in OLDTIMER for inspiration

State and prove Weierstrass Test for Unifrom Convergence of
Series.

(1) /\n:-|N /\x:-A | f[n](x) | <= a[n]

(2) the series a[n] converges

thesis:

(3) the series f[n](x) converges uniformly on A

(4) the series | f[n](x) | converges uniformly on A

(2) the series a[n] converges

thesis:

(3) the series f[n](x) converges uniformly on A

(4) the series | f[n](x) | converges uniformly on A

Let R be the radius of convergence of a power series with
center at 0.

Prove that the series converges uniformly on

{z: |z| <= r} for all r, 0<r<R.

Prove that the series converges uniformly on

{z: |z| <= r} for all r, 0<r<R.

Use Weierstrass Test.

very important hint: r<q<R.

very important hint: r<q<R.

Find the radius of converges of the series ((-1)^n / n) *
z^(n(n+1)).

R=1

hint: a[n(n+1)] = (-1)^n / n

hint: a[n(n+1)] = (-1)^n / n

Let a[n], b[n] be sequences of complex numbers, indexed from
zero.

Suppose that series |a[n]| and series b[n] both converge.

What can we infer from this? (State without proof.)

Suppose that series |a[n]| and series b[n] both converge.

What can we infer from this? (State without proof.)

Let /\n:-|N C[n] = +(k=0 to n) [ a[n-k]*b[k] ] = +(k=0 to n) [
a[k]*b[n-k] ].

Let A = +(n=0 to oo) a[n].

Let B = +(n=0 to oo) b[n].

Then +(n=0 to oo) C[n] = A * B

Let A = +(n=0 to oo) a[n].

Let B = +(n=0 to oo) b[n].

Then +(n=0 to oo) C[n] = A * B

lim x*x * sin(y) / (x*x + y*y)

as (x,y) -> (0,0)

as (x,y) -> (0,0)

= 0

hint: |sin(y)|<=|y|; |xy/(x^2+y^2)| is bounded

hint: |sin(y)|<=|y|; |xy/(x^2+y^2)| is bounded

lim x*x + y*y / (x*x*x*x + y*y*y*y)

as (x,y) -> (0,0)

as (x,y) -> (0,0)

= oo

hint: lim xy/(x+y) = 0 as (x,y) -> (0,0)

better hint: x^4 + y^4 = (x^2 + y^2)^2 - 2*x*x*y*y

hint: lim xy/(x+y) = 0 as (x,y) -> (0,0)

better hint: x^4 + y^4 = (x^2 + y^2)^2 - 2*x*x*y*y

lim x*x / (x*x + y*y)

as (x,y) -> (oo,oo)

as (x,y) -> (oo,oo)

doesn't exist

g : |R x |R -> |R

g(x,y) = sin(xy) / x for (x != 0)

g(0,y) = y

Is g continuous?

g(x,y) = sin(xy) / x for (x != 0)

g(0,y) = y

Is g continuous?

yes

z[n] = x[n] + i*y[n], ( x[n], y[n] are real sequences )

(1) series |z[n]| converges

(2) series |x[n]| and series |y[n]| both converge

Show that (1) <=> (2)

(1) series |z[n]| converges

(2) series |x[n]| and series |y[n]| both converge

Show that (1) <=> (2)

_______

Prove that

series 1 / ( (n+z) * log(n) * log(n) )

converges absolutely for all z:-C.

series 1 / ( (n+z) * log(n) * log(n) )

converges absolutely for all z:-C.

hint1: |z+n|/n -> 1

hint2: series 1/(n*log(n)*log(n)) converges

hint2: series 1/(n*log(n)*log(n)) converges

Show that the series z^n / (n*n) converges uniformly on |z|<=1.

______

Show that the series 2^n / ( z^n + z^(-n) )

converges uniformly on 0 < |z| <= r < 1/2.

converges uniformly on 0 < |z| <= r < 1/2.

hint: |a+b| >= | |a| - |b| |

|z| = 1 and (z != 1)

Show that the series (z^n)/n converges.

Show that the series (z^n)/n converges.

hint: series z^n is bounded

page 15 in OLDTIMER

page 15 in OLDTIMER

Give an example of a power series that converges

for all points |z| = R except one point.

for all points |z| = R except one point.

series (z^n)/n

a>=0, b>0

Prove: sqrt(a/b) <= 1/2 * (a + 1/b).

Prove: sqrt(a/b) <= 1/2 * (a + 1/b).

hint: 0 <= (a*sqrt(b) - 1/sqrt(b))^2

a[n] >= 0

series a[n]*a[n] converges

show that series a[n] / n converges

series a[n]*a[n] converges

show that series a[n] / n converges

hint: sqrt(a/b) <= 1/2 * (a + 1/b)

Prove that in a metric space, if every finitely bound
collection of closed sets has nonempty intersection, then the
space is sequentially compact.

Take any sequence. Suppose that it contains no convergent
subsequence. Let A[n] = {x[n], x[n+1], ...}. {A[n]} is a
sequence of closed sets having the finite intersection
property. Hence there is a point belonging to all A[n]. It must
be a subsequential limit. Contradiction.

Prove that in a topological space, if every finitely bound
collection of closed sets has nonempty intersectiom, then every
open cover of this space contains a finite subcover.

Theorem 57, page 144 in the palace notebook

(The proof is very short.)

(The proof is very short.)

Prove that in a topological space, if every open cover of that
space contains a finite subcover, then every finitely bound
collection of closed sets has nonempty intersection.

Argue by contradiction. This is easy.

Theorem 57, page 144 in the palace notebook

Theorem 57, page 144 in the palace notebook

Let (X,d) be a compact metric space.

Prove that

\/(a,b:-X) /\(x,y:-X) [ d(x,y) <= d(a,b) ]

Prove that

\/(a,b:-X) /\(x,y:-X) [ d(x,y) <= d(a,b) ]

Define g:(X*X) * (X*X) -> R by

g( (x,y), (a,b) ) = d(x,a) + d(y,b).

Show that (X*X,g) is a compact metric space.

Show that d is a continuous function on that space.

Deduce that it attains its maximum.

page 146 in palace

g( (x,y), (a,b) ) = d(x,a) + d(y,b).

Show that (X*X,g) is a compact metric space.

Show that d is a continuous function on that space.

Deduce that it attains its maximum.

page 146 in palace

What is a precompact metric space?

In a precompact metric space every sequence has a Cauchy
subsequence. It is equivalent to total boundedness.

Let a[n] be a Cauchy sequence. Let x[n] be a sequence.

Suppose that d(a[n],x[n]) -> 0.

Show that x[n] is also Cauchy.

Suppose that d(a[n],x[n]) -> 0.

Show that x[n] is also Cauchy.

=

Let (X,d) be a metric space. Let A c X.

Show that: Clo(A) is compact => A is precompact.

Also, show that the converse need not be true.

Show that: Clo(A) is compact => A is precompact.

Also, show that the converse need not be true.

Let A be the whole space X, and let X be non-complete.

Let (X,d) be a metric space. Let A c X.

Show that: A is precompact <=> clo(A) is precompact.

Show that: A is precompact <=> clo(A) is precompact.

To prove (=>) use the fact that if a[n] is Cauchy and

d(a[n],x[n]) -> 0, then x[n] is also Cauchy.

(<=) is trivial.

d(a[n],x[n]) -> 0, then x[n] is also Cauchy.

(<=) is trivial.

Show that a totally bounded metric space can be written as a
finite union of open balls with radius e, for every e>0.

______

Define an interesting metric on P(|N).

The one I picked up from the book "Topological Spaces" during vacation 2000.

In this item, do not prove that it is indeed a metric.

The one I picked up from the book "Topological Spaces" during vacation 2000.

In this item, do not prove that it is indeed a metric.

A,B c |N

if (A = B) then d(A,B) = 0

if (A != B) then d(A,B) = 1/min(A+B)

A+B = (A\B)u(B\A)

page 151 in golden gate

if (A = B) then d(A,B) = 0

if (A != B) then d(A,B) = 1/min(A+B)

A+B = (A\B)u(B\A)

page 151 in golden gate

A,B c N

if (A = B) then d(A,B) = 0

if (A != B) then d(A,B) = 1/min(A+B); A+B = (A\B)u(B\A)

Show that (P(N),d) is a metric space.

if (A = B) then d(A,B) = 0

if (A != B) then d(A,B) = 1/min(A+B); A+B = (A\B)u(B\A)

Show that (P(N),d) is a metric space.

First show that

(*) d(A,B) < 1/m <=> A n [1,m] = B n [1,m].

Using this show the triangle inequality.

page 151 in palace

(*) d(A,B) < 1/m <=> A n [1,m] = B n [1,m].

Using this show the triangle inequality.

page 151 in palace

Let (X,d) and (Y,g) be metric spaces. Let f : X -> Y.

What does it mean that f is an isometry of X into Y?

What does it mean that f is an isometry of X into Y?

/\(a,b:-X) [ d(a,b) = g(f(a),f(b)) ]

(it follows that an isometry is 1-1)

(I som' it rEE)

(it follows that an isometry is 1-1)

(I som' it rEE)

Let (X,d) and (Y,g) be metric spaces.

What does it mean that they are isometric?

What does it mean that they are isometric?

There exists a function f : X -> Y such that

(1) d(a,b) = g(f(a),f(b)) for all a,b from X

(2) f(X) = Y

(this f is bijective)

isometric (I suh met' rik)

(1) d(a,b) = g(f(a),f(b)) for all a,b from X

(2) f(X) = Y

(this f is bijective)

isometric (I suh met' rik)

Let (X,d) be a metric space. Consider the set B(X,|R) of
bounded functions f:X->|R.

Let ||f|| = sup {|f(x)|:x:-X}. Show that B(X,|R) is a metric space with d(f,g) = ||f-g||.

Let ||f|| = sup {|f(x)|:x:-X}. Show that B(X,|R) is a metric space with d(f,g) = ||f-g||.

_______

Let (X,d) be a metric space. Let B(X,|R) be the set of bounded
functions f:X->|R. Equip it with the sup metric. What kind of
metric space is this B(X,|R) ?

it is a complete metric space

page 154 in the palace notebook

page 154 in the palace notebook

Consider the set of all real-valued sequences a[n] such that
/\n 0<=a[n]<=1/n.

Equip this set with the sup metric.

What kind of metric space is this?

Equip this set with the sup metric.

What kind of metric space is this?

It is compact. PROOF: Call it K and show that it is a closed
subset of the set of all bounded real-valued sequences.
Conclude that it is complete. Next show that K is totally
bounded. Here's how. Take any n:-|N.

Let G = {g:-K : /\k:-|N g(k):-{0,1/n,2/n,...,(n-1)/n,1}}.

Show that G is finite.

Show that the collection of balls with center g:-G and ra

Let G = {g:-K : /\k:-|N g(k):-{0,1/n,2/n,...,(n-1)/n,1}}.

Show that G is finite.

Show that the collection of balls with center g:-G and ra

Let X = (0,oo). Define d(x,y) = |x-y| / (x*y). (X,d) is a
metric space.

Is d similar to the Euclidean metric?

Is d similar to the Euclidean metric?

Yes.

Use d(x,y) = |1/x - 1/y|.

Use d(x,y) = |1/x - 1/y|.

Show that in a complete metric space (X,d), A c X:

A is precompact <=> clo(A) is compact.

Demonstrate that completeness is essential here.

A is precompact <=> clo(A) is compact.

Demonstrate that completeness is essential here.

In every metric space

A is precompact <=> clo(A) is precompact

In a complete metric space, a closed set is complete.

Precompact and complete implies compact.

A is precompact <=> clo(A) is precompact

In a complete metric space, a closed set is complete.

Precompact and complete implies compact.

Let K be the set of all functions f:|N->|R such that /\n:-|N
0<=f(n)<=1/n.

Show that if a sequence of functions from K converges pointwise, then it converges uniformly.

Show that if a sequence of functions from K converges pointwise, then it converges uniformly.

page 157 in palace

Prove that every sequence of functions f:|N->[0,1] has a
subsequence that converges pointwise.

page 158 in palace

Let d be a metric on P(N) defined by d(A,B)=1/min(A+B).

Let E = { f:N->R : \/AcN /\x:-N f(x)=1/x * 1A(x) }.

Show that (P(N),d) and (E,sup) are isometric spaces.

Let E = { f:N->R : \/AcN /\x:-N f(x)=1/x * 1A(x) }.

Show that (P(N),d) and (E,sup) are isometric spaces.

page 159 in palace

Let d be a metric on P(N) defined by d(A,B)=1/min(A+B).

Prove that (P(N),d) is compact.

Prove that (P(N),d) is compact.

Let K be the set of all functions f:N->R such that /\n
0<=f[n]<=1/n.

Recall that K is a compact set with respect to the sup metric.

For each AcN, define F.A:N->R by F.A(n) = 1/n * 1A(n).

Let E = {F.A : AcN}. E c K. Show that E is a closed subset of K.

Hence E is compact. Recall that (P(N),d) and (E,sup) are isometric.

Recall that K is a compact set with respect to the sup metric.

For each AcN, define F.A:N->R by F.A(n) = 1/n * 1A(n).

Let E = {F.A : AcN}. E c K. Show that E is a closed subset of K.

Hence E is compact. Recall that (P(N),d) and (E,sup) are isometric.

Give an example of a metric space that is bounded but not
totally bounded.

An infinite discrete metric space is bounded. When you consider
balls of radius 1/7, you see that they are simply single-point
sets. Hence the whole space is not a finite union of sets of
diamater less than 1/7.

Give an example of a metric space that is bounded but not
totally bounded. Don't use the discrete metric. Give an
ambitious example.

Let X = { f:N->R : sup{|f(x)| : x:-N} <= 1 }.

Consider X as a metric space with the sup metric.

It is bounded because it is a closed ball.

X contains all characteristic functions of one-point subsets of N.

The distance between any two such functions equals 1.

Hence X is not a finite union of sets with diameter 1/2.

Consider X as a metric space with the sup metric.

It is bounded because it is a closed ball.

X contains all characteristic functions of one-point subsets of N.

The distance between any two such functions equals 1.

Hence X is not a finite union of sets with diameter 1/2.

Give a definition of the Cantor set.

I = [0,1]. If J = [a,b]

let J(0) = [a , a+(b-a)/3],

let J(1) = [b-(b-a)/3 , b].

If AcN and n:-N, let I(A(n)) = I(1A(1), 1A(2), ..., 1A(n)).

If n:-N then let C[n] = U{I(A(n)) : AcN}.

Let's define C = //\\n:N [ C[n] ].

(Notice that C is an intersection of closed sets.)

let J(0) = [a , a+(b-a)/3],

let J(1) = [b-(b-a)/3 , b].

If AcN and n:-N, let I(A(n)) = I(1A(1), 1A(2), ..., 1A(n)).

If n:-N then let C[n] = U{I(A(n)) : AcN}.

Let's define C = //\\n:N [ C[n] ].

(Notice that C is an intersection of closed sets.)

Let (X,d) , (Y,g) be two metric spaces.

Consider f:X*Y->R.

(1) /\y:-Y f is continuous as a function of x:-X

(2) /\x:-X f is continuous as a function of y:-Y

Does f have to be continuous?

Consider f:X*Y->R.

(1) /\y:-Y f is continuous as a function of x:-X

(2) /\x:-X f is continuous as a function of y:-Y

Does f have to be continuous?

NO.

h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.

h(x,y) = 0 if x=0 or y=0.

f(0,0) = 0

f(x,y) = x*y / (x*x + y*y)

h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.

h(x,y) = 0 if x=0 or y=0.

f(0,0) = 0

f(x,y) = x*y / (x*x + y*y)

Let (X,d) , (Y,g) be two metric spaces.

Consider f:X*Y->R.

(1) /\y:-Y f is uniformly continuous as a function of x:-X

(2) /\x:-X f is uniformly continuous as a function of y:-Y

Does f have to be continuous?

Consider f:X*Y->R.

(1) /\y:-Y f is uniformly continuous as a function of x:-X

(2) /\x:-X f is uniformly continuous as a function of y:-Y

Does f have to be continuous?

NO.

h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.

h(x,y) = 0 if x=0 or y=0.

f(0,0) = 0

f(x,y) = x*y / (x*x + y*y)

h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.

h(x,y) = 0 if x=0 or y=0.

f(0,0) = 0

f(x,y) = x*y / (x*x + y*y)

Prove that the Cantor set has Lebesgue measure zero.

page 168 in the palace notebook

(C,|x-y|) is the Cantor set with the Euclidean metric.

(P(N),d) : d(A,B) = 1/min(A+B).

Show that (P(N),d) and (C,|x-y|) are homeomorphic metric spaces.

(P(N),d) : d(A,B) = 1/min(A+B).

Show that (P(N),d) and (C,|x-y|) are homeomorphic metric spaces.

Show these two:

(1) d(A,B) < 1/m ==> |F(A) - F(B)| <= 3^(-m)

(2) |F(A) - F(B)| < 3^(-m) ==> d(A,B) < 1/m

F is the homeomorphism.

It's defined in the previous item.

page 73 in the palace notebook

(1) d(A,B) < 1/m ==> |F(A) - F(B)| <= 3^(-m)

(2) |F(A) - F(B)| < 3^(-m) ==> d(A,B) < 1/m

F is the homeomorphism.

It's defined in the previous item.

page 73 in the palace notebook

Let a[n] be a decreasing sequence converging to zero.

Show that if |z|=1 and z!=1, then the series a[n]*z^n converges.

Show that if |z|=1 and z!=1, then the series a[n]*z^n converges.

Show that the series z^n is bounded.

Then use the Dirichlet test.

Then use the Dirichlet test.

series (-1)^n / log(n) * z^(3*n-1)

Investigate convergence on |z| = 1.

Investigate convergence on |z| = 1.

For |z|=1,

z*z*z != -1 <=> the series converges

z^3 = -1 <=> the series diverges

z*z*z != -1 <=> the series converges

z^3 = -1 <=> the series diverges

Given a natural number k, show a power series that converges on
its circle except at k points.

series 1/n * z^(k*n)

Let G be an open subset of the complex plane.

Let f : G -> |C , z:-G.

What is the necessary condition for f being diffable at z?

State the condition without proof.

Let f : G -> |C , z:-G.

What is the necessary condition for f being diffable at z?

State the condition without proof.

There exist four limits Ux, Uy, Vx, Vy with

Ux = Vy

Uy = -Vx

See page 18 in OLDTIMER for details.

Ux = Vy

Uy = -Vx

See page 18 in OLDTIMER for details.

G is an open subset of the complex plane. f:G->C. Give a
sufficient condition for f being continuously diffable on G.

1) f has continuous partial derivatives on G

2) Ux(z) = Vy(z) and Uy(z) = -Vx(z), for all z:-G

page 21 in OLDTIMER

2) Ux(z) = Vy(z) and Uy(z) = -Vx(z), for all z:-G

page 21 in OLDTIMER

f(z) = |z|, for all complex z

Show that f is nowhere diffable.

Show that f is nowhere diffable.

hint: the Cauchy-Riemann equations are violated except at (0,0)

Investigate the limit

lim (x+y)/(x*x+y*y) as (x,y) approaches (oo,oo)

lim (x+y)/(x*x+y*y) as (x,y) approaches (oo,oo)

= 0

hint: (x,y) = t * exp(i*a)

hint: (x,y) = t * exp(i*a)

What is a monotone class of sets?

M is a monotone class of sets

iff

1) the union of every increasing sequence of sets from M belongs to M

2) the intersection of every decreasing sequence of sets from M belongs to M

iff

1) the union of every increasing sequence of sets from M belongs to M

2) the intersection of every decreasing sequence of sets from M belongs to M

{ E c |R^n : E is convex }

Is this class of sets monotone?

Is this class of sets monotone?

Yes.

The intersection of convex sets is a convex set.

The union of an increasing sequence of convex sets is convex.

The intersection of convex sets is a convex set.

The union of an increasing sequence of convex sets is convex.

Give the function which shows that the Cantor set is uncountable.

I = [0,1]. If J = [a,b], let J(0) = [a , a+(b-a)/3], let J(1) =
[b-(b-a)/3 , b].

If AcN and n:-N, let I(A(n)) = I(1A(1), 1A(2), ..., 1A(n)).

If n:-N then let C[n] = U{I(A(n)) : AcN}. Let's define C = //\\n:N [ C[n] ].

For AcN, let F(A) be the one point belonging to //\\n:-N I(A(n)).

If AcN and n:-N, let I(A(n)) = I(1A(1), 1A(2), ..., 1A(n)).

If n:-N then let C[n] = U{I(A(n)) : AcN}. Let's define C = //\\n:N [ C[n] ].

For AcN, let F(A) be the one point belonging to //\\n:-N I(A(n)).

Give a description of the Cantor set using series.

1) +(n=1 to oo) [ 2 * 1A(n) * 3^(-n) ]

For every Ac|N, the sum of the series above

uniquely determines an element of the Cantor set.

2) +(n=1 to oo) [ a[n] * 3^(-n) ]

Where a[n] is any sequence such that /\n:-|N [ a[n] :- {0,2} ].

For every Ac|N, the sum of the series above

uniquely determines an element of the Cantor set.

2) +(n=1 to oo) [ a[n] * 3^(-n) ]

Where a[n] is any sequence such that /\n:-|N [ a[n] :- {0,2} ].

Consider the metric space (P(|N),d), d(A,B) = 1/min(A+B).

Let {A[n]} be a sequence contained in P(|N).

(1) lim d(A[n],A) = 0

(2) lim A[n] = A (the lower and upper limits are both equal to A)

Show that (1) <=> (2).

Let {A[n]} be a sequence contained in P(|N).

(1) lim d(A[n],A) = 0

(2) lim A[n] = A (the lower and upper limits are both equal to A)

Show that (1) <=> (2).

page 169 in the palace notebook

Show that every function f:X->|R is the limit of some uniformly
convergent sequence of functions f[n]:X->|R.

f[n] = 1/n * [ n*f(x) ]

[r] denotes the integer part of r:-|R.

r-1 <= [r] <= r

[r] denotes the integer part of r:-|R.

r-1 <= [r] <= r

Can a countable dense subset of |R be written as a countable
intersection of open sets?

NO.

Argue by contradiction.

Use the Baire Category Theorem.

page 48 in OLDTIMER

Argue by contradiction.

Use the Baire Category Theorem.

page 48 in OLDTIMER

Let M be a countable dense subset of |R. Does there a exist a
sequence of continuous functions on |R which converges
pointwise to the characteristic function of M?

NO.

Let there be such a sequence {f[m]}. Put U[m] = f[m]-1 ((1/2,oo)). U[m] is the inverse image of (1/2,oo) under f[m]. U[m] is open because f[m] is continuous. Notice that

M = //\\n:-N \\//(m=n to m=oo) U[m].

M is written as the intersection of countably many open subsets of R. This is impossible. Contradiction. (see the previous item)

Let there be such a sequence {f[m]}. Put U[m] = f[m]-1 ((1/2,oo)). U[m] is the inverse image of (1/2,oo) under f[m]. U[m] is open because f[m] is continuous. Notice that

M = //\\n:-N \\//(m=n to m=oo) U[m].

M is written as the intersection of countably many open subsets of R. This is impossible. Contradiction. (see the previous item)

Let (X,d1) and (X,d2) be metric spaces.

What does it mean that d1 and d2 are similar metrics?

What does it mean that d1 and d2 are similar metrics?

/\x:-X /\{x[n]}cX [ lim d1(x[n],x)=0 <=> lim d2(x[n],x)=0 ]

Let z be a complex number.

Express differently:

|z|*|z| = ?

Express differently:

|z|*|z| = ?

|z|*|z| = z*conjugate(z)

Let X = (0,oo). Define d(x,y) = |x-y| / (x*y).

Show that (X,d) is a metric space.

Show that (X,d) is a metric space.

d(x,y) = |1/x - 1/y|

page 173 in palace

page 173 in palace

Let X = (0,oo). Define d(x,y) = |x-y| / (x*y). (X,d) is a
metric space.

Is is complete?

Is is complete?

NO.

{n} is a Cauchy sequence. It does not converge.

{n} is a Cauchy sequence. It does not converge.

Let X = [0,1). Define d(x,y) = |x/(1-x) - y/(1-y)|.

Show that (X,d) is a metric space.

Show that (X,d) is a metric space.

f : [0,1) -> [0,oo)

f(x) = x / (1-x)

Show that f is 1-1.

Conclude that d is a metric.

f(x) = x / (1-x)

Show that f is 1-1.

Conclude that d is a metric.

Let X = [0,1). Define d(x,y) = |x/(1-x) - y/(1-y)|. (X,d) is a
metric space.

Is (X,d) complete?

Is (X,d) complete?

Yes.

f : [0,1) -> [0,oo)

f(x) = x / (1-x)

Use the fact that f is a bijection.

f : [0,1) -> [0,oo)

f(x) = x / (1-x)

Use the fact that f is a bijection.

Consider an infinitely countable complete metric space.

How many isolated points does it have?

How many isolated points does it have?

Infinitely many.

If you remove an isolated point from a complete metric space,

then the remaining space is still complete.

Use the Baire Category Theorem to obtain one isolated point.

If you remove an isolated point from a complete metric space,

then the remaining space is still complete.

Use the Baire Category Theorem to obtain one isolated point.

Consider a complete metric space which has no isolated points.

Can it be countable?

Can it be countable?

NO.

A countable complete metric space must have an isolated point.

(see the previous item)

A countable complete metric space must have an isolated point.

(see the previous item)

Let A be a closed subset of |R. Suppose that it is infinite and
has no isolated points. Can A be countable?

NO.

A closed subset of |R is complete.

A countable complete set must have an isolated point.

(see the previous item)

A closed subset of |R is complete.

A countable complete set must have an isolated point.

(see the previous item)

Let (X,d) be a metric space. A c X.

What does it mean that A is nowhere-dense?

What does it mean that A is nowhere-dense?

Int(Clo(A)) = O

(the interior of the closure of A is empty)

The closure of A contains no ball.

nowhere-dense = rare

(the interior of the closure of A is empty)

The closure of A contains no ball.

nowhere-dense = rare

What do you call a subset of a metric space whose closure
contains no ball?

1) nowhere-dense

2) rare

2) rare

Let (X,d) be a metric space. A c X.

What does it mean that A is meager?

What does it mean that A is meager?

A meager set is a countable union of rare sets.

(The closure of a rare set contains no ball.)

(Every rare set is meager.)

(The closure of a rare set contains no ball.)

(Every rare set is meager.)

Let (X,d) be a metric space. A c X.

What does it mean that A is rare?

What does it mean that A is rare?

Int(Clo(A)) = O

(the interior of the closure of A is empty)

The closure of A contains no ball.

rare = nowhere-dense

(the interior of the closure of A is empty)

The closure of A contains no ball.

rare = nowhere-dense

What do you call a subset of a metric space that is a countable
union of rare sets? (A rare set is one whose closure contains
no ball.)

1) meager

2) of the first category

2) of the first category

Let (X,d) be a metric space. Let A c X.

What does it mean that A is of the first category?

What does it mean that A is of the first category?

A set of the first category is a countable union of rare sets.

(The closure of a rare set contains no ball.)

(Every rare set is of the first category.)

a set of the first category = a meager set

(The closure of a rare set contains no ball.)

(Every rare set is of the first category.)

a set of the first category = a meager set

Let (X,d) be a metric space. A c B c X.

If B is meager in X, then A is meager in X.

Decide if it's true. Prove your answer.

If B is meager in X, then A is meager in X.

Decide if it's true. Prove your answer.

TRUE

Let (X,d) be a metric space. A c B c X.

If A is meager in X, then B is meager in X.

Decide if it's true. Prove your answer.

If A is meager in X, then B is meager in X.

Decide if it's true. Prove your answer.

FALSE.

Let (X,d) be a metric space. A c B c X.

If A is rare in B, then A is rare in X.

Decide if it's true. Prove your answer.

If A is rare in B, then A is rare in X.

Decide if it's true. Prove your answer.

TRUE

To prove this, use (p->q) <== (-q -> -p).

page 178 in the palace notebook

To prove this, use (p->q) <== (-q -> -p).

page 178 in the palace notebook

Let (X,d) be a metric space. A c B c X.

If A is rare in X, then A is rare in B.

Decide if it's true. Prove your answer.

If A is rare in X, then A is rare in B.

Decide if it's true. Prove your answer.

FALSE.

Let X = |R, and let A be the Cantor.

Then A is rare in X.

Let B=A, then AcB but A is not rare in B.

Let X = |R, and let A be the Cantor.

Then A is rare in X.

Let B=A, then AcB but A is not rare in B.

Let (X,d) be a metric space. A c B c X.

If A is meager in B, then A is meager in X.

Decide if it's true. Prove your answer.

If A is meager in B, then A is meager in X.

Decide if it's true. Prove your answer.

TRUE.

Use: If A is rare in B, then A is rare in X.

Use: If A is rare in B, then A is rare in X.

Let (X,d) be a metric space. A c B c X.

If A is meager in X, then A is meager in B.

Decide if it's true. Prove your answer.

If A is meager in X, then A is meager in B.

Decide if it's true. Prove your answer.

FALSE

Let A=B={1}. Let X=|R.

Then A is meager in X, but A is not meager in B.

Let A=B={1}. Let X=|R.

Then A is meager in X, but A is not meager in B.

Let f[n], g[n] be sequences of functions X->|C. Let f,g : X->|C.

Suppose that f[n] and g[n] converge uniformly to f and g respectively.

Suppose that both the sequences are uniformly bounded.

Is it true that f[n]*g[n] converges uniformly to f*g?

Suppose that f[n] and g[n] converge uniformly to f and g respectively.

Suppose that both the sequences are uniformly bounded.

Is it true that f[n]*g[n] converges uniformly to f*g?

YES

page 178 in the palace notebook

page 178 in the palace notebook

Let (X,d) be a bounded metric space.

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

Prove that (H,h) is a metric space.

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

Prove that (H,h) is a metric space.

This is called the Hausdorff metric space.

page 179 in the palace notebook

page 179 in the palace notebook

Let (X,d) be a bounded metric space.

Let H be the set of all closed subsets of X except the empty set.

What is the definition of the Hausdorff metric space?

(Just recall the definition without proving that it is a metric space.)

Let H be the set of all closed subsets of X except the empty set.

What is the definition of the Hausdorff metric space?

(Just recall the definition without proving that it is a metric space.)

h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

(H,h) is a metric space, see the previous item.

(H,h) is a metric space, see the previous item.

Let (X,d) be a bounded metric space.

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

(H,h) is the Hausdorff metric space.

Prove that d(a,b) = h({a},{b}).

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

(H,h) is the Hausdorff metric space.

Prove that d(a,b) = h({a},{b}).

page 179 in the palace notebook

Let (X,d) be a metric space. f:X->X.

Suppose that f satisfies:

/\x,y:-X [ x!=y ==> d(f(x),f(y)) < d(x,y) ].

Decide if we can conclude that

\/0<K<1 /\x,y:-X d(f(x),f(y)) <= K*d(x,y).

Suppose that f satisfies:

/\x,y:-X [ x!=y ==> d(f(x),f(y)) < d(x,y) ].

Decide if we can conclude that

\/0<K<1 /\x,y:-X d(f(x),f(y)) <= K*d(x,y).

NO.

Consider [2,oo) with the Euclidean metric.

Put f(x) = x + 1/x, for x>=2. Notice that f(x)>=2.

To show that f satisfies the desired condition,

use the fact that f is increasing.

Since f has no fixed point, it cannot be a contraction.

Consider [2,oo) with the Euclidean metric.

Put f(x) = x + 1/x, for x>=2. Notice that f(x)>=2.

To show that f satisfies the desired condition,

use the fact that f is increasing.

Since f has no fixed point, it cannot be a contraction.

Prove that there exists a continuous function f:[0,1]->R that
is neither increasing nor decreasing on any subinterval of [0,1].

Consider the set of all continuous f:[0,1]->R. Equip it with
the sup metric. It's a complete metric space. Let I be a
subinterval of [0,1]. Let A(I) denote the set of all continuous
f:[0,1]->R that are increasing on I. (x<y==>f(x)<=f(y)). A(I)
is closed. Let B(I) be the set of all continuous f:[0,1]->R
that are decreasing on I. B(I) is closed. Let K = A(I) u B(I).
K is closed and Int(K) = 0. Let {I[n]} be a sequenc

Let z be a complex number such that |z|<1. Let m be a natural
number.

Express 1 / (1-z)^m as the sum of a power series.

Express 1 / (1-z)^m as the sum of a power series.

1 / (1-z)^m = +(n=0 to oo) [ z^n * (n+m-1)! / (n!*(m-1)!) ]

page 190 in the palace notebook

page 190 in the palace notebook

Let z be a complex number.

z * conjugate(z) = ?

z * conjugate(z) = ?

z * conjugate(z) = |z|*|z|

Let z be a complex number other than zero.

conjugate(z) / |z|^2 = ?

conjugate(z) / |z|^2 = ?

conjugate(z) / |z|^2 = 1/z

Let z be a complex number other than zero.

z / |z|^2 = ?

z / |z|^2 = ?

z / |z|^2 = 1/conjugate(z)

Let h : |C- > |C be defined h(z)=|z|*|z|.

Where is h diffable?

Where is h diffable?

Only at z = 0.

Show a complex function that is diffable only at one point.

f(z)=|z|^2

f(x,y)=x^2+y^2

f is diffable only at zero

f(x,y)=x^2+y^2

f is diffable only at zero

Consider a complex function defined on an open subset of |C.
Suppose that it is diffable at some point in that set. Does it
have to be diffable in a neighborhood of the point?

NO.

There are complex functions that are diffable only at one point.

1) z->|z|*|z| at 0

2) f(x,y)=xy at 0

There are complex functions that are diffable only at one point.

1) z->|z|*|z| at 0

2) f(x,y)=xy at 0

Let a,b be complex numbers.

exp(a) = exp(b) <=> ????

exp(a) = exp(b) <=> ????

a = b + k*2*pi*i

for some integer k

for some integer k

Describe the set {z:-|C : exp(z)=1}.

{k*2*pi*i : k is an integer}

A good proof is on page 68 in the palace notebook.

A bad proof is on page 24 in OLDTIMER.

Be careful not to repeat the bad proof.

A good proof is on page 68 in the palace notebook.

A bad proof is on page 24 in OLDTIMER.

Be careful not to repeat the bad proof.

Describe the set {z:-C : cos(z)=0}.

{ pi/2 * (2*k+1) : k is an integer}

cos(z)=0 <=> z is real and cos(z)=0

cos(z)=0 <=> z is real and cos(z)=0

exp(z) = u(z) + i*v(z), u=Re(exp), v=Im(exp).

Find u and v.

Find u and v.

exp(x,y) = exp(x)*cos(y) + i*exp(x)*sin(y)

Describe the set {z:-C : sin(z)=0}.

{k*pi : k is an integer}

sin(z)=0 <=> z is real and sin(z)=0

sin(z)=0 <=> z is real and sin(z)=0

Is it possible that z is not real and cos(z)=0 ?

NO.

cos(z)=0 => z is real

cos(z)=0 => z is real

Is it possible that z is not real and sin(z)=0 ?

NO.

sin(z)=0 ==> z is real

sin(z)=0 ==> z is real

Is it possible that z is not real and cos(z)=1 ?

NO.

cos(z)=1 ==> z is real

cos(z)=1 ==> z is real

Is it possible that z is not real and sin(z)=1 ?

NO. sin(z)=1 ==> z is real

Follow this downward chain of reasoning:

1) sin(z) = 1

2) cos(z) = 0

3) z is real

Follow this downward chain of reasoning:

1) sin(z) = 1

2) cos(z) = 0

3) z is real

Is it possible that z is not real and cos(z) = -1 ?

NO.

cos(z)=-1 ==> z is real

cos(z)=-1 ==> z is real

Is it possible that z is not real and sin(z)=-1 ?

NO.

sin(z)=-1 ==> z is real

sin(z)=-1 ==> z is real

Let A[n] c X for all natural n.

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

Prove that for all x:-X

/\x:-X 1(lim_sup A[n])(x) = ???

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

Prove that for all x:-X

/\x:-X 1(lim_sup A[n])(x) = ???

/\x:-X 1(lim_sup A[n])(x) = lim_sup 1(A[n])(x)

The proof is very easy.

The proof is very easy.

Is it possible that z is not real and sin(z) is real?

YES.

cos(Re(z))=0 ==> sin(z) is real

sin(pi/2 + i) is real

cos(Re(z))=0 ==> sin(z) is real

sin(pi/2 + i) is real

Is it possible that z is not real and cos(z) is real?

YES.

cos(i) :- |R

sin(Re(z))=0 ==> cos(z) is real

cos(i) :- |R

sin(Re(z))=0 ==> cos(z) is real

Describe the set {z:-C : cos(z) is real}.

R u {z:-C : Re(z) = k*pi, where k is an integer}

cos(z) is real <=> [ z is real or sin(Re(z))=0 ]

page 27 in OLDTIMER

cos(z) is real <=> [ z is real or sin(Re(z))=0 ]

page 27 in OLDTIMER

Describe the set {z:-C : sin(z) is real}.

R u {z:-C : Re(z) = pi/2 * (2*k+1), where k is an integer}

sin(z) is real <=> [ z is real or cos(Re(z))=0 ]

page 27 in OLDTIMER

sin(z) is real <=> [ z is real or cos(Re(z))=0 ]

page 27 in OLDTIMER

Describe the set {z:-C : Re( cos(z) ) = 0}.

Re( cos(z) ) = 0 <=> cos( Re(z) ) = 0

page 27 in OLDTIMER

page 27 in OLDTIMER

Describe the set {z:-C : Re( sin(z) ) = 0}.

Re( sin(z) ) = 0 <=> sin( Re(z) ) = 0

When cos(z) is purely imaginary, what can we infer about sin(z)?

sin(z) is real

page 27 in OLDTIMER

page 27 in OLDTIMER

When sin(z) is purely imaginary, what can we infer about cos(z)?

cos(z) is real

Follow the downward chain of reasoning:

1) sin(z) is purely imaginary

2) Re(sin(z)) = 0

3) sin(Re(z)) = 0

4) cos(z) is real

Follow the downward chain of reasoning:

1) sin(z) is purely imaginary

2) Re(sin(z)) = 0

3) sin(Re(z)) = 0

4) cos(z) is real

Let (X,d) be a metric space. Let (Y,g) be a complete metric
space. Let A be a dense subset of X. Consider a Lipschitz
function f:A->Y. Can this function be extended to f:X->Y and
still be Lipschitz?

YES

Prove that for every metric space (X,d) there exist:

(1) a complete metric space (X',d')

(2) an isometry f:X->X'

such that f(X) is dense in X'.

(1) a complete metric space (X',d')

(2) an isometry f:X->X'

such that f(X) is dense in X'.

page 53 in golden gate

or

page 138 in the book "Topological Spaces - From Distance to Neighborhood"

or

page 138 in the book "Topological Spaces - From Distance to Neighborhood"

Let {A[n]} be a sequence of sets. Define the lower limit of
this sequence.

\\//(k=1 to k=oo) //\\ (n=k to n=oo) A[n]

x belongs to the lower limit of a sequence of sets

iff

x belongs to almost all sets from this sequence

x belongs to the lower limit of a sequence of sets

iff

x belongs to almost all sets from this sequence

Let {A[n]} be a sequence of sets. Define the upper limit of
this sequence.

//\\(k=1 to k=oo) \\// (n=k to n=oo) A[n]

x belongs to the upper limit of a sequence of sets

iff

x belongs to infinitely many sets from this sequence

x belongs to the upper limit of a sequence of sets

iff

x belongs to infinitely many sets from this sequence

What does it mean that a sequence of sets converges?

The upper limit of this sequence is contained in the lower limit.

For all x, if x belongs to infinitely many sets of this sequence, then it belongs to almost all sets of this sequence.

For all x, if x belongs to infinitely many sets of this sequence, then it belongs to almost all sets of this sequence.

What does it mean that a sequence of sets is increasing?

{A[n]} sequence of sets

/\n:-N [ A[n] c A[n+1] ]

(A constant sequence is increasing.)

/\n:-N [ A[n] c A[n+1] ]

(A constant sequence is increasing.)

What does it mean that a sequence of sets is decreasing?

{A[n]} sequence of sets

/\n:-|N [ A[n+1] c A[n] ]

(A constant sequence is decreasing.)

/\n:-|N [ A[n+1] c A[n] ]

(A constant sequence is decreasing.)

What does it mean that a sequence of sets is monotone?

['mo n.. tOun]

['mo n.. tOun]

It's increasing or it's decreasing.

See the previous two items.

See the previous two items.

A + B = (A\B) u (B\A)

What do you call this operation on sets?

What do you call this operation on sets?

the symmetric difference of two sets A and B

[si 'met rik]

[si 'met rik]

What is a disjoint class of sets?

Every two sets of this class are disjoint.

==========

2001.02.06

This definition is stupid.

==========

2001.02.06

This definition is stupid.

What is a disjoint sequence of sets?

{A[n]} sequence of sets

for all natural m,n, m!=n

A[m] and A[n] are disjoint

for all natural m,n, m!=n

A[m] and A[n] are disjoint

Prove that an increasing sequence of sets converges.

The limit is equal to the union of all the sets from this
sequence.

Prove that a decreasing sequence of sets converges.

The limit is equal to the intersection of all the sets from
this sequence.

Prove that a disjoint sequence of sets converges.

The limit is the empty set.

Let A,B c X.

Let 1(A) denote the characteristic function of the set A.

Prove that the two functions are equal:

1(A+B) = ???

Let 1(A) denote the characteristic function of the set A.

Prove that the two functions are equal:

1(A+B) = ???

1(A+B) = |1(A) - 1(B)|

Let A[n] c X for all natural n.

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

/\x:-X 1(lim_inf A[n])(x) = ???

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

/\x:-X 1(lim_inf A[n])(x) = ???

/\x:-X 1(lim_inf A[n])(x) = lim_inf 1(A[n])(x)

Let A[n] c X for all natural n. Let AcX.

If HcX, then let 1(H) denote the characteristic function of the set H defined on X.

(1) the sequence of sets {A[n]} converges to the set A

(2) the function 1(A[n]) converges pointwise to the function 1(A)

If HcX, then let 1(H) denote the characteristic function of the set H defined on X.

(1) the sequence of sets {A[n]} converges to the set A

(2) the function 1(A[n]) converges pointwise to the function 1(A)

page 13

first measure theory notebook

first measure theory notebook

(A\B) u (B\A) = ???

Express differently.

Express differently.

(A\B) u (B\A) = (A u B) \ (A n B)

(A u B) \ (A n B) = ???

Express differently.

Express differently.

(A u B) \ (A n B) = (A\B) u (B\A)

Is it true that

(A + B) c (A + C) u (C + B)

?

(A + B) c (A + C) u (C + B)

?

Yes.

Is it true that

(A u B) + (C u D) c (A + C) u (B + D)

?

(A u B) + (C u D) c (A + C) u (B + D)

?

Yes.

Is it true that

(A n B) + (C n D) c (A + C) u (B + D)

?

(A n B) + (C n D) c (A + C) u (B + D)

?

Yes.

Is it true that

(A\B) + (C\D) c (A + C) u (B + D)

?

(A\B) + (C\D) c (A + C) u (B + D)

?

Yes.

Is it true that

(A n B) u (C n D) c (A u C) n (B u D)

?

(A n B) u (C n D) c (A u C) n (B u D)

?

Yes.

Is it true that

(A u C) n (B u D) c (A n B) u (C n D)

?

(A u C) n (B u D) c (A n B) u (C n D)

?

No.

Let A and D be empty sets.

Let B = C = {1}.

Let A and D be empty sets.

Let B = C = {1}.

Is it true that

(A + C) u (C + B) c (A + B)

?

(A + C) u (C + B) c (A + B)

?

No.

Let A and B be empty sets.

Let C be non-empty.

Let A and B be empty sets.

Let C be non-empty.

Is it true that

(A + C) u (B + D) c (A u B) + (C u D)

?

(A + C) u (B + D) c (A u B) + (C u D)

?

No.

1) Let A = D = {1}. Let B = C = {2}.

2) Let A = B = C = {1}. Let D = O.

3) A=D={1}, C=O

1) Let A = D = {1}. Let B = C = {2}.

2) Let A = B = C = {1}. Let D = O.

3) A=D={1}, C=O

Is it true that

(A + C) u (B + D) c (A n B) + (C n D)

?

(A + C) u (B + D) c (A n B) + (C n D)

?

No.

Let A = {1}, B = C = D = 0.

The right side is empty.

1 belongs to the left side.

Let A = {1}, B = C = D = 0.

The right side is empty.

1 belongs to the left side.

Is it true that

(A + C) u (B + D) c (A\B) + (C\D)

?

(A + C) u (B + D) c (A\B) + (C\D)

?

No.

Let C=D=0, A=B={1}.

Let C=D=0, A=B={1}.

A[n], B[n] are sequences of subsets of X.

lim_inf ( A[n] n B[n] ) c ( lim_inf A[n] ) n ( lim_inf B[n] )

Is this true?

lim_inf ( A[n] n B[n] ) c ( lim_inf A[n] ) n ( lim_inf B[n] )

Is this true?

Yes.

Even equality holds.

lim_inf ( A[n] n B[n] ) = ( lim_inf A[n] ) n ( lim_inf B[n] )

Even equality holds.

lim_inf ( A[n] n B[n] ) = ( lim_inf A[n] ) n ( lim_inf B[n] )

A[n], B[n] are sequences of subsets of X.

lim_inf (A[n] u B[n]) c lim_inf (A[n]) u lim_inf (B[n])

Is this true?

lim_inf (A[n] u B[n]) c lim_inf (A[n]) u lim_inf (B[n])

Is this true?

No.

Put A[1] = {1}, A[2] = {2}, A[3] = {1}, A[4] = {2}, etc.

Put B[1] = {2}, B[2] = {1}, B[3] = {2}, B[4] = {1}, etc.

The right side is empty.

Put A[1] = {1}, A[2] = {2}, A[3] = {1}, A[4] = {2}, etc.

Put B[1] = {2}, B[2] = {1}, B[3] = {2}, B[4] = {1}, etc.

The right side is empty.

A[n], B[n] are sequences of subsets of X.

lim_inf (A[n]) u lim_inf (B[n]) c lim_inf (A[n] u B[n])

Is this true?

lim_inf (A[n]) u lim_inf (B[n]) c lim_inf (A[n] u B[n])

Is this true?

Yes.

A[n], B[n] are sequences of subsets of X.

lim_sup ( A[n] u B[n] ) c ( lim_sup A[n] ) u ( lim_sup B[n] )

Is this true?

lim_sup ( A[n] u B[n] ) c ( lim_sup A[n] ) u ( lim_sup B[n] )

Is this true?

Yes.

Even equality holds.

lim_sup ( A[n] u B[n] ) = ( lim_sup A[n] ) u ( lim_sup B[n] )

Even equality holds.

lim_sup ( A[n] u B[n] ) = ( lim_sup A[n] ) u ( lim_sup B[n] )

A[n], B[n] are sequences of subsets of X.

( lim_sup A[n] ) u ( lim_sup B[n] ) c lim_sup ( A[n] u B[n] )

Is this true?

( lim_sup A[n] ) u ( lim_sup B[n] ) c lim_sup ( A[n] u B[n] )

Is this true?

Yes.

Even equality holds.

( lim_sup A[n] ) u ( lim_sup B[n] ) = lim_sup ( A[n] u B[n] )

Even equality holds.

( lim_sup A[n] ) u ( lim_sup B[n] ) = lim_sup ( A[n] u B[n] )

A[n], B[n] are sequences of subsets of X. A,B c X.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] u B[n] ) = A u B.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] u B[n] ) = A u B.

A u B c

( lim_inf A[n] ) u ( lim_inf B[n] ) c

lim_inf ( A[n] u B[n] ) c

lim_sup ( A[n] u B[n] ) c

( lim_sup A[n] ) u ( lim_sup B[n] ) c

A u B

( lim_inf A[n] ) u ( lim_inf B[n] ) c

lim_inf ( A[n] u B[n] ) c

lim_sup ( A[n] u B[n] ) c

( lim_sup A[n] ) u ( lim_sup B[n] ) c

A u B

A[n], B[n] are sequences of subsets of X.

( lim_inf A[n] ) n ( lim_inf B[n] ) c lim_inf ( A[n] n B[n] )

Is this true?

( lim_inf A[n] ) n ( lim_inf B[n] ) c lim_inf ( A[n] n B[n] )

Is this true?

Yes.

Even equality holds.

( lim_inf A[n] ) n ( lim_inf B[n] ) = lim_inf ( A[n] n B[n] )

Even equality holds.

( lim_inf A[n] ) n ( lim_inf B[n] ) = lim_inf ( A[n] n B[n] )

A[n], B[n] are sequences of subsets of X.

lim_sup ( A[n] n B[n] ) c ( lim_sup A[n] ) n ( lim_sup B[n] )

Is this true?

lim_sup ( A[n] n B[n] ) c ( lim_sup A[n] ) n ( lim_sup B[n] )

Is this true?

Yes.

A[n], B[n] are sequences of subsets of X.

( lim_sup A[n] ) n ( lim_sup B[n] ) c lim_sup ( A[n] n B[n] )

Is this true?

( lim_sup A[n] ) n ( lim_sup B[n] ) c lim_sup ( A[n] n B[n] )

Is this true?

No.

A[2*n] = {1}; A[2*n+1] = {2}

B[2*n] = {2}; B[2*n+1] = {1}

A[2*n] = {1}; A[2*n+1] = {2}

B[2*n] = {2}; B[2*n+1] = {1}

A[n], B[n] are sequences of subsets of X. A,B c X.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] n B[n] ) = A n B.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] n B[n] ) = A n B.

Use

(1) ( lim_inf A[n] ) n ( lim_inf B[n] ) c lim_inf ( A[n] n B[n] )

(2) lim_sup ( A[n] n B[n] ) c ( lim_sup A[n] ) n ( lim_sup B[n] )

(1) ( lim_inf A[n] ) n ( lim_inf B[n] ) c lim_inf ( A[n] n B[n] )

(2) lim_sup ( A[n] n B[n] ) c ( lim_sup A[n] ) n ( lim_sup B[n] )

B[n] is sequences of subsets of X. B c X.

Given that lim B[n] = B

show that lim X\B[n] = X\B.

Given that lim B[n] = B

show that lim X\B[n] = X\B.

page 19 in 1st measure

A[n], B[n] are sequences of subsets of X. A,B c X.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] \ B[n] ) = A \ B.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] \ B[n] ) = A \ B.

Use:

(1) lim B[n] = B ==> lim X\B[n] = X\B

(2) lim A[n] = A and lim B[n] = B ==> lim ( A[n] n B[n] ) = A n B

(1) lim B[n] = B ==> lim X\B[n] = X\B

(2) lim A[n] = A and lim B[n] = B ==> lim ( A[n] n B[n] ) = A n B

A[n], B[n] are sequences of subsets of X. A,B c X.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] + B[n] ) = A + B.

Given that lim A[n] = A and lim B[n] = B

show that

lim ( A[n] + B[n] ) = A + B.

Use:

(1) lim B[n] = B ==> lim X\B[n] = X\B

(2) lim A[n] = A and lim B[n] = B ==> lim ( A[n] n B[n] ) = A n B

(1) lim B[n] = B ==> lim X\B[n] = X\B

(2) lim A[n] = A and lim B[n] = B ==> lim ( A[n] n B[n] ) = A n B

What does it mean that a complex function satisfies the Cauchy-
Riemann equations at a point in its domain?

z:-A ; A c |C ; A is open ; f:A->|C ; U = Re(f) ; V = Im(f)

(1) all four partial derivatives exist at z: Ux, Uy, Vx, Vy

(2) Ux = Vy and Uy = -Vx

page 18 in OLDTIMER

(1) all four partial derivatives exist at z: Ux, Uy, Vx, Vy

(2) Ux = Vy and Uy = -Vx

page 18 in OLDTIMER

Suppose that a complex function is defined on an open set and
satisfies the Cauchy-Riemann equations at a point. Does it have
to be diffable at that point?

f(x,y) = sqrt( |x*y| ) at (0,0)

C-R satisfied, not diffable

C-R satisfied, not diffable

lim (x,y)->(0,0) [ x*y / sqrt( x*x + y*y) ] = ?

= 0

What is separate continuity?

Let (X,d) , (Y,g) be two metric spaces.

Consider f:X*Y->|R.

(1) /\y:-Y f is continuous as a function of x:-X

(2) /\x:-X f is continuous as a function of y:-Y

(1)and(2) <=> separate continuity

Consider f:X*Y->|R.

(1) /\y:-Y f is continuous as a function of x:-X

(2) /\x:-X f is continuous as a function of y:-Y

(1)and(2) <=> separate continuity

(X,d) is a metric space; a:-X; f:X->|R; f(x)=d(x,a).

Show that f is Lipschitz.

Show that f is Lipschitz.

Use the second triangle inequality.

What is a directed set?

(X,>)

X is a set and > is a binary relation on X such that

(1) /\a,b:-X \/ c:-X [ c>a and c>b ]

(2) /\a,b,c:-X [ a>b and b>c ==> a>c ]

X is a set and > is a binary relation on X such that

(1) /\a,b:-X \/ c:-X [ c>a and c>b ]

(2) /\a,b,c:-X [ a>b and b>c ==> a>c ]

What is a net?

h is a net contained in X

means that

there is a directed set (T,>) (see the previous item)

h:T->X

means that

there is a directed set (T,>) (see the previous item)

h:T->X

Let T be the set of all finite subsets of |N.

For a,b:-T, a>b <=> b is contained in a

Prove that (T,>) is a directed set.

For a,b:-T, a>b <=> b is contained in a

Prove that (T,>) is a directed set.

Very easy.

Let T be the set of all finite subsets of |N.

For a,b:-T, a>b <=> b is contained in a

(T,>) is a directed set.

Let t[n] = (-1)^(n+1) * (1/n), for n:-|N.

For F:-T, let S(F) = the finite sum of t[n] where n:-F.

Does the net S(F) converge?

For a,b:-T, a>b <=> b is contained in a

(T,>) is a directed set.

Let t[n] = (-1)^(n+1) * (1/n), for n:-|N.

For F:-T, let S(F) = the finite sum of t[n] where n:-F.

Does the net S(F) converge?

NO.

page 169 in "Topological Spaces, From Distance to Neighborhood"

page 169 in "Topological Spaces, From Distance to Neighborhood"

What is a singleton?

A singleton is a set that has exactly one element.

{x}

{x}

What do you call a set that has exactly one element?

a singleton ['siN g.l t.n]

Consider the complex function Re(z).

Prove that it is nowhere diffable.

Prove that it is nowhere diffable.

=

a,b:-[-pi,pi] ; cos(a)=cos(b)

Show that a=b or a=(-b).

Show that a=b or a=(-b).

hint: cos is 1-1 on [0,pi] and cos(|a|) = cos(|b|)

Let t[n] be a sequence satisfying -pi<=t[n]<=pi. Let -pi<t<pi.
Suppose that cos(t[n])->cos(t) and sin(t[n])->sin(t). What can
we conclude?

t[n]->t

page 38 in OLDTIMER

page 38 in OLDTIMER

Express differently:

max(x,y) = ???

max(x,y) = ???

max(x,y) = ( x + y + |x - y| ) / 2

Express differently

min(x,y) = ???

min(x,y) = ???

min(x,y) = ( x + y - |x - y| ) / 2

Let A[n] be a sequence of sets contained in X.

Prove that the two conditions are equivalent.

(1) A[n] converges

(2) /\x:-X [ {n:-|N : x:-A[n]} is finite or {n:-|N : not x:- A[n]} is finite} ]

Prove that the two conditions are equivalent.

(1) A[n] converges

(2) /\x:-X [ {n:-|N : x:-A[n]} is finite or {n:-|N : not x:- A[n]} is finite} ]

page 20 first measure theory

Let A[n] be a sequence of sets contained in X. Let A c X.

Prove that the three conditions are equivalent.

(1) A[n] -> A

(2) A[n] + A -> O

(3) lim_sup (A[n] + A) = O

Prove that the three conditions are equivalent.

(1) A[n] -> A

(2) A[n] + A -> O

(3) lim_sup (A[n] + A) = O

page 21 measure theory

Let A[n] be a sequence of sets.

Suppose that it converges to A.

Prove that every subsequence of {A[n]} also converges to A.

Suppose that it converges to A.

Prove that every subsequence of {A[n]} also converges to A.

A = lim_inf A[n] c lim_inf A[k(n)] c lim_sup A[k(n)] c lim_sup
A[n] = A

Prove that a countable intersection of sets from a s-ring
belongs to this s-ring.

//\\(n=1 to n=oo) [ A[n] ] = A[1] \ U(n=1 to n=oo) [ A[1] \
A[n] ]

Side remark:

Notice that this formula can be more general.

Countability is irrelevant.

The choice of the first set is also irrelevant.

Side remark:

Notice that this formula can be more general.

Countability is irrelevant.

The choice of the first set is also irrelevant.

Express the intersection A n B with the difference operation \.

A n B = A \ (A \ B)

A \ (A \ B) = ???

Express differently.

Express differently.

A \ (A \ B) = A n B

What is a ring of sets?

Let X be a set. Let R c P(X).

R is a ring of sets

iff

(1) R is nonempty

(2) A,B :- R => A u B :- R

(3) A,B :- R => A \ B :- R

R is a ring of sets

iff

(1) R is nonempty

(2) A,B :- R => A u B :- R

(3) A,B :- R => A \ B :- R

Let R be a ring of sets. Prove that A,B :- R => A n B :- R.

A n B = A \ (A \ B)

Let R be a ring of sets. Does it have to contain the empty set?

Yes.

In the definition of a ring we required that it is a non-empty collection of sets.

Hence some set belongs to it, say A. Then A\A belongs.

In the definition of a ring we required that it is a non-empty collection of sets.

Hence some set belongs to it, say A. Then A\A belongs.

Let R be a ring of sets. Prove that A,B :- R => A + B :- R.

____

What is a s-ring of sets?

Let X be a set. Let R c P(x).

R is a s-ring of sets

iff

(0) R is nonempty

(1) Countable union of sets from R belongs to R.

(2) A,B :- R => A \ B :- R

R is a s-ring of sets

iff

(0) R is nonempty

(1) Countable union of sets from R belongs to R.

(2) A,B :- R => A \ B :- R

Prove that if R is a s-ring of sets, then every countable
intersection of sets in R again belongs to R.

Use:

//\\(n=1 to n=oo) [ A[n] ] = A[1] \ U(n=1 to n=oo) [ A[1] \ A[n] ]

//\\(n=1 to n=oo) [ A[n] ] = A[1] \ U(n=1 to n=oo) [ A[1] \ A[n] ]

Prove that the intersection of any number of rings of sets is
again a ring of sets.

______

Prove that the intersection of any number of s-rings of sets is
again a s-ring of sets.

_________

Let X be a nonempty set. Let W c P(X). What is "the ring
generated by W"?

The smallest ring containing W.

Such a ring exists.

It is the intersection of all rings containing W.

We denote it R(W).

Such a ring exists.

It is the intersection of all rings containing W.

We denote it R(W).

Let X be a nonempty set. Let W c P(X). What is "the s-ring
generated by W"?

The smallest s-ring containing W.

Such a s-ring exists.

It is the intersection of all s-rings containing W.

We denote it S(W).

Such a s-ring exists.

It is the intersection of all s-rings containing W.

We denote it S(W).

What is an algebra of sets? Give a minimal definition.

Let X be a set. Let R c P(X).

R is an algebra of sets

iff

(1) X :- R

(2) A,B :- R => A u B :- R

(3) B :- R => X \ B :- R

R is an algebra of sets

iff

(1) X :- R

(2) A,B :- R => A u B :- R

(3) B :- R => X \ B :- R

What is a s-algebra of sets?

Let X be a set. Let W c P(X).

W is a s-algebra of sets

iff

(1) X :- W

(2) /\n:-|N A[n]:-W |=> U(n:-|N)_A[n] :- W

(3) B:-W => X\B:-W

W is a s-algebra of sets

iff

(1) X :- W

(2) /\n:-|N A[n]:-W |=> U(n:-|N)_A[n] :- W

(3) B:-W => X\B:-W

Consider a collection of algebras of subsets of X.

X belongs to all the algebras.

Prove that the intersection of this collection is an algebra of subsets of X.

X belongs to all the algebras.

Prove that the intersection of this collection is an algebra of subsets of X.

page 33 in 1st measure

Consider a collection of s-algebras of subsets of X.

X belongs to all the s-algebras.

Prove that the intersection of this collection is a s-algebra of subsets of X.

X belongs to all the s-algebras.

Prove that the intersection of this collection is a s-algebra of subsets of X.

page 34 in first measure

Let X be a set. Let W c P(X).

What is "the algebra generated by W"?

What is "the algebra generated by W"?

The smallest algebra containing W.

Such an algebra exists.

It is the intersection of all algebras containing W.

We denote it K(W).

Such an algebra exists.

It is the intersection of all algebras containing W.

We denote it K(W).

Let X be a set. Let W c P(X).

What is "the s-algebra generated by W"?

What is "the s-algebra generated by W"?

The smallest s-algebra containing W.

Such a s-algebra exists.

It is the intersection of all s-algebras containing W.

We denote it d(W).

Such a s-algebra exists.

It is the intersection of all s-algebras containing W.

We denote it d(W).

A,B c X

Express differently

X \ (A \ B) = ???

Express differently

X \ (A \ B) = ???

X \ (A \ B) = (X \ A) u B

A,B c X

Express differently

(X \ A) u B = ???

Express differently

(X \ A) u B = ???

(X \ A) u B = X \ (A \ B)

Let X be a nonempty set. Let W be a collection of finite
subsets of X. What kind of collection is W?

W is an ideal.

W is a ring of sets.

W is a ring of sets.

Let X be a nonempty set. Let M c X. Let W be the collection of
all A c X such that MnA is finite or M\A is finite. What kind
of collection is W?

It is an algebra of sets.

PROOF:

A,B:-W

If MnA and MnB are both finite, then Mn(AuB) is finite, hence AuB:-W.

If M\A is finite, then M\(AuB) c M\A, hence M\(AuB) is finite, hence AuB:-W.

PROOF:

A,B:-W

If MnA and MnB are both finite, then Mn(AuB) is finite, hence AuB:-W.

If M\A is finite, then M\(AuB) c M\A, hence M\(AuB) is finite, hence AuB:-W.

Let X be a nonempty set. Let W be the collection of all
countable subsets of X. What kind of collection is W?

It is a s-ring of sets.

It is a s-ideal.

It is a s-ideal.

Let X be a nonempty set. Let M c X. Let W be the collection of
all A c X such that MnA is countable or M\A is countable. What
kind of collection is W?

It is a s-algebra of sets.

How do I choose to understand the word "countable" ?

countable = empty or finite or equinumerous with the set of
natural numbers

Is it true that every countable union of sets in a ring may be
written as a disjoint countable union of sets in the ring?

U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ (U(k=1 to k=n-1) A[k]) ]

where A[n] = O

page 41 in 1st measure

where A[n] = O

page 41 in 1st measure

(1) h : |R -> |R is diffable

(2) h(0) = 0

(3) /\x:-|R h'(x) + h(x)*cos(x) = cos(x)

Find the set of all such functions.

(2) h(0) = 0

(3) /\x:-|R h'(x) + h(x)*cos(x) = cos(x)

Find the set of all such functions.

There is one and only one such function.

h(x) = 1 - exp(-sin(x))

hint: multiply both sides of (3) by exp(sin(x))

h(x) = 1 - exp(-sin(x))

hint: multiply both sides of (3) by exp(sin(x))

Let A[n] be a sequence of sets.

Express its union as the union of an increasing sequence of sets.

Express its union as the union of an increasing sequence of sets.

U{A[n]} = U(n:-N) [ U(k=1 to k=n) A[k] ]

Let B[1],B[2],B[3],... be an increasing sequence of sets.

Give an interesting way of writing the union of this sequence.

Give an interesting way of writing the union of this sequence.

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Notice that the equality always holds, even if the sequence is not increasing.

Increasingness is necessary to show the disjointness of the right-hand union.

where B[0] = O.

Notice that the equality always holds, even if the sequence is not increasing.

Increasingness is necessary to show the disjointness of the right-hand union.

Let W be a collection of sets.

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

What more can we say about W?

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

What more can we say about W?

The union of any sequence of sets from W belongs to W.

Let A[n] be a sequence of sets.

Express more simply:

U(n:-N) [ A[n] \ (U(k=1 to n-1) A[k] ) ] = ???

Express more simply:

U(n:-N) [ A[n] \ (U(k=1 to n-1) A[k] ) ] = ???

= U{A[n]}

page 41 in 1st measure

page 41 in 1st measure

Suppose that a collection of sets is monotone and is a ring.
What can we further say about this collection?

It is a s-ring.

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

imply that

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

imply that

Suppose that a collection of sets is monotone and is an
algebra. What can we further say about this collection?

It is a s-algebra.

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

imply that

(1) The union of two sets from W belongs to W.

(2) The union of an inreasing sequence of sets from W belongs to W.

imply that

Let X be a set. Consider a collection of monotone classes of
subsets of X. Does the intersection of this collection have to
be a monotone class?

YES.

page 44 in 1st measure

page 44 in 1st measure

Let X be a nonempty set. Let W c P(X).

What is "the monotone class generated by W"?

What is "the monotone class generated by W"?

The smallest monotone class containing W.

Such a class exists.

It is the intersection of all monotone classes containing W.

page 44 in 1st measure

Such a class exists.

It is the intersection of all monotone classes containing W.

page 44 in 1st measure

Let X be a nonempty set. Let W c P(X).

What are the inclusion relations between R(W),S(W), K(W), d(W), M(W) ?

What are the inclusion relations between R(W),S(W), K(W), d(W), M(W) ?

1) R(W) c S(W)

2) K(W) c d(W)

3) R(W) c K(W)

4) S(W) c d(W)

5) M(W) c S(W)

6) M(W) c d(W)

2) K(W) c d(W)

3) R(W) c K(W)

4) S(W) c d(W)

5) M(W) c S(W)

6) M(W) c d(W)

Let X be a non-empty set. Let W={ {x} : x:-X }.

Describe R(W) - the ring generated by W.

Describe R(W) - the ring generated by W.

R(W) = { A c X : A is finite }

page 45 in 1st measure

page 45 in 1st measure

Let X be a non-empty set. Let W={ {x} : x:-X }.

Describe S(W) - the s-ring generated by W.

Describe S(W) - the s-ring generated by W.

S(W) = { A c X : A is countable }

page 45 in 1st measure

countable = equinumerous with a subset of |N

page 45 in 1st measure

countable = equinumerous with a subset of |N

Let X be a non-empty set. Let W={ {x} : x:-X }.

Describe K(W) - the algebra generated by W.

Describe K(W) - the algebra generated by W.

K(W) = { A c X : A is finite or X\A is finite }

page 45 in 1st measure

page 45 in 1st measure

Let X be a non-empty set. Let W={ {x} : x:-X }.

Describe d(W) - the s-algebra generated by W.

Describe d(W) - the s-algebra generated by W.

d(W) = { A c X : A is countable or X\A is countable }

page 45 in 1st measure

page 45 in 1st measure

Let X be a non-empty set. Let W={ {x} : x:-X }.

Describe M(W) - the monotone class generated by W.

Describe M(W) - the monotone class generated by W.

M(W) = W

Notice that W is itself a monotone class.

page 45 in 1st measure

Notice that W is itself a monotone class.

page 45 in 1st measure

Let D be an open subset of |R^2. Let f:D->|R. Let a,b:-|R, a<b.

(1) the partial derivatives fx, fy exist on D and they are continuous on D

(2) u : ]a,b[ -> U, U c |R

(3) v : ]a,b[ -> V, V c |R

(4) u and v are continuously diffable on ]a,b[.

(5) UxV c D.

(1) the partial derivatives fx, fy exist on D and they are continuous on D

(2) u : ]a,b[ -> U, U c |R

(3) v : ]a,b[ -> V, V c |R

(4) u and v are continuously diffable on ]a,b[.

(5) UxV c D.

F is diffable on ]a,b[.

F'(t) = fx(u(t),v(t))*u'(t) + fy(u(t),v(t))*v'(t)

page 50 in OLDTIMER

F'(t) = fx(u(t),v(t))*u'(t) + fy(u(t),v(t))*v'(t)

page 50 in OLDTIMER

Let X be a non-empty set. Let W c P(X).

Suppose that M(W) is a ring of sets.

What is the relation between M(W) and S(W)?

Suppose that M(W) is a ring of sets.

What is the relation between M(W) and S(W)?

It is always true that M(W) c S(W).

To prove that S(W) c M(W), we show that M(W) is a s-ring. Use:

(1) The union of two sets from K belongs to K.

(2) The union of an increasing sequence of sets from K belongs to K.

imply that

(3) The union of any sequence of sets from K belongs to K.

To prove that S(W) c M(W), we show that M(W) is a s-ring. Use:

(1) The union of two sets from K belongs to K.

(2) The union of an increasing sequence of sets from K belongs to K.

imply that

(3) The union of any sequence of sets from K belongs to K.

Express differently:

A u U(t:-T) B[t] = ???

A u U(t:-T) B[t] = ???

A u U(t:-T) B[t] = U(t:-T) AuB[t]

Express differently:

U(t:-T) AuB[t] = ???

U(t:-T) AuB[t] = ???

U(t:-T) AuB[t] = A u U(t:-T) B[t]

Express differently:

A n U(t:-T) B[t] = ???

A n U(t:-T) B[t] = ???

A n U(t:-T) B[t] = U(t:-T) AnB[t]

Express differently:

U(t:-T) AnB[t] = ???

U(t:-T) AnB[t] = ???

U(t:-T) AnB[t] = A n U(t:-T) B[t]

Express differently:

A n //\\(t:-T) B[t] = ???

A n //\\(t:-T) B[t] = ???

A n //\\(t:-T) B[t] = //\\(t:-T) AnB[t]

Express differently:

//\\(t:-t) AnB[t] = ???

//\\(t:-t) AnB[t] = ???

//\\(t:-T) AnB[t] = A n //\\(t:-T) B[t]

Express differently:

A u //\\(t:-T) B[t] = ???

A u //\\(t:-T) B[t] = ???

A u //\\(t:-T) B[t] = //\\(t:-T) AuB[t]

Express differently:

//\\(t:-T) AuB[t] = ???

//\\(t:-T) AuB[t] = ???

//\\(t:-T) AuB[t] = A u //\\(t:-T) B[t]

Express differently:

A \ U(t:-T) B[t] = ???

A \ U(t:-T) B[t] = ???

A \ U(t:-T) B[t] = //\\(t:-t) A\B[t]

Express differently:

//\\(t:-t) A\B[t] = ???

//\\(t:-t) A\B[t] = ???

//\\(t:-t) A\B[t] = A \ U(t:-T) B[t]

We want to define a metric on the interval [0,1).

We want the metric to be similar to the Euclidean metric.

Is it possible to define such a metric so that [0,1) is a complete metric space?

We want the metric to be similar to the Euclidean metric.

Is it possible to define such a metric so that [0,1) is a complete metric space?

Yes.

d(x,y)=|x/(1-x) - y/(1-y)|

f:[0,1)->[0,oo), f(x)=x/(1-x), f is a continuous bijection

d(x,y)=|x/(1-x) - y/(1-y)|

f:[0,1)->[0,oo), f(x)=x/(1-x), f is a continuous bijection

X set; F:P(X)->[0,oo]; F(O)=0; E,FcX; (G'=X\G)

(a) /\AcX F(A) = F(A n E) + F(A n E')

(b) /\AcX F(A) = F(A n F) + F(A n F')

Prove something interesting for the set E u F.

(a) /\AcX F(A) = F(A n E) + F(A n E')

(b) /\AcX F(A) = F(A n F) + F(A n F')

Prove something interesting for the set E u F.

/\AcX F(A) = F( A n (EuF) ) + F( A n (EuF)' )

page 39 in 2nd measure

page 39 in 2nd measure

Let f[n], g[n] be sequences of functions X->|C. Let f,g : X->|C.

Suppose that f[n] and g[n] converge uniformly to f and g respectively.

Can we conclude that f[n]*g[n] converges uniformly to f*g?

Suppose that f[n] and g[n] converge uniformly to f and g respectively.

Can we conclude that f[n]*g[n] converges uniformly to f*g?

NO.

Let X be |R. Let f[n](x) = 1/n. Let f(x)=0. Let g[n](x) = g(x) = x.

Let X be |R. Let f[n](x) = 1/n. Let f(x)=0. Let g[n](x) = g(x) = x.

Let (X,d) be a metric space. Let A c X.

Consider the set Fr(A) = Clo(A) \ Int(A) = Clo(A) n Clo(X\A).

Is it possible that Fr(A) contains a ball?

Consider the set Fr(A) = Clo(A) \ Int(A) = Clo(A) n Clo(X\A).

Is it possible that Fr(A) contains a ball?

YES.

It can even be an open set.

It can even be the whole space.

Consider |Q as a subset of |R with the Euclidean metric.

Then Fr(|Q) = |R.

It can even be an open set.

It can even be the whole space.

Consider |Q as a subset of |R with the Euclidean metric.

Then Fr(|Q) = |R.

sup { K > 0 : /\a,b:-R K*(|a|+|b|) <= sqrt(a*a+b*b) } = ???

1/sqrt(2)

Let X be a complete metric space, with no isolated points. Let
A be a dense subset of X. Suppose that A is a countable
intersection of open sets. What can we conclude about A?

A is uncountable

page 48 in OLDTIMER

page 48 in OLDTIMER

Let X be a metric space with no isolated points. Let A be a
countable dense subset of X. Suppose that A is a countable
intersection of open sets. What can we conclude about the
metric space X?

X is incomplete

page 48 in OLDTIMER

page 48 in OLDTIMER

Let X be a complete metric space, with no isolated points. Let
A be a countable subset of X, which is a countable intersection
of open sets. What can we conclude about A?

A is not dense

page 48 in OLDTIMER

page 48 in OLDTIMER

Let X be a complete metric space. Let A be an infinitely
countable dense subset of X, which is a countable intersection
of open sets. Is this setup possible?

It is possible. For example:

Let X be {0} u {1/n : n:-N}.

Let A = X.

Let A be a countable intersection of itself.

Or, let A = X\{0}.

Let X be {0} u {1/n : n:-N}.

Let A = X.

Let A be a countable intersection of itself.

Or, let A = X\{0}.

Let X be a complete metric space with no isolated points. Let A
be a countable dense subset of X, which is a countable
intersection of open sets. Is this setup possible?

NO.

page 48 in OLDTIMER

page 48 in OLDTIMER

inf {x + 1/x : x>0} = ???

inf {x + 1/x : x>0} = 2

X set, W c P(X).

(1) X :- W

(2) A,B :- W => A u B :- W

(3) A :- W => X\A :- W

Prove that

(4) A,B :- W => A n B :- W

(1) X :- W

(2) A,B :- W => A u B :- W

(3) A :- W => X\A :- W

Prove that

(4) A,B :- W => A n B :- W

X \ (A n B) = (X\A) u (X\B) belongs, hence (4) is OK.

A \ B = A n (X\B) belongs, hence (5) is OK.

A \ B = A n (X\B) belongs, hence (5) is OK.

Let f be a complex function defined on an open set. Suppose
that this function is diffable at a point. Prove that at this
point function Re(f) has a partial derivative with respect to
the first variable.

page 18 in OLDTIMER

Let f be a complex function defined on an open set. Suppose
that this function is diffable at a point. Prove that at this
point function Im(f) has a partial derivative with respect to
the second variable.

page 18 in OLDTIMER

Let m be a natural number. Let z be a complex number with |z| <
1.

Express differently:

+(n=0 to n=oo) [ z^n * (n+m)! / n!*m! ] = ???

Express differently:

+(n=0 to n=oo) [ z^n * (n+m)! / n!*m! ] = ???

= 1 / (1-z)^(m+1)

page 190 in the palace notebook

page 190 in the palace notebook

Let A,B be subsets of a metric space. Let p be a point in this
space.

Suppose that for every ball with center p,

if this ball is contained in A, then it is not contained in B.

What can we conclude about point p?

Suppose that for every ball with center p,

if this ball is contained in A, then it is not contained in B.

What can we conclude about point p?

p does not belong to Int( A n B )

Let G be an open subset of a metric space X.

Show that for every A c X,

G n Clo(A) is non-empty => G n A is non-empty.

Show that for every A c X,

G n Clo(A) is non-empty => G n A is non-empty.

______

Let G be an open subset of a metric space X.

Show that for every A c X,

G n A is empty => G n Clo(A) is empty.

Show that for every A c X,

G n A is empty => G n Clo(A) is empty.

_____

Let f[n]:E->C, g[n]:E->C be functional sequences.

Suppose that f converges uniformly to the zero function.

Suppose that g[n] is uniformly bounded.

What can we conclude about f[n]*g[n]?

Suppose that f converges uniformly to the zero function.

Suppose that g[n] is uniformly bounded.

What can we conclude about f[n]*g[n]?

f[n]*g[n] converges uniformly to the zero function.

Consider the series x^n / (1 - x^n).

Where does it converge? Consider x complex, |x| != 1.

Where does it converge? Consider x complex, |x| != 1.

|x|!=1 => [ |x|<1 <=> converges ]

X set, M c X.

W = {E c X : M\E is finite}

Describe R(W) - the ring generated by W.

W = {E c X : M\E is finite}

Describe R(W) - the ring generated by W.

R(W) = {E c X : M\E is finite v MnE is finite}

X set, W c P(X).

F : W -> [-oo,oo]

What does it mean that F is subadditive?

F : W -> [-oo,oo]

What does it mean that F is subadditive?

/\A,B:-W [ AuB :- W ==> F(AuB) <= F(A) + F(B) ]

X set, W c P(X).

F : W -> [-oo,oo]

What does it mean that F is finitely subadditive?

F : W -> [-oo,oo]

What does it mean that F is finitely subadditive?

For every natural n, if A[1],...,A[n] belong to W, and if their
union belongs to W,

then F(their union) <= +(k=1 to k=n) [ F(A[k]) ].

then F(their union) <= +(k=1 to k=n) [ F(A[k]) ].

X set, W c P(X).

F : W -> [-oo,oo]

What does it mean that F is countably subadditive?

F : W -> [-oo,oo]

What does it mean that F is countably subadditive?

For every sequence A[n] of sets in W whose union is also in W,

we have F(their countable union) <= +(n=1 to n=oo) [ F(A[n]) ].

we have F(their countable union) <= +(n=1 to n=oo) [ F(A[n]) ].

X set.

What is "the counting measure on X"?

What is "the counting measure on X"?

F : P(X) -> [0,oo]

If A c X is finite, then F(A) = the number of elements of A.

If A c X is infinite, then F(A) = oo.

Notice that it is countably additive!

If A c X is finite, then F(A) = the number of elements of A.

If A c X is infinite, then F(A) = oo.

Notice that it is countably additive!

Let X be an arbitrary set. Does there exist an extended real-
valued function on P(X) that is countably additive?

Yes. The counting measure.

F : P(X) -> [0,oo]

If A c X is finite, then F(A) = the number of elements of A.

If A c X is infinite, then F(A) = oo.

Check that it is countably additive!

F : P(X) -> [0,oo]

If A c X is finite, then F(A) = the number of elements of A.

If A c X is infinite, then F(A) = oo.

Check that it is countably additive!

Let L be a countable set.

Let W = { A c L : L\A is finite }.

Describe M(W) - the monotone class generated by W.

Let W = { A c L : L\A is finite }.

Describe M(W) - the monotone class generated by W.

M(W) = P(L).

page 48 in 1st measure

page 48 in 1st measure

Let A,B be subsets of a metric space.

Suppose that A n Clo(B) is non-empty and A n B is empty.

What can we conclude from this?

Suppose that A n Clo(B) is non-empty and A n B is empty.

What can we conclude from this?

A is not open

X set, let M be a countable subset of X.

W = {E c X : M\E is finite}

Describe S(W) - the s-ring generated by W.

W = {E c X : M\E is finite}

Describe S(W) - the s-ring generated by W.

S(W) = { E c X : MnE is countable or M\E is countable }

Let X be a set, and W c P(X), A c X.

Define "A is W-open" in the language of generalized convergence.

Define "A is W-open" in the language of generalized convergence.

Let Fr(A) denote the boundary of A. (see the prev item)

A is W-open <=> Fr(A) c X\A

page 47 in gen top

A is W-open <=> Fr(A) c X\A

page 47 in gen top

Express differently:

A \ //\\t:T [ B[t] ] = ???

A \ //\\t:T [ B[t] ] = ???

A \ //\\t:T [ B[t] ] = \\//t:-T [ A\B[t] ]

Express differently:

\\//t:-T [ A\B[t] ] = ???

\\//t:-T [ A\B[t] ] = ???

\\//t:-T [ A\B[t] ] = A \ //\\t:T [ B[t] ]

Express differently:

( \\//t:-T B[t] ) \ A = ???

( \\//t:-T B[t] ) \ A = ???

( \\//t:-T B[t] ) \ A = \\//t:-T ( B[t]\A )

Express differently:

\\//t:-T B[t]\A = ???

\\//t:-T B[t]\A = ???

\\//t:-T ( B[t]\A ) = ( \\//t:-T B[t] ) \ A

Express differently:

( //\\t:-T B[t] ) \ A = ???

( //\\t:-T B[t] ) \ A = ???

( //\\t:-T B[t] ) \ A = //\\t:-T ( B[t]\A )

Express differently:

//\\t:-T ( B[t]\A ) = ???

//\\t:-T ( B[t]\A ) = ???

//\\t:-T ( B[t]\A ) = ( //\\t:-T B[t] ) \ A

Express differently:

( Ut:-T A[t] ) u ( Ut:-T B[t] ) = ???

( Ut:-T A[t] ) u ( Ut:-T B[t] ) = ???

( Ut:-T A[t] ) u ( Ut:-T B[t] ) = Ut:T A[t] u B[t]

Express differently:

Ut:T A[t] u B[t] = ???

Ut:T A[t] u B[t] = ???

Ut:T A[t] u B[t] = ( Ut:-T A[t] ) u ( Ut:-T B[t] )

Express differently:

( //\\t:-T A[t] ) n ( //\\t:-T B[t] ) = ???

( //\\t:-T A[t] ) n ( //\\t:-T B[t] ) = ???

( //\\t:-T A[t] ) n ( //\\t:-T B[t] ) = //\\t:-T A[t]nB[t]

Express differently:

//\\t:-T A[t]nB[t] = ???

//\\t:-T A[t]nB[t] = ???

//\\t:-T A[t]nB[t] = ( //\\t:-T A[t] ) n ( //\\t:-T B[t] )

Complete the inclusion:

( Ut:-T A[t] n B[t] ) c ???

( Ut:-T A[t] n B[t] ) c ???

( Ut:-T A[t] n B[t] ) c (Ut:-T A[t]) n (Ut:-T B[t])

Decide if it's true:

Ut:-T A[t] n B[t] c (Ut:-T A[t]) n (Ut:-T B[t])

Ut:-T A[t] n B[t] c (Ut:-T A[t]) n (Ut:-T B[t])

YES.

Decide if it's true:

(Ut:-T A[t]) n (Ut:-T B[t]) c ( Ut:-T A[t] n B[t] )

(Ut:-T A[t]) n (Ut:-T B[t]) c ( Ut:-T A[t] n B[t] )

NO.

Complete the inclusion:

??? c ( (Ut:-T A[t]) n (Ut:-T B[t]) )

??? c ( (Ut:-T A[t]) n (Ut:-T B[t]) )

( Ut:-T A[t] n B[t] ) c ( (Ut:-T A[t]) n (Ut:-T B[t]) )

Complete the inclusion:

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c ???

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c ???

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c //\\t:-T A[t] u
B[t]

Complete the inclusion:

??? c //\\t:-T A[t] u B[t]

??? c //\\t:-T A[t] u B[t]

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c //\\t:-T A[t] u
B[t]

Decide if it's true:

//\\t:-T A[t] u B[t] c ( //\\t:-T A[t] ) u ( //\\t:-T B[t] )

//\\t:-T A[t] u B[t] c ( //\\t:-T A[t] ) u ( //\\t:-T B[t] )

It is false.

Decide if it's true:

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c //\\t:-T A[t] u B[t]

( //\\t:-T A[t] ) u ( //\\t:-T B[t] ) c //\\t:-T A[t] u B[t]

It is true.

Let V be a vector space over |R or over |C.

f : V -> |R

What does it mean that f is a norm?

f : V -> |R

What does it mean that f is a norm?

(1) /\x:-V [ x=O <=> f(x) = 0 ]

(2) /\x,y:-V [ f(x+y) <= f(x) + f(y) ]

(3) /\a:-K /\x:-V [ f(a*x) = |a|*f(x) ]

K is |R or |C

(2) /\x,y:-V [ f(x+y) <= f(x) + f(y) ]

(3) /\a:-K /\x:-V [ f(a*x) = |a|*f(x) ]

K is |R or |C

Let V be a normed vector space.

Prove that /\x:-V ||x|| >= 0.

Prove that /\x:-V ||x|| >= 0.

||x|| = 1/2 * [ ||0-x|| + ||x-0|| ] >= 1/2 * ||0-0|| = 0

Let V be a normed vector space.

Prove that /\x,y,z:-V ||x-y|| <= ||x-z|| + ||z-y||.

Prove that /\x,y,z:-V ||x-y|| <= ||x-z|| + ||z-y||.

hint: x-y = (x-z)+(z-y)

Let V be a normed vector space.

Prove that /\x:-V ||-x|| = ||x||.

Prove that /\x:-V ||-x|| = ||x||.

Recall that -x = (-1)*x

Let V be a normed vector space.

Prove that /\x,y:-V ||x-y|| <= ||x|| + ||y||.

Prove that /\x,y:-V ||x-y|| <= ||x|| + ||y||.

||x-y|| = ||x+(-y)|| <= ||x|| + ||-y|| =

= ||x|| + ||(-1)*y|| = ||x|| + |-1|*||y|| = ||x|| + ||y||

= ||x|| + ||(-1)*y|| = ||x|| + |-1|*||y|| = ||x|| + ||y||

Define the Euclidean norm on |R^n.

x = (x[1],x[2],...,x[n])

||x|| = sqrt( +(k=1 to k=n) x[k]*x[k] )

(In the proof that it is a norm we have to use the Schwarz inequality.)

page 39 in "Topologia", the first/old notebook

||x|| = sqrt( +(k=1 to k=n) x[k]*x[k] )

(In the proof that it is a norm we have to use the Schwarz inequality.)

page 39 in "Topologia", the first/old notebook

Define the max norm on |R^n and prove that it is a norm.

x = (x[1],x[2],...,x[n])

||x|| = max { |x[k]| : k:-{1,2,...,n} }.

||x|| = max { |x[k]| : k:-{1,2,...,n} }.

Define the taxi norm on |R^n and prove that it is a norm.

x = (x[1],x[2],...,x[n])

||x|| = +(k=1 to k=n) |x[k]|

||x|| = +(k=1 to k=n) |x[k]|

What does it mean that two norms are equivalent?

Let W be a vector space. Let f,g be two norms on W.

\/A>0 \/B>0 /\x:-W A*f(x) <= g(x) <= B*f(x)

It is an equivalence relation.

\/A>0 \/B>0 /\x:-W A*f(x) <= g(x) <= B*f(x)

It is an equivalence relation.

Let a,b be real numbers.

sqrt(a*a + b*b) <= ???

sqrt(a*a + b*b) <= ???

sqrt(a*a + b*b) <= |a| + |b|

Let D be a connected subset of |C.

Let f,g:D->|C be continuous.

Suppose that /\z:-D exp(f(z)) = exp(g(x)).

What can we say about functions f and g?

Let f,g:D->|C be continuous.

Suppose that /\z:-D exp(f(z)) = exp(g(x)).

What can we say about functions f and g?

\/k:-|Z /\z:-D f(z) = g(z) + k*2*PI*i

page 37 in OLDTIMER

page 37 in OLDTIMER

A,M,B are arbitrary sets.

Express differently in three ways:

(M\A)\B = ??? = ??? = ???

Express differently in three ways:

(M\A)\B = ??? = ??? = ???

(M\A)\B = M\(AuB) = (M\A)n(M\B) = (M\B)\A

A,M,B are arbitrary sets.

Express it differently in three ways:

M\(AuB) = ??? = ??? = ???

Express it differently in three ways:

M\(AuB) = ??? = ??? = ???

M\(AuB) = (M\A)n(M\B) = (M\A)\B = (M\B)\A

A,M,B are arbitrary sets.

Express it differently in three ways:

(M\A)n(M\B) = ??? = ??? = ???

Express it differently in three ways:

(M\A)n(M\B) = ??? = ??? = ???

(M\A)n(M\B) = (M\A)\B = M\(AuB) = (M\B)\A

Let A,B subsets of X. If GcX, then 1(G) denotes the
characteristic function of G.

Express differently:

1(AuB) = ???

Express differently:

1(AuB) = ???

1(AuB) = max {1(A),1(B)}

Let A,B be subsets of X. If GcX, then 1(G) denotes the
characteristic function of G.

Express differently:

max {1(A),1(B)} = ???

Express differently:

max {1(A),1(B)} = ???

max {1(A),1(B)} = 1(AuB)

Let A,B subsets of X. If GcX, then 1(G) denotes the
characteristic function of G.

Express differently:

A c B <=> ???

Express differently:

A c B <=> ???

A c B <=> 1(A)<=1(B)

Let A,B be subsets of X. If GcX, then 1(G) denotes the
characteristic function of G.

Express differently:

( 1(A)<=1(B) ) <=> ???

Express differently:

( 1(A)<=1(B) ) <=> ???

( 1(A)<=1(B) ) <=> A c B

Let x>0, y>0.

Complete without proof:

Arg(x+iy) = arctan(....)

Complete without proof:

Arg(x+iy) = arctan(....)

Arg(x+iy) = arctan(y/x)

page 91 in OLDTIMER

page 91 in OLDTIMER

Let x>0, y>0.

Complete without proof:

Arg(x+iy) = arccos( ??? )

Complete without proof:

Arg(x+iy) = arccos( ??? )

Arg(x+iy) = arccos(x / sqrt(x*x+y*y))

You can find the proof on pages 91,95 in OLDTIMER.

You can find the proof on pages 91,95 in OLDTIMER.

Show a function that satisfies:

(1) f:D->|R; D is an open subset of |R^2

(2) partial derivatives fxy and fyx exist at (x0,y0)

(3) fxy(x0,y0) != fyx(x0,y0)

(1) f:D->|R; D is an open subset of |R^2

(2) partial derivatives fxy and fyx exist at (x0,y0)

(3) fxy(x0,y0) != fyx(x0,y0)

f(0,0) = 0

f(x,y) = xy(x^2-y^2) / (x^2+y^2)

fxy(0,0) = -1

fyx(0,0) = 1

f(x,y) = xy(x^2-y^2) / (x^2+y^2)

fxy(0,0) = -1

fyx(0,0) = 1

Let R c P(X) be a ring of sets. Let A c X.

M(R) denotes the monotone class generated by R.

Show that { B:-M(R) : AuB,A\B,B\A :- M(R) } is a monotone class.

M(R) denotes the monotone class generated by R.

Show that { B:-M(R) : AuB,A\B,B\A :- M(R) } is a monotone class.

page 51 in 1st measure

Let R c P(X) be a ring of sets. Let A:-R.

M(R) denotes the monotone class generated by R.

Prove that R is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

M(R) denotes the monotone class generated by R.

Prove that R is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

page 52 in 1st measure

Let R c P(X) be a ring of sets. Let A:-R.

M(R) denotes the monotone class generated by R.

Prove that M(R) is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

M(R) denotes the monotone class generated by R.

Prove that M(R) is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

Step1: R is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

Step2: { B:-M(R) : AuB,A\B,B\A :- M(R) } is a monotone class.

See the two previous items.

page 53 in 1st measure

Step2: { B:-M(R) : AuB,A\B,B\A :- M(R) } is a monotone class.

See the two previous items.

page 53 in 1st measure

Let R c P(X) be a ring of sets.

M(R) denotes the monotone class generated by R.

Let A belong to M(R).

Prove that M(R) is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

M(R) denotes the monotone class generated by R.

Let A belong to M(R).

Prove that M(R) is contained in { B:-M(R) : AuB,A\B,B\A :- M(R) }.

First show that R is contained in { B:-M(R) : AuB,A\B,B\A :-
M(R) }.

You have to use the previous item to do this.

page 53 in 1st measure

You have to use the previous item to do this.

page 53 in 1st measure

Solve the equation: exp(z) = 2 + 2*i.

Re(z) = log(2*sqrt(2))

Im(z) = PI/4 + k*2*PI

Im(z) = PI/4 + k*2*PI

Let R c P(X) be a ring of sets.

M(R) denotes the monotone class generated by R.

S(R) denotes the s-ring generated by R.

Prove that M(R) = S(R).

M(R) denotes the monotone class generated by R.

S(R) denotes the s-ring generated by R.

Prove that M(R) = S(R).

0) Define for AcX, M[A] = { B:-M(R) : AuB,A\B,B\A :- M(R) }.

1) /\AcX M[A] is a monotone class.

2) /\A:-R R c M[A]

3) /\A:-R M(R) c M[A]

4) /\A:-M(R) R c M[A]

5) /\A:-M(R) M(R) c M[A]

6) Conclude that M(R) is a ring.

1) /\AcX M[A] is a monotone class.

2) /\A:-R R c M[A]

3) /\A:-R M(R) c M[A]

4) /\A:-M(R) R c M[A]

5) /\A:-M(R) M(R) c M[A]

6) Conclude that M(R) is a ring.

X set, K c P(X), W c P(X), K and W are monotone, A c X

What can we say about { B:-K : AuB:-W } ?

What can we say about { B:-K : AuB:-W } ?

It is monotone.

X set, K c P(X), W c P(X), K and W are monotone, A c X

What can we say about { B:-K : A\B:-W } ?

What can we say about { B:-K : A\B:-W } ?

It is monotone.

X set, K c P(X), W c P(X), K and W are monotone, A c X

What can we say about { B:-K : AnB:-W } ?

What can we say about { B:-K : AnB:-W } ?

It is monotone.

Let z be a complex number.

Re(i*z) = ???

Re(i*z) = ???

Re(i*z) = (-1)*Im(z)

Let z be a complex number.

Im(i*z) = ???

Im(i*z) = ???

Im(i*z) = Re(z)

Let z be a complex number.

| Re(z) | <= ???

| Re(z) | <= ???

| Re(z) | <= |z|

Let z be a complex number.

| Im(z) | <= ???

| Im(z) | <= ???

| Im(z) | <= |z|

Let a[n] be a sequence of complex numbers, indexed from n=0 to
n=oo.

Suppose lim_sup |a[n]|^(1/n) <= 1.

Let S[n] = a[0] + a[1] + ... + a[n].

Let z be a complex number with |z|<1.

What can we say about +(n=0 to n=oo) [ S[n]*z^n ] ?

Suppose lim_sup |a[n]|^(1/n) <= 1.

Let S[n] = a[0] + a[1] + ... + a[n].

Let z be a complex number with |z|<1.

What can we say about +(n=0 to n=oo) [ S[n]*z^n ] ?

+(n=0 to n=oo) [ S[n]*z^n ] = 1/(1-z) * ( +(n=0 to n=oo) [
a[n]*z^n ] )

Hint: use the theorem about multiplying series.

page 14 in OLDTIMER

Hint: use the theorem about multiplying series.

page 14 in OLDTIMER

Let A be an connected subset of |R.

Let g : A -> |C diffable.

Suppose that g'(x)=0 for all x:-A.

What can we say about function g?

Let g : A -> |C diffable.

Suppose that g'(x)=0 for all x:-A.

What can we say about function g?

It is constant.

page 29 in OLDTIMER

page 29 in OLDTIMER

Let D be an open connected subset of |C.

Let g : D -> |C diffable.

Suppose that g'(z)=0 for all z:-D.

What can we say about function g?

Let g : D -> |C diffable.

Suppose that g'(z)=0 for all z:-D.

What can we say about function g?

It is constant.

page 30 in OLDTIMER

page 30 in OLDTIMER

Let (X,dX), (Y,dY) be metric spaces. Let f : XxY -> |R.

(1) /\x0:-X /\y0:-Y /\E>0 \/A>0 /\y:-Y [ dY(y,y0)<A ==> |f(x0,y)-f(x0,y0)|<E ]

(2) /\x0:-X /\E>0 \/A>0 /\x:-X /\y:-Y [ dX(x,x0)<A ==> |f(x,y)- f(x0,y)|<E ]

What can we say about function f?

(1) /\x0:-X /\y0:-Y /\E>0 \/A>0 /\y:-Y [ dY(y,y0)<A ==> |f(x0,y)-f(x0,y0)|<E ]

(2) /\x0:-X /\E>0 \/A>0 /\x:-X /\y:-Y [ dX(x,x0)<A ==> |f(x,y)- f(x0,y)|<E ]

What can we say about function f?

It is continuous. The proof is swift.

Let z,w be complex numbers, z != 0.

exp(w) = z <=> ???

exp(w) = z <=> ???

Re(w)=log|z| and Im(w)=Arg(z)+k*2*PI, for some integer k

w = log|z| + i*Arg(z) + k+2*PI*i, for some integer k

page 60,36 in OLDTIMER

w = log|z| + i*Arg(z) + k+2*PI*i, for some integer k

page 60,36 in OLDTIMER

Arg : |C \ {0} -> ]-PI,PI] ; z = |z|*exp(i*Arg(z)).

Where is Arg continuous?

Where is Arg continuous?

|C \ {z<=0}

page 38 in OLDTIMER

page 38 in OLDTIMER

Let D be an open subset of |R^2.

Let f : D -> |R.

Suppose that f has bounded partial derivatives.

Does f have to be continuous?

Let f : D -> |R.

Suppose that f has bounded partial derivatives.

Does f have to be continuous?

Yes.

page 40 in OLDTIMER

page 40 in OLDTIMER

Let D be an open subset of |R^2.

Let f : D -> |R. Suppose that the partial derivative fy exists on D.

(1) /\(x0,y0):-D /\E>0 \/A>0 /\(x,y0):-D |x-x0|<A ==> |f(x,y0)-f(x0,y0)|<E

(2) \/M>0 /\(x,y):-D |fy(x,y)|<M

Let f : D -> |R. Suppose that the partial derivative fy exists on D.

(1) /\(x0,y0):-D /\E>0 \/A>0 /\(x,y0):-D |x-x0|<A ==> |f(x,y0)-f(x0,y0)|<E

(2) \/M>0 /\(x,y):-D |fy(x,y)|<M

Yes.

page 41 in OLDTIMER

page 41 in OLDTIMER

Let D,E be open subsets of |C.

f : D -> |C; g : E -> |C; f(D) c E

(1) f is continuous on D

(2) g is diffable on f(D)

(3) /\z:-D g'(f(z)) != 0

(4) /\z:-D g(f(z))=z

f : D -> |C; g : E -> |C; f(D) c E

(1) f is continuous on D

(2) g is diffable on f(D)

(3) /\z:-D g'(f(z)) != 0

(4) /\z:-D g(f(z))=z

It is diffable and /\z:-D f'(z)=1/g'(f(z)).

page 42 in OLDTIMER

page 42 in OLDTIMER

D = |C \ {z<=0}. f : D -> |C. k is an integer.

f(z) = log|z| + i*(Arg(z) + 2*k*PI)

Is f diffable?

f(z) = log|z| + i*(Arg(z) + 2*k*PI)

Is f diffable?

Yes.

/\z:-D f'(z) = 1/z

page 43 in OLDTIMER

/\z:-D f'(z) = 1/z

page 43 in OLDTIMER

E = |C \ {z>=0}. Let w : E -> |C.

(1) /\z:-E exp(w(z)) = z

(2) w is continuous

Does such a function exist?

(1) /\z:-E exp(w(z)) = z

(2) w is continuous

Does such a function exist?

Yes.

If Im(z)>=0, then w(z) = log|z| + i*Arg(z).

If Im(z)<0, then w(z) = log|z| + i*Arg(z) + i*2*PI.

page 43 in OLDTIMER

If Im(z)>=0, then w(z) = log|z| + i*Arg(z).

If Im(z)<0, then w(z) = log|z| + i*Arg(z) + i*2*PI.

page 43 in OLDTIMER

E = |C \ {0}. Let w : E -> |C.

(1) /\z:-E exp(w(z)) = z

(2) w is continuous

Does such a function exist?

(1) /\z:-E exp(w(z)) = z

(2) w is continuous

Does such a function exist?

No.

page 45 in OLDTIMER

page 45 in OLDTIMER

Let A be an open subset of |R. Let h : A -> |C.

Suppose that h is diffable.

What can we say about function Re(h)?

Suppose that h is diffable.

What can we say about function Re(h)?

It is diffable and [Re(h)]' = Re(h').

page 46 in OLDTIMER

page 46 in OLDTIMER

Let A be an open subset of |R. Let h : A -> |C.

Suppose that h is diffable.

What can we say about function Im(h)?

Suppose that h is diffable.

What can we say about function Im(h)?

It is diffable and [Im(h)]' = Im(h').

page 46 in OLDTIMER

page 46 in OLDTIMER

Let D be an open subset of |C.

Let a,b :- |C and the interval [a,b] c D.

Let f : D -> |C diffable.

We have /\z:-D |f'(z)|<=M.

Prove that |f(b)-f(a)| <= ???.

Let a,b :- |C and the interval [a,b] c D.

Let f : D -> |C diffable.

We have /\z:-D |f'(z)|<=M.

Prove that |f(b)-f(a)| <= ???.

|f(b)-f(a)| <= 2*M*|b-a|

page 47 in OLDTIMER

page 47 in OLDTIMER

Let M be an open connected subset of |C.

Let h : M -> |C be diffable.

Suppose that Re(h) is constant on M.

What can we say about function h?

Let h : M -> |C be diffable.

Suppose that Re(h) is constant on M.

What can we say about function h?

It is constant.

page 49 in OLDTIMER

page 49 in OLDTIMER

Let M be an open connected subset of |C.

Let h : M -> |C be diffable.

Suppose that Im(h) is constant on M.

What can we say about function h?

Let h : M -> |C be diffable.

Suppose that Im(h) is constant on M.

What can we say about function h?

It is constant.

page 49 in OLDTIMER

page 49 in OLDTIMER

Let D be an open subset of |R^2. Let (a,b):-D. Let f : D -> |R.

(1) the second partial derivatives fxy and fyx exist on D

(2) fxy and fyx are continuous at (a,b).

Is it true that fxy(a,b) = fyx(a,b)?

(1) the second partial derivatives fxy and fyx exist on D

(2) fxy and fyx are continuous at (a,b).

Is it true that fxy(a,b) = fyx(a,b)?

Yes.

page 54 in OLDTIMER

page 54 in OLDTIMER

Let a,b:-|R. Let f : [a,b] -> |C.

Suppose that Re(f) and Im(f) are Riemann-integrable on [a,b].

Define the complex integral of f on [a,b].

Suppose that Re(f) and Im(f) are Riemann-integrable on [a,b].

Define the complex integral of f on [a,b].

Integral([a,b],f) := Integral([a,b],Re(f)) +
i*Integral([a,b],Im(f))

Let a,b:-|R. Let f : [a,b] -> |C. Let f be integrable.

Prove that

/\w:-|C Integral([a,b],w*f) = w * Integral([a,b],f)

Prove that

/\w:-|C Integral([a,b],w*f) = w * Integral([a,b],f)

page 56 in OLDTIMER

Think about complexification of real vector spaces.

page 154 in the algebra notebook

Think about complexification of real vector spaces.

page 154 in the algebra notebook

Let a,b:-|R. Let f,g : [a,b] -> |C. Let f,g be integrable.

Prove that

Integral([a,b],f+g) = Integral([a,b],f) + Integral([a,b],g).

Prove that

Integral([a,b],f+g) = Integral([a,b],f) + Integral([a,b],g).

Very easy.

Think about complexification of real vector spaces.

page 154 in the algebra notebook

Think about complexification of real vector spaces.

page 154 in the algebra notebook

Integral( 0, 4, 1/(1+sqrt(x)), dx ) = ???

4 - 2*log(3)

Integral( 0, 4, 1/(1+sqrt(x)), dx ) = 4 - 2*log(3)

hint: g(t)=t*t; later: log(1+t)

Integral( 0, 4, 1/(1+sqrt(x)), dx ) = 4 - 2*log(3)

hint: g(t)=t*t; later: log(1+t)

Integral( 0, log(2), sqrt( exp(x)-1 ), dx ) = ???

Integral( 0, log(2), sqrt( exp(x)-1 ), dx ) = 2 - pi/2

hint: g(t)=log(1+t*t); later: arctan

hint: g(t)=log(1+t*t); later: arctan

f:[0,1]->|R, f(x)=x*sqrt(1+x)

Calculate Integral([0,1],f)

Calculate Integral([0,1],f)

4*(sqrt(2)+1) / 15

hint1: g:[1,sqrt(2)]->|R, g(t)=t*t-1

hint2: t=sqrt(1+x)

hint1: g:[1,sqrt(2)]->|R, g(t)=t*t-1

hint2: t=sqrt(1+x)

g : |R -> |C

h : |C -> |C

h(z) = g(Re(z))

Suppose that g is diffable. Does h have to be diffable?

h : |C -> |C

h(z) = g(Re(z))

Suppose that g is diffable. Does h have to be diffable?

NO.

Choose g(x)=x.

Then h(z)=Re(z), and h is nowhere diffable!

Choose g(x)=x.

Then h(z)=Re(z), and h is nowhere diffable!

Let A be a connected subset of |R.

Let g : A -> |R.

Suppose that g is diffable and /\x:-A g'(x)=0.

What can we say about function g?

Let g : A -> |R.

Suppose that g is diffable and /\x:-A g'(x)=0.

What can we say about function g?

It is constant.

Use Lagrange's mean value theorem.

Use Lagrange's mean value theorem.

Let X be a metric space.

Let x[n] be a sequence in X.

Let B = \\//(k=1 to oo) //\\(n=k to oo) B(x[n], 1/n).

Suppose that B is not empty.

What can we conclude?

Let x[n] be a sequence in X.

Let B = \\//(k=1 to oo) //\\(n=k to oo) B(x[n], 1/n).

Suppose that B is not empty.

What can we conclude?

sequence x[n] converges

Let X be a set, and let W c P(X).

Define W-convergence in X.

Define W-convergence in X.

Let (T,>) be a directed set. Let x:T->X be a net in X. Let k:-X.

The net x W-converges to k

iff

/\B:-W [ k:-B ==> \/t0:-T /\t>t0 x(t):-B ]

page 44 in the General Topology notebook

The net x W-converges to k

iff

/\B:-W [ k:-B ==> \/t0:-T /\t>t0 x(t):-B ]

page 44 in the General Topology notebook

Let S be a set. Give a collection of sets, W c P(|R^S), such
that pointwise convergence of functions S->|R is equivalent to
W-convergence.

If s:-S, and Uc|R is open, define U(s) = {f:-|R^S : f(s):-U}.

Let W = {U(s): s:-S and Uc|R open}.

page 44 in the General Topology notebook

Let W = {U(s): s:-S and Uc|R open}.

page 44 in the General Topology notebook

Let X be a set, and W1,W2 c P(X).

Prove that if W1 c W2, then W2-convergence implies W1- convergence.

Prove that if W1 c W2, then W2-convergence implies W1- convergence.

page 45 in gen top

Let X be a set, and W c P(X). Let d be a metric on X.

Define what it means that W is compatible with d.

Define what it means that W is compatible with d.

W-convergence is the same as d-convergence

page 46 in the Gen Top notebook

page 46 in the Gen Top notebook

Let X be a set, and W c P(X), A c X, p:-X.

Define "p is W-adherent to A".

Define "p is W-adherent to A".

\/(T,>)directed set \/a:T->A [ net a W-converges to p ]

Notice the formal difficulty here.

page 46 in the Gen Top notebook

"p is W-adherent to A" =

= "there exists a net contained in A that converges to p"

Notice the formal difficulty here.

page 46 in the Gen Top notebook

"p is W-adherent to A" =

= "there exists a net contained in A that converges to p"

Let X be a set, and W c P(X), A c X.

Define "the boundary of A" in the language of generalized convergence.

Define "the boundary of A" in the language of generalized convergence.

the set of all points which are W-adherent to A and W-adherent
to X\A

page 46 in the Gen Top notebook

Let Fr(A) denote the boundary of A.

page 46 in the Gen Top notebook

Let Fr(A) denote the boundary of A.

Let X be a set, and W c P(X), A c X.

Define "A is W-closed" in the language of generalized convergence.

Define "A is W-closed" in the language of generalized convergence.

Let Fr(A) denote the boundary of A. (see the prev item)

A is W-closed <=> Fr(A) c A

page 47 in gen top

A is W-closed <=> Fr(A) c A

page 47 in gen top

Let X be a set, W c P(X), A c X.

Prove that

A is W-open <=> X\A is W-closed

Prove that

A is W-open <=> X\A is W-closed

Use Fr(A) = Fr(X\A)

page 47 in the gen top notebook

page 47 in the gen top notebook

Let X be a set, W c P(X), A c X.

Prove that

A u Fr(A) = {x:-X : x is W-adherent to A}

Prove that

A u Fr(A) = {x:-X : x is W-adherent to A}

Notice that if x:-B, then x is W-adherent to B.

page 47 in the gen top notebook

page 47 in the gen top notebook

Let X be a set, W c P(X), A c X.

(1) A is W-closed

(2) Every point W-adherent to A belongs to A.

Prove that (1)<=>(2).

(1) A is W-closed

(2) Every point W-adherent to A belongs to A.

Prove that (1)<=>(2).

page 48 in the gen top notebook

Let X be a set, and W c P(X).

What does W* mean in the language of generalized convergence?

What does W* mean in the language of generalized convergence?

W* is the set of all W-open sets

W* c P(X)

page 48 in Gen Top

W* c P(X)

page 48 in Gen Top

Let X be a set, and W c P(X).

Prove that every set belonging to W, is W-open.

Prove that every set belonging to W, is W-open.

page 48 in gen top

Let X be a set, and W c P(X).

Let (T,>) be a directed set.

Let x:T->X be a net in X, which W-converges to k:-X.

Let p:T->T such that /\t:-T p(t)>t.

What can we say about the net {x(p(t))}t:-T ?

Let (T,>) be a directed set.

Let x:T->X be a net in X, which W-converges to k:-X.

Let p:T->T such that /\t:-T p(t)>t.

What can we say about the net {x(p(t))}t:-T ?

It also W-converges to k.

page 49 in gen top

page 49 in gen top

Let X be a set, and W c P(X).

Prove that W-convergence and W*-convergence are the same.

Prove that W-convergence and W*-convergence are the same.

page 49 in gen top

Let X be a set. Is it possible to find W c P(X) such that every
net in X W-converges to every point in X?

Yes. Let W = {X}. Or let W = {O}.

Prove that

/\x,y:-|R x*exp( x*(y^2+1)+1 ) >= -1.

/\x,y:-|R x*exp( x*(y^2+1)+1 ) >= -1.

First show that /\x:-|R x*exp(x+1) >= -1.

Let W c P(X) be an algebra of sets. Let M c X.

Find the algebra generated by W u {M}.

K(W u {M}) = ???

Find the algebra generated by W u {M}.

K(W u {M}) = ???

K( W u {M} ) = { (A n M) u (B \ M) : A,B :- W }

page 51 in 2nd measure

page 51 in 2nd measure

Define the generalized Cartesian product.

Let J be a non-empty set. Let X be a set.

Let I : J -> P(X). Now {I(j)}j:-J is a collection of subsets of X.

The Cartesian product of sets I(j) indexed with j:-J is defined as follows:

{ f: f : J -> Uj:-J I(j) and /\j:-J f(j):-I(j) }

Let I : J -> P(X). Now {I(j)}j:-J is a collection of subsets of X.

The Cartesian product of sets I(j) indexed with j:-J is defined as follows:

{ f: f : J -> Uj:-J I(j) and /\j:-J f(j):-I(j) }

Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:-
I(j) A[i,j] be a set.

Complete:

\\//j:-J //\\i:-I(j) A[i,j] = ???

Complete:

\\//j:-J //\\i:-I(j) A[i,j] = ???

Let K be the Cartesian product of sets I(j) indexed with j:-J.

\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]

page 55 in 1st measure

\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]

page 55 in 1st measure

Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:-
I(j) A[i,j] be a set.

Let K be the Cartesian product of sets I(j) indexed with j:-J.

Complete:

//\\f:-K \\//j:-J A[f(j),j] = ???

Let K be the Cartesian product of sets I(j) indexed with j:-J.

Complete:

//\\f:-K \\//j:-J A[f(j),j] = ???

//\\f:-K \\//j:-J A[f(j),j] = \\//j:-J //\\i:-I(j) A[i,j]

page 55 in 1st measure

page 55 in 1st measure

Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:-
I(j) A[i,j] be a set.

Complete:

//\\j:-J \\//i:-I(j) A[i,j] = ???

Complete:

//\\j:-J \\//i:-I(j) A[i,j] = ???

Let K be the Cartesian product of sets I(j) indexed with j:-J.

//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]

page 55 in 1st measure

//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]

page 55 in 1st measure

Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:-
I(j) A[i,j] be a set.

Let K be the Cartesian product of sets I(j) indexed with j:-J.

Complete:

\\//f:-K //\\j:-J A[f(j),j] = ???

Let K be the Cartesian product of sets I(j) indexed with j:-J.

Complete:

\\//f:-K //\\j:-J A[f(j),j] = ???

\\//f:-K //\\j:-J A[f(j),j] = //\\j:-J \\//i:-I(j) A[i,j]

page 55 in 1st measure

page 55 in 1st measure

Let (Y,g) be a complete metric space. Let f : X -> Y be 1-1 and
onto.

Let d : XxX -> |R be defined by d(x,u) = g(f(x),f(u)).

What can we conclude?

Let d : XxX -> |R be defined by d(x,u) = g(f(x),f(u)).

What can we conclude?

(X,d) is a complete metric space

page 67 in OLDTIMER

page 67 in OLDTIMER

If W is a collection of sets,

let Ws denote the collection of all finite unions of sets from W,

let Wd denote the collection of all finite intersections of sets from W.

Prove that Wsd c Wds.

let Ws denote the collection of all finite unions of sets from W,

let Wd denote the collection of all finite intersections of sets from W.

Prove that Wsd c Wds.

Use

//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]

where K is the Cartesian product of sets I(j) indexed with j:-J.

page 57 in 1st measure

//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]

where K is the Cartesian product of sets I(j) indexed with j:-J.

page 57 in 1st measure

If W is a collection of sets,

let Ws denote the collection of all finite unions of sets from W,

let Wd denote the collection of all finite intersections of sets from W.

Prove that Wds c Wsd.

let Ws denote the collection of all finite unions of sets from W,

let Wd denote the collection of all finite intersections of sets from W.

Prove that Wds c Wsd.

Use

\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]

where K is the Cartesian product of sets I(j) indexed with j:-J.

page 57 in 1st measure

\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]

where K is the Cartesian product of sets I(j) indexed with j:-J.

page 57 in 1st measure

Let A be a collection of sets.

Can we conclude that M(R(A)) = S(A) ?

Can we conclude that M(R(A)) = S(A) ?

Yes.

1) R(A) c S(A)

2) S(R(A)) c S(A)

3) S(R(A)) = S(A)

4) M(R(A)) = S(R(A)) page 54 in 1st measure

5) M(R(A)) = S(A)

1) R(A) c S(A)

2) S(R(A)) c S(A)

3) S(R(A)) = S(A)

4) M(R(A)) = S(R(A)) page 54 in 1st measure

5) M(R(A)) = S(A)

Let D be an open subset of |R^2.

Let f : D -> |R.

Suppose that the partial derivatives of f exist on D.

Let (a,b) be a point in D.

Suppose that fx(a,b)=fy(a,b)=0.

Does it mean that f has an extremum at (a,b)?

Let f : D -> |R.

Suppose that the partial derivatives of f exist on D.

Let (a,b) be a point in D.

Suppose that fx(a,b)=fy(a,b)=0.

Does it mean that f has an extremum at (a,b)?

NO.

f(x,y)=x^2-y^2 at (0,0)

f(x,y)=x^2-y^2 at (0,0)

Let X be a set, and W c P(X).

Decide and prove which is true:

W* is the _________ collection of sets that generates W- convergence.

(1) largest

(2) smallest

Decide and prove which is true:

W* is the _________ collection of sets that generates W- convergence.

(1) largest

(2) smallest

W* is the largest collection of sets that generates W-
convergence.

page 51 in gen top

page 51 in gen top

Let X be a set, and W c P(X).

Prove that (W*)* c W*.

Prove that (W*)* c W*.

Use the fact that W-convergence implies W*-convergence.

page 52 in gen top

page 52 in gen top

X, W c P(X)

What does it mean that W is a semi-ring?

What does it mean that W is a semi-ring?

(1) W is non-empty

(2) A,B:-W => AnB:-W

(3) A,B:-W => A\B is a finite disjoint union of sets from W

page 60 in 1st measure

(2) A,B:-W => AnB:-W

(3) A,B:-W => A\B is a finite disjoint union of sets from W

page 60 in 1st measure

Let W c P(X) be a semi-ring.

Does W have to contain the empty set?

Does W have to contain the empty set?

Yes.

W is non-empty, hence A:-W.

Then A\A is a finite union of sets from W.

Hence W has to contain the empty set.

page 60 in 1st measure

W is non-empty, hence A:-W.

Then A\A is a finite union of sets from W.

Hence W has to contain the empty set.

page 60 in 1st measure

X, J c P(X)

J is a semi-ring.

Prove that Js = R(J).

Js denotes the collection of all finite unions of sets from J.

R(J) denotes the ring generated by J.

J is a semi-ring.

Prove that Js = R(J).

Js denotes the collection of all finite unions of sets from J.

R(J) denotes the ring generated by J.

Show that Js is a ring.

HINTS:

A\B :- Jsds

Jd = J

Jsds = Jdss = Jds = Js

HINTS:

A\B :- Jsds

Jd = J

Jsds = Jdss = Jds = Js

Let W c P(X) satisfy:

(1) /\B,A:-W B u A :- W

(2) /\B,A:-W B n A :- W

Does W have to be a ring?

(1) /\B,A:-W B u A :- W

(2) /\B,A:-W B n A :- W

Does W have to be a ring?

NO.

W = { {1} }

W = { {4}, {4,5} }

W = { O, {6}, {6,7} }

W = { {1} }

W = { {4}, {4,5} }

W = { O, {6}, {6,7} }

sup { sqrt(a/b) / (a + 1/b) : a,b>0 } = ???

= 1/2

Let 0 < b < a.

log(a/b) < ???

log(a/b) < ???

0 < b < a ==> log(a/b) < (a-b)/b

page 153 in golden gate

page 153 in golden gate

Let 0 < b < a.

log(a/b) > ???

log(a/b) > ???

0 < b < a ==> log(a/b) > (a-b)/a

page 153 in golden gate

page 153 in golden gate

Let 0 < b < a.

(a-b)/a < ??? < (a-b)/b

(a-b)/a < ??? < (a-b)/b

(a-b)/a < log(a/b) < (a-b)/b

Use Lagrange's Mean Value theorem.

page 153 in golden gate

Use Lagrange's Mean Value theorem.

page 153 in golden gate

Let X be a set, and J c P(X).

Let J be a semi-ring. Prove that every finite union of sets from J can be written as a disjoint finite union of sets from J.

Let J be a semi-ring. Prove that every finite union of sets from J can be written as a disjoint finite union of sets from J.

page 63 in 1st measure

page 66 in 2nd measure

page 66 in 2nd measure

Let X be a set, and J c P(X). Let J be a semi-ring.

Let H denote the set of all disjoint finite unions of sets from J.

What are the two steps that lead to the conclusion that R(J) = H?

Let H denote the set of all disjoint finite unions of sets from J.

What are the two steps that lead to the conclusion that R(J) = H?

1) R(J) = Js

2) Js = H

page 65 in 1st measure

2) Js = H

page 65 in 1st measure

Express differently:

( U(t:-T) A[t] ) x B = ???

( U(t:-T) A[t] ) x B = ???

( U(t:-T) A[t] ) x B = U(t:-T) ( A[t] x B )

Express differently:

U(t:-T) ( A[t] x B ) = ???

U(t:-T) ( A[t] x B ) = ???

U(t:-T) ( A[t] x B ) = ( U(t:-T) A[t] ) x B

Express differently:

A x U(t:-T) B[t] = ???

A x U(t:-T) B[t] = ???

A x U(t:-T) B[t] = U(t:-T) A x B[t]

Express differently:

U(t:-T) A x B[t] = ???

U(t:-T) A x B[t] = ???

U(t:-T) A x B[t] = A x U(t:-T) B[t]

Express differently:

A x //\\(t:-T) B[t] = ???

A x //\\(t:-T) B[t] = ???

A x //\\(t:-T) B[t] = //\\(t:-T) A x B[t]

Express differently:

//\\(t:-T) A x B[t] = ???

//\\(t:-T) A x B[t] = ???

//\\(t:-T) A x B[t] = A x //\\(t:-T) B[t]

Express differently:

( //\\(t:-T) A[t] ) x B = ???

( //\\(t:-T) A[t] ) x B = ???

( //\\(t:-T) A[t] ) x B = //\\(t:-T) ( A[t] x B )

Express differently:

//\\(t:-T) ( A[t] x B ) = ???

//\\(t:-T) ( A[t] x B ) = ???

//\\(t:-T) ( A[t] x B ) = ( //\\(t:-T) A[t] ) x B

Express differently:

( \\//(t:-T) A[t] ) x ( \\//g:-G B[g] ) = ???

( \\//(t:-T) A[t] ) x ( \\//g:-G B[g] ) = ???

( \\//(t:-T) A[t] ) x ( \\//g:-G B[g] ) = \\//(t:-T)
\\//(g:-G) A[t] x B[g]

Express differently:

\\//(t:-T) \\//(g:-G) A[t] x B[g] = ???

\\//(t:-T) \\//(g:-G) A[t] x B[g] = ???

\\//(t:-T) \\//(g:-G) A[t] x B[g] = ( \\//(t:-T) A[t] ) x (
\\//g:-G B[g] )

Express differently:

( //\\(t:-T) A[t] ) x ( //\\g:-G B[g] ) = ???

( //\\(t:-T) A[t] ) x ( //\\g:-G B[g] ) = ???

( //\\(t:-T) A[t] ) x ( //\\g:-G B[g] ) = //\\(t:-T)
//\\(g:-G) A[t] x B[g]

Express differently:

//\\(t:-T) //\\(g:-G) A[t] x B[g] = ???

//\\(t:-T) //\\(g:-G) A[t] x B[g] = ???

//\\(t:-T) //\\(g:-G) A[t] x B[g] = ( //\\(t:-T) A[t] ) x (
//\\g:-G B[g] )

Express differently:

(A x M) n (B x N) = ???

(A x M) n (B x N) = ???

(A x M) n (B x N) = (A n B) x (M n N)

Express differently:

(A n B) x (M n N) = ???

(A n B) x (M n N) = ???

(A n B) x (M n N) = (A x M) n (B x N)

Express this as a disjoint union:

(A x M) \ (B x N) = ???

(A x M) \ (B x N) = ???

(A x M) \ (B x N) = (A\B)x(MnN) u (A\B)x(M\N) u (AnB)x(M\N)

Let A c X and B c Y.

A x B c ???

A x B c ???

A x B c X x Y

(A x M) u (B x N) c ???

(A x M) u (B x N) c (A u B) x (M u N)

??? c (A u B) x (M u N)

(A x M) u (B x N) c (A u B) x (M u N)

Is this generally true?

(A u B) x (M u N) c (A x M) u (B x N)

(A u B) x (M u N) c (A x M) u (B x N)

No.

A = N = {1}

M = B = O

A = N = {1}

M = B = O

X, W c P(X), W is a semi-ring in X

Y, V c P(Y), V is a semi-ring in Y

Prove that { A x B : A:-W , B:-V } is a semi-ring in XxY.

Y, V c P(Y), V is a semi-ring in Y

Prove that { A x B : A:-W , B:-V } is a semi-ring in XxY.

page 67 in 1st measure

Let A,B be two semi-rings in X.

Does their intersection (A n B) have to be a semi-ring in X?

Does their intersection (A n B) have to be a semi-ring in X?

NO.

A = { O, {1,2,3}, {1}, {2,3} }

B = { O, {1,2,3}, {1}, {2}, {3} }

page 69 in 1st measure

A = { O, {1,2,3}, {1}, {2,3} }

B = { O, {1,2,3}, {1}, {2}, {3} }

page 69 in 1st measure

In the extended real number system, how is multiplication
defined?

(1) /\x,y:-|R* x*y = y*x

(2) 0*oo = 0

(3) /\x:-|R* [ x>0 ==> x*oo = oo ]

(4) /\a,b,c:-|R* (a*b)*c = a*(b*c)

(5) (-oo) = (-1)*oo

page 70 in 1st measure

(2) 0*oo = 0

(3) /\x:-|R* [ x>0 ==> x*oo = oo ]

(4) /\a,b,c:-|R* (a*b)*c = a*(b*c)

(5) (-oo) = (-1)*oo

page 70 in 1st measure

In the extended real number system,

why can't we define: oo+(-oo)=0?

why can't we define: oo+(-oo)=0?

Because then we could prove that oo+(-oo)=7 by using
associativity.

1) oo+(-oo)=0

2) 7+[ oo+(-oo) ] = 7 + 0

3) [7 + oo] + (-oo) = 7

4) oo + (-oo) = 7

1) oo+(-oo)=0

2) 7+[ oo+(-oo) ] = 7 + 0

3) [7 + oo] + (-oo) = 7

4) oo + (-oo) = 7

Let X be a set, and W c P(X).

What does it mean that function T:W->[-oo,oo] is additive?

What does it mean that function T:W->[-oo,oo] is additive?

/\B,A:-W [ BuA:-W and BnA=O ==> T(B u A) = T(B) + T(A) ]

Notice that if T is additive on W and O:-W,

then T(O)=0 or T(O)=oo or T(O)=-oo.

Notice that if T is additive on W and O:-W,

then T(O)=0 or T(O)=oo or T(O)=-oo.

Let X be a set, W c P(X), W is a ring.

Let T:W->|R* be additive.

Is it possible that T(A)=oo and T(B)=-oo for some A,B:-W ?

Let T:W->|R* be additive.

Is it possible that T(A)=oo and T(B)=-oo for some A,B:-W ?

NO.

page 71 in 1st measure

page 71 in 1st measure

f : |C -> |C

(1) f(0) = 1

(2) /\w,z:-|C f(w+z) = f(w)*f(z)

(3) \/a:-|C lim(z->0) (f(z)-1) / z = a

What can we conclude about function f?

(1) f(0) = 1

(2) /\w,z:-|C f(w+z) = f(w)*f(z)

(3) \/a:-|C lim(z->0) (f(z)-1) / z = a

What can we conclude about function f?

f is diffable and

/\z:-|C f'(z) = a*f(z)

page 75 in OLDTIMER

/\z:-|C f'(z) = a*f(z)

page 75 in OLDTIMER

A,B c X

1) A n B = O <=> ???

2) A n B = O <=> ???

1) A n B = O <=> ???

2) A n B = O <=> ???

1) A n B = O <=> A c X\B

2) A n B = O <=> B c X\A

2) A n B = O <=> B c X\A

A,B c X

1) A c X\B <=> ???

2) A c X\B <=> ???

1) A c X\B <=> ???

2) A c X\B <=> ???

1) A c X\B <=> B c X\A

2) A c X\B <=> B n A = O

2) A c X\B <=> B n A = O

A,B c X

1) A n (X\B) = O <=> ???

2) A n (X\B) = O <=> ???

1) A n (X\B) = O <=> ???

2) A n (X\B) = O <=> ???

1) A n (X\B) = O <=> A c B

2) A n (X\B) = O <=> X\B c X\A

2) A n (X\B) = O <=> X\B c X\A

A,B c X

1) A c B <=> ???

2) A c B <=> ???

3) A c B <=> ???

1) A c B <=> ???

2) A c B <=> ???

3) A c B <=> ???

1) A c B <=> X\B c X\A

2) A c B <=> A n (X\B) = O

3) A c B <=> A u B = B

2) A c B <=> A n (X\B) = O

3) A c B <=> A u B = B

A,B c X

A \ (X\B) = ???

A \ (X\B) = ???

A \ (X\B) = A n B = B n A = B \ (X\A)

Let X = {x:|N->|C : +(n=1 to oo) |x(n)| < oo}.

If x:-X, let g(x) = +(n=1 to oo) |x(n)|.

Is g a norm on X?

If x:-X, let g(x) = +(n=1 to oo) |x(n)|.

Is g a norm on X?

Yes.

Does this series converge?

1/n * (-1)^n * i^( n*(n+1) )

1/n * (-1)^n * i^( n*(n+1) )

YES.

a[n] = (-1)^n * i^( n*(n+1) )

a[1]=1 a[2]=-1 a[3]=-1 a[4]=1

a[n+4] = a[n]

Hence the series a[n] is bounded.

a[n] = (-1)^n * i^( n*(n+1) )

a[1]=1 a[2]=-1 a[3]=-1 a[4]=1

a[n+4] = a[n]

Hence the series a[n] is bounded.

Let D be an open subset of |R^2.

Let g : D -> |R. Suppose that the partial derivatives gx, gy both exist on D.

Let (a,b) be a point in D.

Suppose that /\(x,y):-D g(x,y) <= g(a,b).

What can we conclude?

Let g : D -> |R. Suppose that the partial derivatives gx, gy both exist on D.

Let (a,b) be a point in D.

Suppose that /\(x,y):-D g(x,y) <= g(a,b).

What can we conclude?

gx(a,b) = gy(a,b) = 0

Prove that

/\x:-|R \/k:-|Z [ -PI < x + k*2*PI <= PI ]

/\x:-|R \/k:-|Z [ -PI < x + k*2*PI <= PI ]

hint: k-1 < (x-PI)/(2*PI) <= k

page 62 in OLDTIMER

page 62 in OLDTIMER

-PI <= x,y <= PI

cos(x) = cos(y)

What can we conclude?

cos(x) = cos(y)

What can we conclude?

x = -y or x = y

Let x,y be real numbers such that cos(x)=cos(y).

What can we conclude?

What can we conclude?

\/k:-|Z x = y + k*2*PI or x = -y + k*2*PI

page 62 in OLDTIMER

page 62 in OLDTIMER

Let A be a collection of sets.

S(R(A)) = ???

S(R(A)) = ???

R(A) c S(A), hence S(R(A) c S(A)

A c R(A) c S(R(A)), hence S(A) c S(R(A))

S(R(A)) = S(A)

see page 35 in 1st measure

A c R(A) c S(R(A)), hence S(A) c S(R(A))

S(R(A)) = S(A)

see page 35 in 1st measure

Let W c P(X). Let u : W -> |R* be additive.

(1) A, B, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(B)?

(1) A, B, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(B)?

Answer: u(B) :- |R

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

Let W c P(X). Let u : W -> |R* be additive.

(1) B, A, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(A\B)?

(1) B, A, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(A\B)?

Answer: u(A\B):-|R and u(A\B) = u(A) - u(B)

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

4) u(A) - u(B) = u(A\B)

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

4) u(A) - u(B) = u(A\B)

Let A[n] be a decreasing sequence of sets.

Prove that for every n:-|N

A[n] = A[1] \ ( U(k=1 to k=n-1) A[k]\A[k+1] )

Prove that for every n:-|N

A[n] = A[1] \ ( U(k=1 to k=n-1) A[k]\A[k+1] )

2) page 53 in 2nd measure notebook

1) page 76 in 1st measure notebook

2) is better than 1)

1) page 76 in 1st measure notebook

2) is better than 1)

x does not belong to A\B <=> [ ??? => ??? ]

x does not belong to A\B <=> [ x:-A => x:-B ]

inspiring line:{x does not belong to A\B <=> x:-AcB}

inspiring line:{x does not belong to A\B <=> x:-AcB}

Let A[n] be a sequence of sets.

Prove that

//\\(n:-|N) A[n] = A[1] \ ( U(n:-|N) A[n] \ A[n+1] )

Prove that

//\\(n:-|N) A[n] = A[1] \ ( U(n:-|N) A[n] \ A[n+1] )

page 78 in 1st measure

Let W c P(X) be a s-ring.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let /\n:-|N A[n] :- W.

Suppose that the series y(A[n]) converges.

What can we conclude about the value of y(lim_sup A[n]) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let /\n:-|N A[n] :- W.

Suppose that the series y(A[n]) converges.

What can we conclude about the value of y(lim_sup A[n]) ?

y(lim_sup A[n]) = 0

page 54 in 2nd measure

page 54 in 2nd measure

Let a[n] be a sequence of complex numbers.

Suppose that the series a[n] converges.

What can we immediately conclude?

Suppose that the series a[n] converges.

What can we immediately conclude?

1) a[n] tends to zero

2) lim(k->oo) +(n=k to n=oo) a[n] = 0

2) lim(k->oo) +(n=k to n=oo) a[n] = 0

Let W c P(X) be a s-ring.

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the value y( U(n:-|N) A[n] ) ?

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the value y( U(n:-|N) A[n] ) ?

y( U(n:-|N) A[n] ) = lim(n->oo) y(A[n])

(this limit is proven to exist in the course of the proof)

page 157 in 1st measure

(this limit is proven to exist in the course of the proof)

page 157 in 1st measure

Let W c P(X) be a s-ring.

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the limit of y(A[n]) ?

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the limit of y(A[n]) ?

It exists and

lim(n->oo) y(A[n]) = y( U(n:-|N) A[n] )

hint: A[n] is increasing, hence A[n] = \\*//(k=1 to k=n) A[k]\A[k-1].

page 157 in 1st measure

lim(n->oo) y(A[n]) = y( U(n:-|N) A[n] )

hint: A[n] is increasing, hence A[n] = \\*//(k=1 to k=n) A[k]\A[k-1].

page 157 in 1st measure

Let W c P(X) be a s-ring.

Let y : W -> [-oo,oo] be countably additive and y(O)=0.

Let A[n] be a decreasing sequence of sets from W.

Suppose that y(A[k]) :- |R, for some k:-|N.

What can we conclude about the value of y( //\\(n:-|N) A[n] ) ?

Let y : W -> [-oo,oo] be countably additive and y(O)=0.

Let A[n] be a decreasing sequence of sets from W.

Suppose that y(A[k]) :- |R, for some k:-|N.

What can we conclude about the value of y( //\\(n:-|N) A[n] ) ?

y( //\\(n:-|N) A[n] ) = lim(n->oo) y(A[n])

the limit is shown to exist in the course of the proof

In the proof, we use the previous item.

page 158 in 1st measure

page 55 in the 2nd measure notebook

the limit is shown to exist in the course of the proof

In the proof, we use the previous item.

page 158 in 1st measure

page 55 in the 2nd measure notebook

Let W c P(X) be a ring.

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = oo.

What can we conclude?

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = oo.

What can we conclude?

y(B) = oo

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(A) = oo, so is y(B).

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(A) = oo, so is y(B).

Let W c P(X) be a ring.

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = -oo.

What can we conclude?

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = -oo.

What can we conclude?

y(B) = -oo

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + Y(B\A).

Since y(A) = -oo, so is y(B).

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + Y(B\A).

Since y(A) = -oo, so is y(B).

Let W c P(X) be a ring.

Let y : W -> [0,oo] be additive.

Let A,B :- W.

Suppose that A c B.

What can we conclude?

Let y : W -> [0,oo] be additive.

Let A,B :- W.

Suppose that A c B.

What can we conclude?

y(A) <= y(B)

Since A c B, B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(B\A)>=0, we get y(A) <= y(A) + y(B\A) = y(B).

Hence y(A) <= y(B).

Since A c B, B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(B\A)>=0, we get y(A) <= y(A) + y(B\A) = y(B).

Hence y(A) <= y(B).

Let W be a ring of sets.

Let y : W -> [0,oo] be countably additive.

Can we conclude that y is countably subadditive on W?

Let y : W -> [0,oo] be countably additive.

Can we conclude that y is countably subadditive on W?

Yes.

hint: U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ U(k=1 to n-1)_A[k] ]

page 84 in 1st measure

hint: U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ U(k=1 to n-1)_A[k] ]

page 84 in 1st measure

Let W be a ring of sets.

Let y : W -> [0,oo[ be additive.

Prove that

/\B,A:-W |y(A) - y(B)| <= y(A+B)

Let y : W -> [0,oo[ be additive.

Prove that

/\B,A:-W |y(A) - y(B)| <= y(A+B)

We will use [ E c F => y(E) <= y(F). ]

y(A) <= y(A u B) = y( (A+B) u (AnB) ) = y(A+B) + y(AnB) <= y(A+B) + y(B)

Hence y(A) <= y(A+B) + y(B).

Since the values are real numbers, we get y(A) - y(B) <= y(A+B).

y(A) <= y(A u B) = y( (A+B) u (AnB) ) = y(A+B) + y(AnB) <= y(A+B) + y(B)

Hence y(A) <= y(A+B) + y(B).

Since the values are real numbers, we get y(A) - y(B) <= y(A+B).

Let W be a semi-ring of sets. Let y : W -> [-oo,oo] be finitely
additive.

Let E be written, in two different ways, as a finite disjoint union of sets in W.

What can we conclude?

Let E be written, in two different ways, as a finite disjoint union of sets in W.

What can we conclude?

Let E = U(i=1 to n) C[i] = U(j=1 to m) D[j]. And C[i],D[j] :- W.

We can conclude:

+(i=1 to n) y(C[i]) = +(j=1 to m) y(D[j])

page 86 in 1st measure

Notice here that finite additivity is required.

We can conclude:

+(i=1 to n) y(C[i]) = +(j=1 to m) y(D[j])

page 86 in 1st measure

Notice here that finite additivity is required.

Let J be a semi-ring of sets.

Let y : J -> [-oo,oo] be finitely additive.

Extend function y so that y : R(J) -> [-oo,oo] is additive.

Let y : J -> [-oo,oo] be finitely additive.

Extend function y so that y : R(J) -> [-oo,oo] is additive.

page 87, 88 in 1st measure

Notice that finite additivity is required.

Notice that finite additivity is required.

Let a,b,c belong to [0,oo].

Prove that (a+b)*c = (a*c) + (b*c).

Prove that (a+b)*c = (a*c) + (b*c).

___

Let C and D be sets.

Suppose that

1) (a,b) :- C x D

2) (x,y) :- C x D

What can we conclude about (a,y) and (x,b) ?

Suppose that

1) (a,b) :- C x D

2) (x,y) :- C x D

What can we conclude about (a,y) and (x,b) ?

(a,y) :- C x D

(x,b) :- C x D

(x,b) :- C x D

Let A,B,C,D,E,F be non-empty sets.

Suppose that (AxC) u (BxD) = ExF.

Suppose that A n B = O.

What can we conclude?

Suppose that (AxC) u (BxD) = ExF.

Suppose that A n B = O.

What can we conclude?

C = D

Take any d:-D. Choose some b:-B. Now (b,d):-BxD. Hence (b,d):- ExF.

Choose some (a,c):-AxC. Now (a,c):-ExF. We conclude that (a,d):- ExF.

Hence (a,d):-AxC or (a,d):-BxD. But a:-A and AnB=O, h

Take any d:-D. Choose some b:-B. Now (b,d):-BxD. Hence (b,d):- ExF.

Choose some (a,c):-AxC. Now (a,c):-ExF. We conclude that (a,d):- ExF.

Hence (a,d):-AxC or (a,d):-BxD. But a:-A and AnB=O, h

Let A,B,C,D be non-empty sets.

Prove that if A=B or C=D, then

(A x C) u (B x D) = (A u B) x (C u D).

Prove that if A=B or C=D, then

(A x C) u (B x D) = (A u B) x (C u D).

___

Let J1 be a semi-ring in X, and J2 a semi-ring in Y.

Let u : J1 -> [0,oo] and y : J2 -> [0,oo] be additive.

Let J = { E x F : E:-J1 and F:-J2 }.

Let q : J -> |R*, q(E x F) = u(E) * y(F).

What can we conclude about q?

Let u : J1 -> [0,oo] and y : J2 -> [0,oo] be additive.

Let J = { E x F : E:-J1 and F:-J2 }.

Let q : J -> |R*, q(E x F) = u(E) * y(F).

What can we conclude about q?

q is additive

page 89 in 1st measure

page 89 in 1st measure

Let W be a ring of sets.

Let y : W -> |R* be additive.

Prove that for every A,B :- W

y(A u B) + y(A n B) = ???

Let y : W -> |R* be additive.

Prove that for every A,B :- W

y(A u B) + y(A n B) = ???

y(A u B) + y(A n B) = y(A) + y(B)

1) A u B = A u (B\A) disjointly

2) y(A u B) = y(A) + y(B\A)

3) y(A u B) + y(A n B) = y(A) + y(B\A) + y(B n A)

4) y(A u B) + y(A n B) = y(A) + y(B)

1) A u B = A u (B\A) disjointly

2) y(A u B) = y(A) + y(B\A)

3) y(A u B) + y(A n B) = y(A) + y(B\A) + y(B n A)

4) y(A u B) + y(A n B) = y(A) + y(B)

Let W be a ring of sets.

Let y : W -> |R* be additive.

Prove that for every A,B :- W

y(A) + y(B) = ???

Let y : W -> |R* be additive.

Prove that for every A,B :- W

y(A) + y(B) = ???

y(A) + y(B) = y(A u B) + y(A n B)

PROOF:

0) A u B = A u (B\A) disjointly

1) y(A u B) + y(A n B) = y(A) + y(B\A) + y(B n A)

2) y(A u B) + y(A n B) = y(A) + y(B)

PROOF:

0) A u B = A u (B\A) disjointly

1) y(A u B) + y(A n B) = y(A) + y(B\A) + y(B n A)

2) y(A u B) + y(A n B) = y(A) + y(B)

Let (T,>) be a directed set and let p:-T.

Let P = { t:-T : t > p }.

What can we say about (P,>) ?

Let P = { t:-T : t > p }.

What can we say about (P,>) ?

(P,>) is also a directed set

Let W c P(X). Let A,B c X, x :- X.

Suppose that x is W-adherent to A u B.

What can we conclude?

Suppose that x is W-adherent to A u B.

What can we conclude?

x is W-adherent to A or x is W-adherent to B

page 53 in gen top

page 53 in gen top

Let W c P(X), A,B c X.

Suppose that A,B are W-closed.

What can we conclude?

Suppose that A,B are W-closed.

What can we conclude?

1) A u B is W-closed

2) A n B is W-closed

page 54 in gen top

2) A n B is W-closed

page 54 in gen top

Let W c P(X), A,B c X.

Suppose that A,B are W-open.

What can we conclude?

Suppose that A,B are W-open.

What can we conclude?

A n B is W-open, page 55 in gen top

A u B is W-open, page 56 in gen top

A u B is W-open, page 56 in gen top

Let W c P(X). Let M c P(X) be a collection of W-closed sets.

What can we conclude?

What can we conclude?

//\\M is W-closed

page 55 in gen top

page 55 in gen top

Let W c P(X). Let M c P(X) be a collection of W-open sets.

What can we conclude?

What can we conclude?

\\//M is W-open

page 56 in gen top

page 56 in gen top

Let X c P(X).

Which of these statements are true?

(1) the empty set is W-closed

(2) the empty set is W-open

(3) X is W-closed

(4) X is W-open

Which of these statements are true?

(1) the empty set is W-closed

(2) the empty set is W-open

(3) X is W-closed

(4) X is W-open

all are true

Let W c P(X) and M c P(Y). Let f : X -> Y.

(1) /\a:-X [ if net x in X W-converges to a, then net f(x) M- converges to f(a) ]

(2) /\BcY [ B is M-closed ==> f-1(B) is W-closed ]

Prove that (1)=>(2).

(1) /\a:-X [ if net x in X W-converges to a, then net f(x) M- converges to f(a) ]

(2) /\BcY [ B is M-closed ==> f-1(B) is W-closed ]

Prove that (1)=>(2).

page 57 in gen top

Let W c P(X) and M c P(Y). Let f : X -> Y.

(1) /\BcY [ B is M-open ==> f-1(B) is W-open ]

(2) /\a:-X [ if net x in X W-converges to a, then net f(x) M- converges to f(a) ]

Prove that (1)=>(2).

(1) /\BcY [ B is M-open ==> f-1(B) is W-open ]

(2) /\a:-X [ if net x in X W-converges to a, then net f(x) M- converges to f(a) ]

Prove that (1)=>(2).

page 57 in gen top

Let W c P(X) and M c P(Y). Let f : X -> Y.

(1) /\BcY [ B is M-closed ==> f-1(B) is W-closed ]

(2) /\BcY [ B is M-open ==> f-1(B) is W-open ]

Prove that (1)=>(2).

(1) /\BcY [ B is M-closed ==> f-1(B) is W-closed ]

(2) /\BcY [ B is M-open ==> f-1(B) is W-open ]

Prove that (1)=>(2).

page 57 in gen top

Let W c P(X), A c X, x:-X.

(1) x is W-adherent to A

(2) /\FcX [ F is W-closed and A c F ==> x:-F ]

Prove that (1)=>(2).

(1) x is W-adherent to A

(2) /\FcX [ F is W-closed and A c F ==> x:-F ]

Prove that (1)=>(2).

page 58 in gen top

Let W c P(X), A c X, x:-X.

(1) /\FcX [ F is W-closed and A c F ==> x:-F ]

(2) /\GcX [ G is W-open and x:-G ==> AnG != O ]

Prove that (1)=>(2).

(1) /\FcX [ F is W-closed and A c F ==> x:-F ]

(2) /\GcX [ G is W-open and x:-G ==> AnG != O ]

Prove that (1)=>(2).

page 58 in gen top

Let W c P(X), A c X, x:-X.

(1) /\GcX [ G is W-open and x:-G ==> AnG != O ]

(2) x is W-adherent to A

Prove that (1)=>(2).

(1) /\GcX [ G is W-open and x:-G ==> AnG != O ]

(2) x is W-adherent to A

Prove that (1)=>(2).

page 58 in gen top

Let W c P(X), a:-X.

Suppose /\B,A:-W [ B n A :- W ].

Let K = {A:-W : a:-A} be non-empty.

Let f : K -> X satisfy /\B:-K f(B):-B.

What interesting fact can we conclude from this setup?

Suppose /\B,A:-W [ B n A :- W ].

Let K = {A:-W : a:-A} be non-empty.

Let f : K -> X satisfy /\B:-K f(B):-B.

What interesting fact can we conclude from this setup?

1) (K,c) is a directed set, where c is the inclusion relation

2) the net f W-converges to a

page 60 in gen top

2) the net f W-converges to a

page 60 in gen top

Let X be a set. Let W,G c P(X).

What does it mean that W is a base for G ?

What does it mean that W is a base for G ?

/\E:-G \/BcW E = \\//B

Every set from G can be written as a union of sets from W.

(a union of any cardinality, that is)

page 61 in gen top

Every set from G can be written as a union of sets from W.

(a union of any cardinality, that is)

page 61 in gen top

Let X be a set, and W c P(X).

(1) X :- W

(2) /\B,A:-W A n B :- W

Prove that W is a base for W*.

(1) X :- W

(2) /\B,A:-W A n B :- W

Prove that W is a base for W*.

page 61 in gen top

Let X be a set, and W c P(X).

What does it mean that W is a topology?

What does it mean that W is a topology?

1) O :- W

2) X :- W

3) /\B,A:-W [ B n A :- W ]

4) /\McW [ \\//M :- W ]

2) X :- W

3) /\B,A:-W [ B n A :- W ]

4) /\McW [ \\//M :- W ]

Let X be a set, and W c P(X).

If W is a topology, then W* = ???

If W is a topology, then W* = ???

W* = W

HINT:

From ( /\B,A:-W AnB:-W ) we conclude that W is a base for W*.

That is the subject matter of another item.

page 63 in gen top

HINT:

From ( /\B,A:-W AnB:-W ) we conclude that W is a base for W*.

That is the subject matter of another item.

page 63 in gen top

Let X be a set. Let A,B c P(X).

Prove that A c B ==> A* c B*.

Prove that A c B ==> A* c B*.

page 63 in gen top

Let W c P(X).

Let Wd denote the collection of all finite intersections of sets from W.

Consider the set Wd u {X}.

What interesting feature does it have?

Let Wd denote the collection of all finite intersections of sets from W.

Consider the set Wd u {X}.

What interesting feature does it have?

It is a base for W*.

page 64 in gen top

page 64 in gen top

Let W c P(X).

Is it always true that W* is the smallest topology containing W?

Is it always true that W* is the smallest topology containing W?

Yes.

Recall that W* always is a topology.

Recall that if AcBcP(X), then A*cB*.

Recall that if Y is a topology, then Y*=Y.

Put it all together.

page 64,66 in gen top

Recall that W* always is a topology.

Recall that if AcBcP(X), then A*cB*.

Recall that if Y is a topology, then Y*=Y.

Put it all together.

page 64,66 in gen top

Let d be metric on |R.

Is it possible that (|R,d) is incomplete?

Is it possible that (|R,d) is incomplete?

Yes.

d(x,y) = | arctan(x) - arctan(y) |

lim(n->oo) arctan(n) = PI/2

Hence the sequence {arctan(n)} is Cauchy with respect to the Euclidean metric.

Hence the sequence {n} is Cauchy with respect to d.

d(x,y) = | arctan(x) - arctan(y) |

lim(n->oo) arctan(n) = PI/2

Hence the sequence {arctan(n)} is Cauchy with respect to the Euclidean metric.

Hence the sequence {n} is Cauchy with respect to d.

(1) X c |R is incomplete in |R with respect to the Euclidean
metric.

(2) (X,d) is a complete metric space

(3) (|R,r) is an incomplete metric space

(4) (|R,r) and (X,d) are homeomorphic

Is this setup possible?

(2) (X,d) is a complete metric space

(3) (|R,r) is an incomplete metric space

(4) (|R,r) and (X,d) are homeomorphic

Is this setup possible?

Let X = ] -PI/2 , PI/2 [.

Let d(y,x) = |tan(y) - tan(x)|.

Now (X,d) is a complete metric space.

Let r(x,y) = |arctan(x) - arctan(y)|.

Now (|R,r) is an incomplete metric space.

Let f : |R -> X, f(x) = arctan(x).

Now f is a homeomorphism.

Let d(y,x) = |tan(y) - tan(x)|.

Now (X,d) is a complete metric space.

Let r(x,y) = |arctan(x) - arctan(y)|.

Now (|R,r) is an incomplete metric space.

Let f : |R -> X, f(x) = arctan(x).

Now f is a homeomorphism.

If X is a set, what is an "outer measure" on X?

Y : P(X) -> [0,oo]

(1) Y(O) = 0

(2) A c B c X => Y(A) <= Y(B)

(3) Y is countably subadditive

(1) Y(O) = 0

(2) A c B c X => Y(A) <= Y(B)

(3) Y is countably subadditive

If X is a set, what is a "pseudometric" on X?

d : X x X -> |R

(1) /\a:-X d(a,a) = 0

(2) /\a,b:-X d(a,b) = d(b,a)

(3) /\a,b,c:-X d(a,c) <= d(a,b) + d(b,c)

REMARK: Maybe it's convenient to allow +oo as a possible value d(x,y)=oo.

(1) /\a:-X d(a,a) = 0

(2) /\a,b:-X d(a,b) = d(b,a)

(3) /\a,b,c:-X d(a,c) <= d(a,b) + d(b,c)

REMARK: Maybe it's convenient to allow +oo as a possible value d(x,y)=oo.

Let Y be an outer measure on X.

Define the corresponding pseudometric on P(X).

Prove that it is a pseudometric.

Define the corresponding pseudometric on P(X).

Prove that it is a pseudometric.

If A,B c X, then let d(A,B) = Y(A+B).

To prove the triangle inequality, use

A + B c (A + C) u (C + B).

REMARK:

This d can assume plus infinity, hence it is controversial

To prove the triangle inequality, use

A + B c (A + C) u (C + B).

REMARK:

This d can assume plus infinity, hence it is controversial

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

d(A u B , C u D) <= ???

and let d be the corresponding pseudometric on P(X).

d(A u B , C u D) <= ???

d(A u B , C u D) <= d(A,C) + d(B,D)

page 94 in 1st measure

page 94 in 1st measure

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

d(A n B , C n D) <= ???

and let d be the corresponding pseudometric on P(X).

d(A n B , C n D) <= ???

d(A n B , C n D) <= d(A,C) + d(B,D)

page 94 in 1st measure

page 94 in 1st measure

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

d(A \ B , C \ D) <= ???

and let d be the corresponding pseudometric on P(X).

d(A \ B , C \ D) <= ???

d(A \ B , C \ D) <= d(A,C) + d(B,D)

page 94 in 1st measure

page 94 in 1st measure

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

d(A + B , C + D) <= ???

and let d be the corresponding pseudometric on P(X).

d(A + B , C + D) <= ???

d(A + B , C + D) <= d(A,C) + d(B,D)

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

In this context,

what does it mean that a sequence of sets converges to a set?

and let d be the corresponding pseudometric on P(X).

In this context,

what does it mean that a sequence of sets converges to a set?

/\n:-|N A[n] c X and A c X

A[n] -> A <=> lim d(A[n],A) = 0 <=> lim Y(A[n] + A) = 0

A[n] -> A <=> lim d(A[n],A) = 0 <=> lim Y(A[n] + A) = 0

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] u B[n] -> A u B.

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] u B[n] -> A u B.

use:

d(A[n] u B[n] , A u B) <= d(A[n],A) + d(B[n],B)

page 95 in 1st measure

d(A[n] u B[n] , A u B) <= d(A[n],A) + d(B[n],B)

page 95 in 1st measure

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] n B[n] -> A n B.

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] n B[n] -> A n B.

use:

d(A[n] n B[n] , A n B) <= d(A[n],A) + d(B[n],B)

d(A[n] n B[n] , A n B) <= d(A[n],A) + d(B[n],B)

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] \ B[n] -> A \ B.

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] \ B[n] -> A \ B.

use:

d(A[n] \ B[n] , A \ B) <= d(A[n],A) + d(B[n],B)

d(A[n] \ B[n] , A \ B) <= d(A[n],A) + d(B[n],B)

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] + B[n] -> A + B.

and let d be the corresponding pseudometric on P(X).

Let A[n] -> A and B[n] -> B.

Prove that A[n] + B[n] -> A + B.

use:

d(A[n] + B[n] , A + B) <= d(A[n],A) + d(B[n],B)

d(A[n] + B[n] , A + B) <= d(A[n],A) + d(B[n],B)

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Let B,A c X and Y(A) < oo , Y(B) < oo.

|Y(A) - Y(B)| <= ???

and let d be the corresponding pseudometric on P(X).

Let B,A c X and Y(A) < oo , Y(B) < oo.

|Y(A) - Y(B)| <= ???

|Y(A) - Y(B)| <= d(A,B)

|Y(A) - Y(B)| <= Y(A+B)

way (1): use d(A,O) <= d(A,B) + d(B,O).

way (2): use Y(A) <= Y(AuB) <= Y(A+B) + Y(AnB) <= Y(A+B) + Y(B)

|Y(A) - Y(B)| <= Y(A+B)

way (1): use d(A,O) <= d(A,B) + d(B,O).

way (2): use Y(A) <= Y(AuB) <= Y(A+B) + Y(AnB) <= Y(A+B) + Y(B)

Let Y be an outer measure on X,

and let d be the corresponding pseudometric on P(X).

Suppose that /\n:-|N Y(A[n]) < oo , Y(A) < oo.

Suppose that A[n] -> A, ( lim(n->oo) Y(A[n]+A) = 0 ).

What can we conclude?

and let d be the corresponding pseudometric on P(X).

Suppose that /\n:-|N Y(A[n]) < oo , Y(A) < oo.

Suppose that A[n] -> A, ( lim(n->oo) Y(A[n]+A) = 0 ).

What can we conclude?

lim Y(A[n]) = Y(A)

hint: use |Y(A[n]) - Y(A)| <= d(A[n],A)

page 96 in 1st measure

hint: use |Y(A[n]) - Y(A)| <= d(A[n],A)

page 96 in 1st measure

Let a : |N x |N -> [0,oo].

Consider this line:

+(n=1 to n=oo) +(k=1 to k=oo) a(n,k) = +(k=1 to k=oo) +(n=1 to n=oo) a(n,k)

Prove this line supposing that infinity does not occur in

Consider this line:

+(n=1 to n=oo) +(k=1 to k=oo) a(n,k) = +(k=1 to k=oo) +(n=1 to n=oo) a(n,k)

Prove this line supposing that infinity does not occur in

page 59 in 2nd measure

page 97 in 1st measure

page 97 in 1st measure

Let X[n] be a sequence of sets. Let X be the union of all X[n].

Let S[n] be a s-ring in X[n], for every n:-|N.

Let S = { E c X : /\n:-|N ( E n X[n] :- S[n] ) }.

What can we conclude about S?

Let S[n] be a s-ring in X[n], for every n:-|N.

Let S = { E c X : /\n:-|N ( E n X[n] :- S[n] ) }.

What can we conclude about S?

S is a s-ring

page 99 in 1st measure

page 99 in 1st measure

Let X[n] be a sequence of sets. Let X be the union of all X[n].

Let S[n] be a s-ring in X[n], for every n:-|N.

Let y[n] : S[n] -> [0,oo] be countably additive, for every n:-|N.

Let S = { E c X : /\n:-|N ( E n X[n] :- S[n] ) }.

Let Y : S -> [0,oo] be defined by Y(E) = +(n=1 to n=oo) y[n](E n X[n]).

Let S[n] be a s-ring in X[n], for every n:-|N.

Let y[n] : S[n] -> [0,oo] be countably additive, for every n:-|N.

Let S = { E c X : /\n:-|N ( E n X[n] :- S[n] ) }.

Let Y : S -> [0,oo] be defined by Y(E) = +(n=1 to n=oo) y[n](E n X[n]).

ANSWER: Y is countably additive.

page 99 in 1st measure

By the way, S is a s-ring in X.

But that is the subject matter of the previous item.

page 99 in 1st measure

By the way, S is a s-ring in X.

But that is the subject matter of the previous item.

Consider the set of all bounded intervals on the real line,
including singletons and the empty set. Is it a ring?

NO.

[1,7] \ [2,6] does not belong

[1,7] \ [2,6] does not belong

Consider the set of all bounded intervals on the real line,
including singletons and the empty set. Is it a semi-ring?

Yes.

page 60 in 1st measure

page 60 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

In the context of measure theory,

how do we define the term "interval" in |R^p ?

In the context of measure theory,

how do we define the term "interval" in |R^p ?

An interval in |R is a bounded connected subset of |R,
including the empty set.

If {P[n]}(n=1 to n=p) is a collection of intervals in |R,

then the Cartesian product P[1] x P[2] x ... x P[p] is an interval in |R^p.

If {P[n]}(n=1 to n=p) is a collection of intervals in |R,

then the Cartesian product P[1] x P[2] x ... x P[p] is an interval in |R^p.

Consider the p-dimensional Euclidean space |R^p.

In the context of measure theory,

consider the collection of all intervals in |R^p.

Is it a ring?

In the context of measure theory,

consider the collection of all intervals in |R^p.

Is it a ring?

NO.

[1,7] \ [2,4] does not belong

[1,7] \ [2,4] does not belong

Consider the p-dimensional Euclidean space |R^p.

In the context of measure theory,

consider the collection of all intervals in |R^p.

Is it a semi-ring?

In the context of measure theory,

consider the collection of all intervals in |R^p.

Is it a semi-ring?

Yes.

Recall that the intervals in |R form a semi-ring.

Then recall that if X,Y are semi-rings,

then { A x B : A:-X, B:-Y } is also a semi-ring.

page 101 in 1st measure

Recall that the intervals in |R form a semi-ring.

Then recall that if X,Y are semi-rings,

then { A x B : A:-X, B:-Y } is also a semi-ring.

page 101 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

Define an additive function on W

which will eventually lead to the Lebesgue measure on |R^p.

Let W be the collection of all intervals in |R^p.

Define an additive function on W

which will eventually lead to the Lebesgue measure on |R^p.

For -oo < a <= b < oo,

if <a,b> is an interval in |R, then let m(<a,b>) = b - a.

If U = U[1] x ... x U[p], where U[k] is an interval in |R,

then let m(U) = m(U[1]) * ... * m(U[p]).

to see that this function is additive,

if <a,b> is an interval in |R, then let m(<a,b>) = b - a.

If U = U[1] x ... x U[p], where U[k] is an interval in |R,

then let m(U) = m(U[1]) * ... * m(U[p]).

to see that this function is additive,

Consider the p-dimensional Euclidean space |R^p.

In the context of measure theory,

what are the elementary sets?

In the context of measure theory,

what are the elementary sets?

A subset of |R^p is elementary iff

it is a finite union of intervals in |R^p.

Let W be the collection of all intervals in |R^p.

Let E denote the collection of all elementary sets.

Then E = Ws.

page 102 in first measure

it is a finite union of intervals in |R^p.

Let W be the collection of all intervals in |R^p.

Let E denote the collection of all elementary sets.

Then E = Ws.

page 102 in first measure

Consider the p-dimensional Euclidean space |R^p.

In the context of measure theory,

what are the properties of the collection of elementary sets

that follow immediately from its definition?

In the context of measure theory,

what are the properties of the collection of elementary sets

that follow immediately from its definition?

E is a ring of sets.

If W denotes the collection of all intervals in |R^p,

and E denotes the collection of all elementary sets,

then E = Ws = R(W) = the set of all finite disjoint unions of sets from W

If W denotes the collection of all intervals in |R^p,

and E denotes the collection of all elementary sets,

then E = Ws = R(W) = the set of all finite disjoint unions of sets from W

Let A be a complex number.

What can we say about the limit of the sequence {cos(n*A)} ?

What can we say about the limit of the sequence {cos(n*A)} ?

the sequence cos(n*A) does not tend to zero

page 189 in the palace notebook

page 189 in the palace notebook

Let A be an arbitrary set.

Let T be a proposition.

Let Y(a) be a proposition for every a:-A.

Find an equivalent formula for:

[ \/a:-A Y(a) ] => T

Let T be a proposition.

Let Y(a) be a proposition for every a:-A.

Find an equivalent formula for:

[ \/a:-A Y(a) ] => T

[ \/a:-A Y(a) ] => T <=> /\a:-A [ Y(a) => T ]

Let A be an arbitrary set.

Let T be a proposition.

Let Y(a) be a proposition for every a:-A.

Find an equivalent formula for:

/\a:-A [ Y(a) => T ]

Let T be a proposition.

Let Y(a) be a proposition for every a:-A.

Find an equivalent formula for:

/\a:-A [ Y(a) => T ]

/\a:-A [ Y(a) => T ] <=> [ \/a:-A Y(a) ] => T

Let X be a set. Let M be a subset of X.

Let W = { A c X : M c A }.

Describe R(W).

Let W = { A c X : M c A }.

Describe R(W).

R(W) = { E c X : M c E or M n E = O }

page 56 in the 2nd measure notebook

page 56 in the 2nd measure notebook

Let X be a set. Let M be a subset of X.

Let W = { A c X : M c A }.

Describe S(W) - the s-ring generated by W.

Let W = { A c X : M c A }.

Describe S(W) - the s-ring generated by W.

S(W) = { E c X : M c E or M n E = O }

See page 142 in 2nd measure.

(page 57 in the 2nd measure notebook)

See page 142 in 2nd measure.

(page 57 in the 2nd measure notebook)

Let X be a set and let M be a countable subset of X.

Let W = { E c X : M \ E is finite }.

Describe S(W) - the s-ring generated by W.

Let W = { E c X : M \ E is finite }.

Describe S(W) - the s-ring generated by W.

S(W) = P(X)

page 58 in 2nd measure

page 58 in 2nd measure

(1) +(n=1 to n=oo) a[n] = oo

(2) +(n=1 to n=oo) b[n] = b, b is a real number

What can we conclude?

(2) +(n=1 to n=oo) b[n] = b, b is a real number

What can we conclude?

+(n=1 to n=oo) a[n]+b[n] = oo

page 193 in 1st measure

page 193 in 1st measure

a : |N x |N -> [0,oo]

Consider this line:

+(n=1 to n=oo) +(k=1 to k=oo) a(n,k) = +(k=1 to k=oo) +(n=1 to n=oo) a(n,k)

Suppose that infinity is assumed by one of the values pre

Consider this line:

+(n=1 to n=oo) +(k=1 to k=oo) a(n,k) = +(k=1 to k=oo) +(n=1 to n=oo) a(n,k)

Suppose that infinity is assumed by one of the values pre

page 59 in 2nd measure

Determine if it is true:

/\x,y:-|R\{0} |x/y + y/x| >= 2

/\x,y:-|R\{0} |x/y + y/x| >= 2

YES.

Investigate

lim (x,y)->(0,0) x*y / (x + x^2 + y^2)

lim (x,y)->(0,0) x*y / (x + x^2 + y^2)

doesn't exist

smart hint: -y^2 = x

1) x[n] = 1/n, y[n] = 1/n, lim = 0

2) x[n] = -1/n^2, y[n] = -1/n, lim = oo

page 68 in OLDTIMER

smart hint: -y^2 = x

1) x[n] = 1/n, y[n] = 1/n, lim = 0

2) x[n] = -1/n^2, y[n] = -1/n, lim = oo

page 68 in OLDTIMER

Let s be a complex number other than -1.

s / (1+s) = ???

s / (1+s) = ???

s / (1+s) = 1 - 1/(1+s)

Let s be a complex number other than -1.

1 - 1/(1+s) = ???

1 - 1/(1+s) = ???

1 - 1/(1+s) = s / (1+s)

Let y be a nonnegative function on P(X), such that

(1) if A c B c X, then y(A)<=y(B);

(2) if A is a disjoint countable union of sets A[n] c X,

then y(A) <= y(A[1]) + y(A[2]) + ... + y(A[n]) + ...

What further can we conclude about function y?

(1) if A c B c X, then y(A)<=y(B);

(2) if A is a disjoint countable union of sets A[n] c X,

then y(A) <= y(A[1]) + y(A[2]) + ... + y(A[n]) + ...

What further can we conclude about function y?

It is countably subadditive.

page 43 in 2nd measure

page 43 in 2nd measure

Let M,B,A be arbitrary sets.

(A n M) \ (B n M) = ???

(A n M) \ (B n M) = ???

(A n M) \ (B n M) = (A \ B) n M

Let W c P(X) be a ring in X.

Let y : W -> [0,oo] be additive.

y(A u B) <= ???

Let y : W -> [0,oo] be additive.

y(A u B) <= ???

y(A u B) <= y(A) + y(B)

Recall that y(A u B) + y(A n B) = y(A) + y(B).

Notice that y(A n B) >= 0.

Hence y(A u B) <= y(A) + y(B).

page 104 in 1st measure

Recall that y(A u B) + y(A n B) = y(A) + y(B).

Notice that y(A n B) >= 0.

Hence y(A u B) <= y(A) + y(B).

page 104 in 1st measure

Let A be a bounded connected subset of |R. Let E>0.

Prove that there exists a closed interval F c A such that d(A) <= d(F) + E.

d(A) and d(F) denote the length of those intervals.

Prove that there exists a closed interval F c A such that d(A) <= d(F) + E.

d(A) and d(F) denote the length of those intervals.

page 105 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Let A:-W. Let E>0.

There exists a closed F:-W with FcA and m(A)<=m(F)+E.

How can that be proven?

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Let A:-W. Let E>0.

There exists a closed F:-W with FcA and m(A)<=m(F)+E.

How can that be proven?

The proof is tedious.

page 106 in 1st measure

page 106 in 1st measure

Let A be a bounded connected subset of |R. Let E>0.

Prove that there exists an open interval G c |R

such that A c G and d(G) <= d(A) + E.

Here d(A) and d(G) denote the length of those intervals.

Prove that there exists an open interval G c |R

such that A c G and d(G) <= d(A) + E.

Here d(A) and d(G) denote the length of those intervals.

If A = <a,b>, let G = ]a-E/2 , b+E/2[.

page 108 in 1st measure

page 108 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Let A:-W. Let E>0.

There exists an open G:-W with AcG and m(G)<=m(A)+E.

How can that be proven?

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Let A:-W. Let E>0.

There exists an open G:-W with AcG and m(G)<=m(A)+E.

How can that be proven?

The proof is tedious.

page 109 in 1st measure

page 109 in 1st measure

Let X be a metric space. Let W c P(X).

Let y : W -> |R*.

How did I choose to define that "y is regular" ?

Let y : W -> |R*.

How did I choose to define that "y is regular" ?

/\A:-W /\E>0 \/F:-W \/G:-W

F is closed and G is open and F c A c G

and y(G\A)<E and y(A\F)<E

Originally, the item contained:

and y(G)-E <= y(A) <= y(F)+E

F is closed and G is open and F c A c G

and y(G\A)<E and y(A\F)<E

Originally, the item contained:

and y(G)-E <= y(A) <= y(F)+E

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all elementary subsets of |R^p.

Let m be the "volume" function on W.

Prove that m is regular, that is:

/\A:-W /\E>0 \/F:-W \/G:-W

F is closed and G is open and F c A c G

Let W be the collection of all elementary subsets of |R^p.

Let m be the "volume" function on W.

Prove that m is regular, that is:

/\A:-W /\E>0 \/F:-W \/G:-W

F is closed and G is open and F c A c G

page 112 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all elementary subsets of |R^p.

Let y : W -> [0,oo) be additive and regular.

Extend y into y* so that y* is an outer measure on |R^p.

Prove that it is an outer measure.

Make a wise comment about what's going on.

Let W be the collection of all elementary subsets of |R^p.

Let y : W -> [0,oo) be additive and regular.

Extend y into y* so that y* is an outer measure on |R^p.

Prove that it is an outer measure.

Make a wise comment about what's going on.

The proof that y* is an outer measure can be found

on pages 113,116 in 1st measure.

That proof requires neither additivity nor regularity of y.

The proof is conducted in a broader setting on page 95 in 2nd measure.

The proof that y*=y on W requires both additivity

on pages 113,116 in 1st measure.

That proof requires neither additivity nor regularity of y.

The proof is conducted in a broader setting on page 95 in 2nd measure.

The proof that y*=y on W requires both additivity

Let 0 <= a[n] <= oo, 0 <= b[n] <= oo.

+(n=1 to n=oo) a[n]+b[n] = ???

+(n=1 to n=oo) a[n]+b[n] = ???

+(n=1 to oo) a[n]+b[n] = +(n=1 to oo)_a[n] + ( +(n=1 to oo)
b[n] )

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all elementary subsets of |R^p.

Let y : W -> [0,oo) be additive and regular.

Let y* : P(|R^p) -> [0,oo] be defined as follows:

y*(E) = inf { +(n=1 to n=oo) y(A[n]) : A[n]:-W, A[n] is open, E c U(n:-|N)A[n] }.

Let W be the collection of all elementary subsets of |R^p.

Let y : W -> [0,oo) be additive and regular.

Let y* : P(|R^p) -> [0,oo] be defined as follows:

y*(E) = inf { +(n=1 to n=oo) y(A[n]) : A[n]:-W, A[n] is open, E c U(n:-|N)A[n] }.

/\A:-W y*(A) = y(A)

Recall a more general theorem from which this one follows.

Do not prove the theorem in this item.

page 114 in 1st measure

page 91 in 2nd measure

Recall a more general theorem from which this one follows.

Do not prove the theorem in this item.

page 114 in 1st measure

page 91 in 2nd measure

Let f be a monotonic real-valued function defined on a closed
interval in |R.

Does it have to be Riemann-integrable?

Does it have to be Riemann-integrable?

Yes.

page 65 in OLDTIMER

page 65 in OLDTIMER

Let [a,b] be a bounded interval in |R.

Let f : [a,b] -> |R have both one-sided limits at every point in [a,b].

What can we conclude?

Let f : [a,b] -> |R have both one-sided limits at every point in [a,b].

What can we conclude?

f is bounded,

we exploit the sequential compactness of [a,b]

page 63 in OLDTIMER

we exploit the sequential compactness of [a,b]

page 63 in OLDTIMER

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets in W.

What can we say about the value of y( lim_inf A[n] ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets in W.

What can we say about the value of y( lim_inf A[n] ) ?

y( lim_inf A[n] ) <= lim_inf y( A[n] )

page 159 in 1st measure

page 159 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_inf A[n] ) < lim_inf y( A[n] )

Is this possible?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_inf A[n] ) < lim_inf y( A[n] )

Is this possible?

Let X = {1,2}. Let y be the counting measure on {1,2}.

Let A[2*n] = {1} and A[2*n+1] = {2}.

Now, y( lim_inf A[n] ) = y( O ) = 0.

But, lim_inf y( A[n] ) = 1.

see page 159 in 1st measure

Let A[2*n] = {1} and A[2*n+1] = {2}.

Now, y( lim_inf A[n] ) = y( O ) = 0.

But, lim_inf y( A[n] ) = 1.

see page 159 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

lim_inf y( A[n] ) < y( lim_inf A[n] )

Is this possible?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

lim_inf y( A[n] ) < y( lim_inf A[n] )

Is this possible?

NO.

lim_inf y( A[n] ) >= y( lim_inf A[n] )

page 159 in 1st measure

lim_inf y( A[n] ) >= y( lim_inf A[n] )

page 159 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

What can we say about the value of lim_sup y( A[n] ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

What can we say about the value of lim_sup y( A[n] ) ?

lim_sup y( A[n] ) <= y( lim_sup A[n] )

page 160 in 1st measure

page 160 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_sup A[n] ) < lim_sup y( A[n] )

Is this possible?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_sup A[n] ) < lim_sup y( A[n] )

Is this possible?

YES.

Let y be the counting measure on |N.

Let A[n] = {n}. Then lim_sup A[n] = O. Hence y( lim_sup A[n] ) = 0.

But lim_sup y( A[n] ) = 1.

Let y be the counting measure on |N.

Let A[n] = {n}. Then lim_sup A[n] = O. Hence y( lim_sup A[n] ) = 0.

But lim_sup y( A[n] ) = 1.

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose y( U{A[n]} ) < oo.

y( lim_sup A[n] ) < lim_sup y( A[n] )

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose y( U{A[n]} ) < oo.

y( lim_sup A[n] ) < lim_sup y( A[n] )

NO.

If y( U{A[n]} ) < oo, then

y( lim_sup A[n] ) >= lim_sup y( A[n] ).

page 160 in 1st measure

If y( U{A[n]} ) < oo, then

y( lim_sup A[n] ) >= lim_sup y( A[n] ).

page 160 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

lim_sup y( A[n] ) < y( lim_sup A[n] )

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

lim_sup y( A[n] ) < y( lim_sup A[n] )

Let y be the counting measure on {2,3}.

Let A[2*n] = {2}, and A[2*n+1] = {3}.

Now, lim_sup y( A[n] ) = 1.

And, y( lim_sup A[n] ) = y( {2,3} ) = 2.

see page 160 in 1st measure

Let A[2*n] = {2}, and A[2*n+1] = {3}.

Now, lim_sup y( A[n] ) = 1.

And, y( lim_sup A[n] ) = y( {2,3} ) = 2.

see page 160 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

Let A[n] converge to A.

What can we say about the sequence y( A[n] ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

Let A[n] converge to A.

What can we say about the sequence y( A[n] ) ?

lim(n->oo) y( A[n] ) = y(A)

page 161 in 1st measure

page 161 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

What can we say about the value of y( lim_sup A[n] ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

What can we say about the value of y( lim_sup A[n] ) ?

y( lim_sup A[n] ) >= lim_sup y( A[n] )

page 160 in 1st measure

page 160 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

What can we say about the value of lim_inf y( A[n] ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

What can we say about the value of lim_inf y( A[n] ) ?

lim_inf y( A[n] ) >= y( lim_inf A[n] )

page 159 in 1st measure

page 159 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

Let A[n] converge to A.

What can we say about the sequence y( A[n] + A ) ?

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

Suppose that y( U{A[n]} ) < oo.

Let A[n] converge to A.

What can we say about the sequence y( A[n] + A ) ?

lim(n->oo) y( A[n] + A ) = 0

page 63 in 2nd measure

page 63 in 2nd measure

Let W c P(X), A c X.

Let A^ be the set of all points which are W-adherent to A.

Let Clo(A) = //\\ {FcX: F is W-closed and A c F}.

What can we say about these two sets?

Let A^ be the set of all points which are W-adherent to A.

Let Clo(A) = //\\ {FcX: F is W-closed and A c F}.

What can we say about these two sets?

A^ = Clo(A)

page 65 in gen top

page 65 in gen top

Let W c P(X), A c X, x :- X.

What does it mean that x is sequentially W-adherent to A ?

What does it mean that x is sequentially W-adherent to A ?

There exists a sequence in A, which W-converges to x.

Let W c P(X), A c X.

What does it mean that A is sequentially W-closed ?

What does it mean that A is sequentially W-closed ?

(1) Every point, that is sequentially W-adherent to A, belongs
to A.

(2) If a sequence contained in A W-converges to x, then x belongs to A.

These two are equivalent.

page 66 in gen top

(2) If a sequence contained in A W-converges to x, then x belongs to A.

These two are equivalent.

page 66 in gen top

Let X be a set, and W c P(X), and A c X.

Let As denote the set of all points which are sequentially W- adherent to A.

Does As have to be sequentially W-closed?

Let As denote the set of all points which are sequentially W- adherent to A.

Does As have to be sequentially W-closed?

NO.

page 67 in gen top

page 67 in gen top

Let W be a ring. Let y : W -> |R* be additive. Let A,B :- W.

Suppose that B c A and y(A):-|R.

Is it possible that y(B) = oo or that y(B) = -oo ?

Suppose that B c A and y(A):-|R.

Is it possible that y(B) = oo or that y(B) = -oo ?

NO.

1) Since B c A, we have A = B u (A\B) disjointly.

2) Since W is a ring and y is additive on W, we have y(A) = y(B) + y(A\B).

3) Since y(A):-|R, both y(B) and y(A\B) must be real.

4) Hence y(B) :- |R.

page 75 in 1st measure

1) Since B c A, we have A = B u (A\B) disjointly.

2) Since W is a ring and y is additive on W, we have y(A) = y(B) + y(A\B).

3) Since y(A):-|R, both y(B) and y(A\B) must be real.

4) Hence y(B) :- |R.

page 75 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Define the collection MF(y) of all finitely y-measurable subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Define the collection MF(y) of all finitely y-measurable subsets of |R^p.

MF(y) = { A c |R^p : \/{A[n]}cE lim(n->oo) y(A[n]+A) = 0 }

page 117 in 1st measure

page 117 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Prove that it is a ring.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Prove that it is a ring.

You can use the theorem from the next item.

page 118 in 1st measure

page 64 in 2nd measure

page 118 in 1st measure

page 64 in 2nd measure

Let X be a set, and let Y : P(X) -> [0,oo] satisfy:

(1) Y(O) = 0.

(2) A c B c X ==> Y(A) <= Y(B).

(3) Y(A u B) <= Y(A) + Y(B).

Let M be a ring in X.

Let F = { A c X : \/{A[n]}cM lim(n->oo) Y(A[n] + A) = 0 }.

Can we conclude that F is a ring?

(1) Y(O) = 0.

(2) A c B c X ==> Y(A) <= Y(B).

(3) Y(A u B) <= Y(A) + Y(B).

Let M be a ring in X.

Let F = { A c X : \/{A[n]}cM lim(n->oo) Y(A[n] + A) = 0 }.

Can we conclude that F is a ring?

F is a ring

page 64 in 2nd measure

page 118 in 1st measure

page 64 in 2nd measure

page 118 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

/\B,A:-MF(y) y*(A u B) + y*(A n B) = ???

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

/\B,A:-MF(y) y*(A u B) + y*(A n B) = ???

/\B,A:-MF(y) y*(A u B) + y*(A n B) = y*(A) + y*(B)

page 118 in 1st measure

page 118 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Define the collection M(y) of all y-measurable subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Define the collection M(y) of all y-measurable subsets of |R^p.

M(y) = the collection of countable unions of sets from MF(y)

page 117 in 1st measure

page 117 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Prove that y* is additive on MF(y).

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Prove that y* is additive on MF(y).

page 119 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Recall that MF(y) is a ring.

Use the fact that every countable union of sets in a ring can be written as a disjoint countable union of sets in that ring.

page 120 in 1st measure

Use the fact that every countable union of sets in a ring can be written as a disjoint countable union of sets in that ring.

page 120 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

y*(A) = +(n=1 to n=oo) y*(A[n])

Recall that MF(y) is a ring and y* is additive on MF(y)

and use this in the proof.

page 121 in 1st measure

Recall that MF(y) is a ring and y* is additive on MF(y)

and use this in the proof.

page 121 in 1st measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_inf A[n] ) < lim_inf y( A[n] ) < lim_sup y( A[n] ) < y( lim_sup A[n] )

Let y : W -> [0,oo] be countably additive and y(O)=0.

Let A[n] be a sequence of sets from W.

y( lim_inf A[n] ) < lim_inf y( A[n] ) < lim_sup y( A[n] ) < y( lim_sup A[n] )

Let X = {1,2,3}.

Let y be the counting measure on X.

Let A[2n] = {1}, A[2n+1] = {2,3}.

We have 0 < 1 < 2 < 3.

see pages 159,160 in 1st measure

Let y be the counting measure on X.

Let A[2n] = {1}, A[2n+1] = {2,3}.

We have 0 < 1 < 2 < 3.

see pages 159,160 in 1st measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

y*(A) < oo

PROOF:

Since A:-MF(y), there is B:-E such that y*(A+B)<7, and y*(B)<oo.

Notice that A c (A+B) u B. Hence y*(A) <= y*(A+B) + y*(B) < oo.

page 122 in 1st measure

PROOF:

Since A:-MF(y), there is B:-E such that y*(A+B)<7, and y*(B)<oo.

Notice that A c (A+B) u B. Hence y*(A) <= y*(A+B) + y*(B) < oo.

page 122 in 1st measure

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

A :- MF(y)

For the proof, use the fact below.

If A is a countable disjoint union of sets in MF(y), say A = U(n:-|N)_A[n],

then y*(A) = +(n:-|N)_y*(A[n]).

For the proof, use the fact below.

If A is a countable disjoint union of sets in MF(y), say A = U(n:-|N)_A[n],

then y*(A) = +(n:-|N)_y*(A[n]).

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

YES.

In order to prove this, we use:

(1) MF(y) is a ring

(2) A:-M(y) ==> [ y*(A)<oo <=> A:-MF(y) ]

page 124 in 1st measure

In order to prove this, we use:

(1) MF(y) is a ring

(2) A:-M(y) ==> [ y*(A)<oo <=> A:-MF(y) ]

page 124 in 1st measure

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Use the fact that y* is countably additive on MF(y).

page 125 in 1st measure

Take any A[n],A :- M(y) such that \\*//(n:-|N)_A[n] = A.

If \/k:-|N y(A[k])=oo, then y(A) = oo = +(n=1 to oo) y(A[n]).

If /\k:-|N y(A[k])<oo, then /\k:-|N A[k] :- MF(y)

and use Theorem 86 on page 121 in 1st measure.

page 125 in 1st measure

Take any A[n],A :- M(y) such that \\*//(n:-|N)_A[n] = A.

If \/k:-|N y(A[k])=oo, then y(A) = oo = +(n=1 to oo) y(A[n]).

If /\k:-|N y(A[k])<oo, then /\k:-|N A[k] :- MF(y)

and use Theorem 86 on page 121 in 1st measure.

( U(t:-T) G[t] ) \ ( U(t:-T) E[t] ) c ???

( U(t:-T) G[t] ) \ ( U(t:-T) E[t] ) c U(t:-T) G[t]\E[t]

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

page 127 in 1st measure

page 79 in 2nd measure

page 79 in 2nd measure

Let A,B,M be arbitrary sets.

(A n B) u (A n M) u (B \ M) = ???

(A n B) u (A n M) u (B \ M) = ???

(A n B) u (A n M) u (B \ M) = (A n M) u (B \ M)

Let A[n] c X for all natural n.

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

/\x:-X lim_inf 1(A[n]) = ???

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

/\x:-X lim_inf 1(A[n]) = ???

lim_inf 1(A[n]) = 1(lim_inf A[n])

page 11 in 1st measure

page 11 in 1st measure

Let G c |R satisfy:

(1) /\a:-G /\n:-|N n*a :- G

(2) /\a,b:-G a-b :- G

(3) G has a cluster point

What can we conclude about G ?

(1) /\a:-G /\n:-|N n*a :- G

(2) /\a,b:-G a-b :- G

(3) G has a cluster point

What can we conclude about G ?

G is dense in |R

page 70 in OLDTIMER

page 70 in OLDTIMER

Let A be an irrational number.

Consider the set {n*A - [n*A] : n:-|N}.

What can we conclude about this set?

Consider the set {n*A - [n*A] : n:-|N}.

What can we conclude about this set?

It is infinite, hence: it has a cluster point in [0,1].

page 72 in OLDTIMER

page 72 in OLDTIMER

Let A be an irrational number.

Consider the set {n*A + m : n,m are integers}.

What can we conclude about this set?

Consider the set {n*A + m : n,m are integers}.

What can we conclude about this set?

It is dense in |R.

page 72 in OLDTIMER

page 72 in OLDTIMER

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Yes.

|R^p = U(n:-|N) [-n,n] x ... x [-n,n]

the Cartesian product of [-n,n] p times

see also page 129 in 1st measure for a different proof

|R^p = U(n:-|N) [-n,n] x ... x [-n,n]

the Cartesian product of [-n,n] p times

see also page 129 in 1st measure for a different proof

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Use:

/\E:-M(y) /\e>0 \/G:-M(y) [ G is open and E c G and y*(G\E) < e ]

page 130 in 1st measure

/\E:-M(y) /\e>0 \/G:-M(y) [ G is open and E c G and y*(G\E) < e ]

page 130 in 1st measure

Prove that every open subset of |R^p is a countable union of
bounded open intervals.

page 131 in 1st measure

Prove that every open subset of |R^p is a countable union of
open balls.

a similar proof is on page 131 in 1st measure

Prove that every open subset of |R^p is a countable union of
closed balls.

a similar proof is on page 131 in 1st measure

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

YES.

Every open subset of |R^p is a countable union of bounded open intervals.

page 131 in 1st measure

Every open subset of |R^p is a countable union of bounded open intervals.

page 131 in 1st measure

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

YES.

Every open subset of |R^p is a countable union of bounded open intervals.

Hence M(y) contains all open sets.

And since M(y) is an algebra, it also contains all closed sets.

page 131 in 1st measure

Every open subset of |R^p is a countable union of bounded open intervals.

Hence M(y) contains all open sets.

And since M(y) is an algebra, it also contains all closed sets.

page 131 in 1st measure

What are Borel sets?

Let (X,G) be a topological space.

The s-algebra generated by G is called the s-algebra of Borel sets.

The s-algebra generated by G is called the s-algebra of Borel sets.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Yes.

Let G denote the collection of all open subsets of |R^p.

Then G c M(y).

M(y) is a s-algebra, hence s(G) c M(y).

s(G) is the s-algebra generated by G

s(G) is the collection of all Borel sets

Let G denote the collection of all open subsets of |R^p.

Then G c M(y).

M(y) is a s-algebra, hence s(G) c M(y).

s(G) is the s-algebra generated by G

s(G) is the collection of all Borel sets

Let F be a s-ring in X.

Let f : X -> |R*.

(1) /\a:-|R {x:-X : f(x)>a} :- F

(2) /\a:-|R {x:-X : f(x)>=a} :- F

Prove that (1) => (2).

Let f : X -> |R*.

(1) /\a:-|R {x:-X : f(x)>a} :- F

(2) /\a:-|R {x:-X : f(x)>=a} :- F

Prove that (1) => (2).

page 132 in 1st measure

Let F be a s-algebra in X.

Let f : X -> |R*.

(2) /\a:-|R {x:-X : f(x)>=a} :- F

(3) /\a:-|R {x:-X : f(x)<a} :- F

Prove that (2) => (3).

Let f : X -> |R*.

(2) /\a:-|R {x:-X : f(x)>=a} :- F

(3) /\a:-|R {x:-X : f(x)<a} :- F

Prove that (2) => (3).

page 132 in 1st measure

Let F be a s-ring in X.

Let f : X -> |R*.

(3) /\a:-|R {x:-X : f(x)<a} :- F

(4) /\a:-|R {x:-X : f(x)<=a} :- F

Prove that (3)=>(4).

Let f : X -> |R*.

(3) /\a:-|R {x:-X : f(x)<a} :- F

(4) /\a:-|R {x:-X : f(x)<=a} :- F

Prove that (3)=>(4).

page 132 in 1st measure

Let F be a s-algebra in X.

Let f : X -> |R*.

(4) /\a:-|R {x:-X : f(x)<=a} :- F

(1) /\a:-|R {x:-X : f(x)>a} :- F

Prove that (4)=>(1).

Let f : X -> |R*.

(4) /\a:-|R {x:-X : f(x)<=a} :- F

(1) /\a:-|R {x:-X : f(x)>a} :- F

Prove that (4)=>(1).

page 132 in 1st measure

Let F be a s-algebra in X.

Let f : X -> |R*.

What does it mean that "f is measurable" ?

Let f : X -> |R*.

What does it mean that "f is measurable" ?

(1) /\a:-|R {x:-X : f(x)>a} :- F

(2) /\a:-|R {x:-X : f(x)>=a} :- F

(3) /\a:-|R {x:-X : f(x)<a} :- F

(4) /\a:-|R {x:-X : f(x)<=a} :- F

these four are equivalent

page 132,133 in 1st measure

(2) /\a:-|R {x:-X : f(x)>=a} :- F

(3) /\a:-|R {x:-X : f(x)<a} :- F

(4) /\a:-|R {x:-X : f(x)<=a} :- F

these four are equivalent

page 132,133 in 1st measure

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

What can we conclude about function |f| ?

Let f : X -> |R* be measurable.

What can we conclude about function |f| ?

|f| is measurable

page 133 in 1st measure

page 133 in 1st measure

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

Prove:

/\a:-|R {x:-X : f(x)=a} :- F

Let f : X -> |R* be measurable.

Prove:

/\a:-|R {x:-X : f(x)=a} :- F

page 134 in 1st measure

Let F be a s-algebra in X.

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = sup(n:-|N) f[n](x).

What can we conclude about function g?

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = sup(n:-|N) f[n](x).

What can we conclude about function g?

g is measurable

page 134 in 1st measure

page 134 in 1st measure

Let F be a s-algebra in X.

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = inf(n:-|N) f[n](x).

What can we conclude about function g?

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = inf(n:-|N) f[n](x).

What can we conclude about function g?

g is measurable

page 134 in 1st measure

page 134 in 1st measure

Let F be a s-algebra in X.

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = lim_inf(n->oo) f[n](x).

What can we conclude about function g?

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = lim_inf(n->oo) f[n](x).

What can we conclude about function g?

g is measurable

page 135 in 1st measure

page 135 in 1st measure

Let F be a s-algebra in X.

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = lim_sup(n->oo) f[n](x).

What can we conclude about function g?

Let f[n] be a sequence of measurable functions X -> |R*.

Let g : X->|R* be defined by g(x) = lim_sup(n->oo) f[n](x).

What can we conclude about function g?

g is measurable

page 135 in 1st measure

page 135 in 1st measure

Decide if it's true:

/\a,b:-|R \/x,y:-|R [ x+y = a and x-y = b ]

/\a,b:-|R \/x,y:-|R [ x+y = a and x-y = b ]

true

x = (a+b)/2

y = (a-b)/2

x = (a+b)/2

y = (a-b)/2

f : |C -> |C

(1) f is diffable

(2) \/A:-|C /\z:-|C f'(z) = A*f(z)

(3) f(0) = 1

What can we conclude?

(1) f is diffable

(2) \/A:-|C /\z:-|C f'(z) = A*f(z)

(3) f(0) = 1

What can we conclude?

(4) /\a,b:-|C f(a+b) = f(a)*f(b)

hint: fix w and define g(z)=f(z)*f(w-z)

page 78 in OLDTIMER

hint: fix w and define g(z)=f(z)*f(w-z)

page 78 in OLDTIMER

Let A,B,C,D,E,F be non-empty sets.

Decide if it's true:

(A x C) u (B x D) = E x F ==> A=B or C=D

Decide if it's true:

(A x C) u (B x D) = E x F ==> A=B or C=D

NO.

A={1}; B={1,2}; C={7}; D={7,8}; E={1,2}; F={7,8}

(A x C) u (B x D) = E x F and A!=B and C!=D

A={1}; B={1,2}; C={7}; D={7,8}; E={1,2}; F={7,8}

(A x C) u (B x D) = E x F and A!=B and C!=D

Let x > 0.

(x+2) / (x+1)^2 < ???

(x+2) / (x+1)^2 < ???

(x+2) / (x+1)^2 < 1/x

Let F be a s-algebra in X.

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : \/A:-|R lim(n->oo) f[n](x) = A}

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : \/A:-|R lim(n->oo) f[n](x) = A}

it belongs to F

page 83 in 2nd measure

page 83 in 2nd measure

Suppose that 0 < x < y.

x/y ??? (x+1)/(y+1)

x/y ??? (x+1)/(y+1)

x/y < (x+1)/(y+1)

Let X be a set. Let d : XxX -> [0,oo].

(1) /\x,y,z:-X d(x,z) <= d(x,y) + d(z,y)

Let u:-X, 0 < r <= oo.

Let U = {x:-X : d(u,x) < r}.

What can we conclude about set U?

(1) /\x,y,z:-X d(x,z) <= d(x,y) + d(z,y)

Let u:-X, 0 < r <= oo.

Let U = {x:-X : d(u,x) < r}.

What can we conclude about set U?

/\a:-U \/e>0 /\x:-X d(x,a)<e => x:-U

page 69 in gen top

page 69 in gen top

Let K be a set and /\k:-K d[k] : XxX -> [0,oo].

Let W = { U c X : \/k:-K /\a:-U \/e>0 /\x:-X d[k](x,a)<e => x:-U }.

Let x:T->X be a net in X, and let u:-X.

(1) /\k:-K /\e>0 \/p:-T /\t>p d[k](x[t] , u) < e

What can we conclude?

Let W = { U c X : \/k:-K /\a:-U \/e>0 /\x:-X d[k](x,a)<e => x:-U }.

Let x:T->X be a net in X, and let u:-X.

(1) /\k:-K /\e>0 \/p:-T /\t>p d[k](x[t] , u) < e

What can we conclude?

net x W-converges to u

page 71 in gen top

page 71 in gen top

Let K be a set and /\k:-K d[k] : XxX -> [0,oo].

(1) /\k:-K /\x:-X d[k](x,x)=0

(2) /\k:-K /\x,y:-X d[k](x,y) = d[k](y,x)

(3) /\k:-K /\x,y,z:-X d(x,z) <= d(x,y) + d(x,y)

Let W = { U c X : \/k:-K /\a:-U \/e>0 /\x:-X d[k](x,a)<e => x:-U }.

Let x:T->X be a net in X, which converges to u:-X.

(1) /\k:-K /\x:-X d[k](x,x)=0

(2) /\k:-K /\x,y:-X d[k](x,y) = d[k](y,x)

(3) /\k:-K /\x,y,z:-X d(x,z) <= d(x,y) + d(x,y)

Let W = { U c X : \/k:-K /\a:-U \/e>0 /\x:-X d[k](x,a)<e => x:-U }.

Let x:T->X be a net in X, which converges to u:-X.

page 71 in gen top

Let W be a s-ring in X.

Let y : W -> [0,oo] countably additive, y(O)=0.

In this context, what are the y-zero sets?

The symbol for the collection of all such sets,

let it be N(y).

Let y : W -> [0,oo] countably additive, y(O)=0.

In this context, what are the y-zero sets?

The symbol for the collection of all such sets,

let it be N(y).

N(y) = { EcX : \/A:-W EcA and y(A)=0 }

Recall that it is a s-ideal.

That is the subject matter of another item.

page 72 in 2nd measure

Recall that it is a s-ideal.

That is the subject matter of another item.

page 72 in 2nd measure

Imagine that a rectangle is written as a finite union of
rectangles, which can overlap only along the edges. Decide if
it must be true: The union of some two of these rectangles
forms a rectangle.

FALSE

ADD

AOC

BBC

ADD

AOC

BBC

Let f be a real-valued function defined on a connected subset
of |R.

Suppose that it is differentiable and that its derivative is bounded.

What can we conclude about this function?

Suppose that it is differentiable and that its derivative is bounded.

What can we conclude about this function?

It satisfies a Lipschitz condition.

Use Lagrange's mean value theorem to prove it.

page 156 in golden gate

Use Lagrange's mean value theorem to prove it.

page 156 in golden gate

Let f be a real-valued function defined on an open subset of |R.

Suppose that it is differentiable and Lipschitz.

What can we conclude about its derivative?

Suppose that it is differentiable and Lipschitz.

What can we conclude about its derivative?

Its derivative is bounded.

Notice that f may be defined on a disconnected subset of |R.

page 156 in golden gate

Notice that f may be defined on a disconnected subset of |R.

page 156 in golden gate

Let f be a real-valued function defined on an open subset of |R.

Suppose that it is differentiable and that its derivative is bounded.

Does this function have to be Lipschitz?

Suppose that it is differentiable and that its derivative is bounded.

Does this function have to be Lipschitz?

If it's defined on a connected subset, then yes.

But generally, no.

See page 80 in OLDTIMER for an example.

But generally, no.

See page 80 in OLDTIMER for an example.

Let R be a ring of sets in X.

Let W = { EcX : E:-R or X\E:-R }.

What can we conclude about W?

Let W = { EcX : E:-R or X\E:-R }.

What can we conclude about W?

W is an algebra of sets in X

and W = K(R)

page 67 in 2nd measure

and W = K(R)

page 67 in 2nd measure

Let F be a s-ring in X. Let f : X -> |R*. Let M be dense in |R.

(1) /\a:-M {x:-X : f(x) > a} :- F

What can we conclude?

(1) /\a:-M {x:-X : f(x) > a} :- F

What can we conclude?

(2) /\a:-|R {x:-X : f(x) > a} :- F

page 65 in 2nd measure

page 65 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Prove that y is additive on M.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Prove that y is additive on M.

Let A:-M and AnB=O.

Then y(AuB) = y( (AuB)nA ) + y( (AuB)\A ).

Notice that y( (AuB)nA ) = y(A).

Since AnB=O, we get y( (AuB)\A ) = y(B).

Hence y(AuB) = y(A) + y(B). This is enough to prove additivity.

page 41 in 2nd measure

Then y(AuB) = y( (AuB)nA ) + y( (AuB)\A ).

Notice that y( (AuB)nA ) = y(A).

Since AnB=O, we get y( (AuB)\A ) = y(B).

Hence y(AuB) = y(A) + y(B). This is enough to prove additivity.

page 41 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E) + y(A n F)

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E) + y(A n F)

/\AcX y( A n (E u F) ) = y(A n E) + y(A n F)

Take any AcX.

Since E:-M, we have y(An(EuF)) = y(An(EuF) n E) + y(An(EuF) \ E).

Since EnF=O, we get y(An(EuF)) = y(A n E) + y(A n F).

page 42 in 2nd measure

Take any AcX.

Since E:-M, we have y(An(EuF)) = y(An(EuF) n E) + y(An(EuF) \ E).

Since EnF=O, we get y(An(EuF)) = y(A n E) + y(A n F).

page 42 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E n F) + y(A n E n F') + y(A n E' n F)

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E n F) + y(A n E n F') + y(A n E' n F)

/\AcX y(A n (E u F) ) = y(A n E n F) + y(A n E n F') + y(A n E'
n F')

page 40 in 2nd measure

the line following two stars: (**)

page 40 in 2nd measure

the line following two stars: (**)

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[1],E[2],...,E[n] } be a finite disjoint collection of sets in M.

Prove that:

/\AcX y( ??? ) = +(k=1 to k=n) y(A n E[k])

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[1],E[2],...,E[n] } be a finite disjoint collection of sets in M.

Prove that:

/\AcX y( ??? ) = +(k=1 to k=n) y(A n E[k])

/\AcX y( A n E ) = +(k=1 to k=n) y(A n E[k])

where E = E[1] u E[2] u ... u E[n]

page 43 in 2nd measure

where E = E[1] u E[2] u ... u E[n]

page 43 in 2nd measure

Let y : P(X) -> |R* such that

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

page 44 in 2nd measure

Let y : P(X) -> |R* such that

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[n] }(n:-|N) be a disjoint collection of sets in M.

Let E be the union of this collection.

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[n] }(n:-|N) be a disjoint collection of sets in M.

Let E be the union of this collection.

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n])
+ ...

Conclude from this that y is countably additive on M.

page 94, 47 in 2nd measure

Conclude from this that y is countably additive on M.

page 94, 47 in 2nd measure

Let y:P(X)->|R* be such that

(2) if A c B c X, then y(A)<=y(B)

(3) y is subadditive

Let E be a subset of X such that y(E) = 0.

What can we conclude?

(2) if A c B c X, then y(A)<=y(B)

(3) y is subadditive

Let E be a subset of X such that y(E) = 0.

What can we conclude?

/\AcX y(A) = y(AnE) + y(A\E)

PROOF

A = (A n E) u (A \ E)

y(A) <= y(A n E) + y(A \ E) <= y(E) + y(A) = y(A).

We showed y(A) = y(A n E) + y(A \ E).

page 49 in 2nd measure

PROOF

A = (A n E) u (A \ E)

y(A) <= y(A n E) + y(A \ E) <= y(E) + y(A) = y(A).

We showed y(A) = y(A n E) + y(A \ E).

page 49 in 2nd measure

Let y be an outer measure on X.

Define the collection of all measurable subsets of X.

State without proof four important theorems about this collection.

Define the collection of all measurable subsets of X.

State without proof four important theorems about this collection.

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

By definition, E is measurable <=> E:-M.

(1) M is a s-algebra

(2) If { E[n] : n:-|N } is a disjoint collection of sets in M,

and if E is the union of this collection, then

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n]) + ...

By definition, E is measurable <=> E:-M.

(1) M is a s-algebra

(2) If { E[n] : n:-|N } is a disjoint collection of sets in M,

and if E is the union of this collection, then

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n]) + ...

+(n=1 to n=oo) (-1)^(n+1) / n = ???

+(n=1 to n=oo) (-1)^(n+1) / n = log(2)

page 83 in OLDTIMER

page 83 in OLDTIMER

What is an "ideal"?

Let W be a collection of sets.

W is an ideal

iff

(0) W is non-empty

(1) B,A:-W => AuB :- W

(2) B:-W and A c B => A:-W

W is an ideal

iff

(0) W is non-empty

(1) B,A:-W => AuB :- W

(2) B:-W and A c B => A:-W

What is a "s-ideal"?

Let W be a collection of sets.

W is a s-ideal

iff

(0) W is non-empty

(1) [ /\n:-|N A[n]:-W ] => U(n:-|N)_A[n] :- W

(2) B:-W and A c B => A:-W

W is a s-ideal

iff

(0) W is non-empty

(1) [ /\n:-|N A[n]:-W ] => U(n:-|N)_A[n] :- W

(2) B:-W and A c B => A:-W

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably addive, y(O)=0.

Let N(y) = {BcX : \/A:-W BcA and y(A)=0}.

What can we conclude about N(y)?

Let y : W -> [0,oo] be countably addive, y(O)=0.

Let N(y) = {BcX : \/A:-W BcA and y(A)=0}.

What can we conclude about N(y)?

N(y) is a s-ideal

page 68 in 2nd measure

page 68 in 2nd measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably addive, y(O)=0.

Let W^ = {EcX : \/A,B:-W AcEcB and y(B\A)=0}.

What can we conclude about W^ ?

Let y : W -> [0,oo] be countably addive, y(O)=0.

Let W^ = {EcX : \/A,B:-W AcEcB and y(B\A)=0}.

What can we conclude about W^ ?

W^ is a s-ring

page 69 in 2nd measure

page 69 in 2nd measure

Let W be a ring of sets. Let y:W->|R* be additive.

Prove that if A,B:-W and AcB and y(B\A):-|R then

[ y(B):-|R <=> ??? ]

Prove that if A,B:-W and AcB and y(B\A):-|R then

[ y(B):-|R <=> ??? ]

[ y(B):-|R <=> y(A):-|R ]

page 70 in 2nd measure

page 70 in 2nd measure

Let W be a ring of sets. Let y:W->|R* be additive.

Prove that if A,B:-W and AcB and y(B\A):-|R then

y(A):-|R <=> ???

Prove that if A,B:-W and AcB and y(B\A):-|R then

y(A):-|R <=> ???

y(A):-|R <=> y(B):-|R

page 70 in 2nd measure

page 70 in 2nd measure

Let x,y be real numbers.

Suppose that |x+y| <= |x+iy|.

What can we conclude?

Suppose that |x+y| <= |x+iy|.

What can we conclude?

x*y <= 0

Let x,y be real numbers.

Suppose that |x+iy| <= |x+y|.

What can we conclude?

Suppose that |x+iy| <= |x+y|.

What can we conclude?

x*y >= 0

Let W be a ring in X. Let y : W -> [0,oo] be additive.

Let A1,A2,B1,B2 be sets in W, and let E c X.

Suppose:

A1 c E c B1 and y(B1 \ A1) = 0,

A2 c E c B2 and y(B2 \ A1) = 0.

What can we conclude?

Let A1,A2,B1,B2 be sets in W, and let E c X.

Suppose:

A1 c E c B1 and y(B1 \ A1) = 0,

A2 c E c B2 and y(B2 \ A1) = 0.

What can we conclude?

y(A1) = y(B1) = y(A2) = y(B2)

page 70 in 2nd measure

page 70 in 2nd measure

Let W be a s-ring in X.

Let y : W -> [0,oo] countably additive, y(O)=0.

Let W^ = { EcX : \/A,B:-W AcEcB and y(B\A)=0 }.

Define a countably additive function on W^, that is an extension of y.

Prove that it is countably additive.

(In this SM-collection, this function will be denoted y^.)

Let y : W -> [0,oo] countably additive, y(O)=0.

Let W^ = { EcX : \/A,B:-W AcEcB and y(B\A)=0 }.

Define a countably additive function on W^, that is an extension of y.

Prove that it is countably additive.

(In this SM-collection, this function will be denoted y^.)

y^(E) = y(A) = y(B),

where A,B:-W and AcEcB and y(B\A)=0

page 71 in 2nd measure

where A,B:-W and AcEcB and y(B\A)=0

page 71 in 2nd measure

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive, y(O)=0.

Define a s-ring in X that contains W.

Don't prove that it is a s-ring, in this item.

(In this SM collection, it will be denoted W^.)

Let y : W -> [0,oo] be countably additive, y(O)=0.

Define a s-ring in X that contains W.

Don't prove that it is a s-ring, in this item.

(In this SM collection, it will be denoted W^.)

Let W^ = { EcX : \/A,B:-W AcEcB and y(B\A)=0 }.

page 72 in 2nd measure

page 72 in 2nd measure

Let W be a s-ring in X.

Let y : W -> [0,oo] countably additive, y(O)=0.

Prove that N(y^) c W^.

Let y : W -> [0,oo] countably additive, y(O)=0.

Prove that N(y^) c W^.

page 73 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

What can we conclude about the sets:

{x:-X : f(x) = oo}, {x:-X : f(x) = -oo} ?

Let f : X -> |R* be measurable.

What can we conclude about the sets:

{x:-X : f(x) = oo}, {x:-X : f(x) = -oo} ?

They belong to F.

page 73 in 2nd measure

page 73 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

What can we conclude about the set

{x:-X : -oo < f(x) < oo } ?

Let f : X -> |R* be measurable.

What can we conclude about the set

{x:-X : -oo < f(x) < oo } ?

It belongs to F.

page 73 in 2nd measure

page 73 in 2nd measure

Let F be a s-algebra in X.

Suppose that F is not equal to P(X).

Prove that there exists a function X -> |R*

that is not measurable.

Suppose that F is not equal to P(X).

Prove that there exists a function X -> |R*

that is not measurable.

page 74 in 2nd measure

Let F be a s-algebra in X. Let E:-F.

Let f : X -> |R* be measurable.

Prove that /\a:-|R {x:-E : f(x)>a} ???.

Let f : X -> |R* be measurable.

Prove that /\a:-|R {x:-E : f(x)>a} ???.

/\a:-|R {x:-E : f(x)>a} :- F

page 74 in 2nd measure

page 74 in 2nd measure

Let F be a s-algebra in X.

Let /\n:-|N f[n] : X - > |R* be measurable.

Let X be a countable disjoint union of sets in F, X = \\//(n:- |N) E[n].

Let f : X -> |R* be defined as follows: x:-E[n] ==> f(x) := f[n](x).

What can we conclude about function f?

Let /\n:-|N f[n] : X - > |R* be measurable.

Let X be a countable disjoint union of sets in F, X = \\//(n:- |N) E[n].

Let f : X -> |R* be defined as follows: x:-E[n] ==> f(x) := f[n](x).

What can we conclude about function f?

It is measurable.

Notice here that X could be written as a finite disjoint union.

page 75 in 2nd measure

page 84 in 2nd measure

Notice here that X could be written as a finite disjoint union.

page 75 in 2nd measure

page 84 in 2nd measure

Let W be a ring in X. Let y:W->|R* be a additive.

Suppose that whenever {A[n]} is a decreasing sequence of sets in W,

such that its intersection is empty, then lim(n->oo) y(A[n]) = 0.

What can we conclude from this?

Suppose that whenever {A[n]} is a decreasing sequence of sets in W,

such that its intersection is empty, then lim(n->oo) y(A[n]) = 0.

What can we conclude from this?

y is countably additive on W

page 75 in 2nd measure

page 75 in 2nd measure

Let W be a ring in X. Let y:W->|R* be countably additive.

Let A[n] be a decreasing sequence of sets in W, having empty intersection.

Suppose that y(A[1]) :- |R.

What can we conclude from this?

Let A[n] be a decreasing sequence of sets in W, having empty intersection.

Suppose that y(A[1]) :- |R.

What can we conclude from this?

lim(n->oo) y(A[n]) = 0

page 77 in 2nd measure

page 77 in 2nd measure

Let W be a ring in X. Let y : W -> |R*.

(1) /\B:-W /\AcB [ y(B) = 0 ==> A:-W ]

(2) /\EcX ???

Find an equivalent condition.

(1) /\B:-W /\AcB [ y(B) = 0 ==> A:-W ]

(2) /\EcX ???

Find an equivalent condition.

(2) /\EcX [ \/A,B:-W AcEcB and y(B\A)=0 ] => E:-W

page 78 in 2nd measure

page 78 in 2nd measure

Let T be a proposition. Let A be a set.

Let Y(a) be a proposition for every a:-A.

Find an equivalent condition for:

(1) T => [ /\a:-A Y(a) ].

Let Y(a) be a proposition for every a:-A.

Find an equivalent condition for:

(1) T => [ /\a:-A Y(a) ].

(2) /\a:-A [ T => Y(a) ]

Let T be a proposition. Let A be a set.

Let Y(a) be a proposition for every a:-A.

Find an equivalent condition for:

(1) /\a:-A [ T => Y(a) ]

Let Y(a) be a proposition for every a:-A.

Find an equivalent condition for:

(1) /\a:-A [ T => Y(a) ]

(2) T => [ /\a:-A Y(a) ]

Let W be a s-ring in X.

Let y : W -> [0,oo] be countably additive, y(O)=0.

Prove that N(y) c W <=> ???.

Let y : W -> [0,oo] be countably additive, y(O)=0.

Prove that N(y) c W <=> ???.

N(y) c W <=> W^ = W

In the proof use a theorem which works in a simpler setting.

Simply recall that theorem.

page 78 in 2nd measure

In the proof use a theorem which works in a simpler setting.

Simply recall that theorem.

page 78 in 2nd measure

Let W be a ring in X. Let y : W -> |R*.

(1) /\EcX [ \/A,B:-W AcEcB and y(B\A)=0 ] => E:-W

(2) /\B:-W ???

Find an equivalent condition.

(1) /\EcX [ \/A,B:-W AcEcB and y(B\A)=0 ] => E:-W

(2) /\B:-W ???

Find an equivalent condition.

(2) /\B:-W /\AcB [ y(B) = 0 ==> A:-W ]

page 78 in 2nd measure

page 78 in 2nd measure

Let W be a s-ring in X.

Let y : W -> [0,oo] countably additive, y(O)=0.

Prove that W^ = W <=> ???.

Let y : W -> [0,oo] countably additive, y(O)=0.

Prove that W^ = W <=> ???.

W^ = W <=> N(y) c W

In the proof use a theorem which works in a simpler setting.

Simply recall that theorem.

page 78 in 2nd measure

In the proof use a theorem which works in a simpler setting.

Simply recall that theorem.

page 78 in 2nd measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let A c |R^p and y*(A) = 0.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let A c |R^p and y*(A) = 0.

A :- MF(y)

page 79 in 2nd measure

page 79 in 2nd measure

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

For every A:-M(y) there exist two Borel sets F,G

such that F c A c G and y(G\F)=y(G\A)=y(A\F)=0.

Hence M(y) c B^, if B denotes the set of all Borel sets

and B^ as on page 72 in 2nd measure.

page 81 in 2nd measure

such that F c A c G and y(G\F)=y(G\A)=y(A\F)=0.

Hence M(y) c B^, if B denotes the set of all Borel sets

and B^ as on page 72 in 2nd measure.

page 81 in 2nd measure

Let X be a complete metric space. Let W be a ring in X.

Let y : W -> [0,oo) be additive and regular.

Let A:-W and suppose that A is totally bounded.

How can we express y(A) in a very interesting way?

Prove the answer.

Let y : W -> [0,oo) be additive and regular.

Let A:-W and suppose that A is totally bounded.

How can we express y(A) in a very interesting way?

Prove the answer.

y(A) = inf { +(n=1 to oo) y(E[n]) : E[n] is an open subset of W
and A c \\//E[n] }.

page 91 in 2nd measure

page 91 in 2nd measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Let |B denote the set of all Borel subsets of |R^p.

Prove that

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Let |B denote the set of all Borel subsets of |R^p.

Prove that

page 81 in 2nd measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Every y-measurable set is a disjoint union of _______ and

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Every y-measurable set is a disjoint union of _______ and

Every y-measurable set is a disjoint union of a Borel set and
of a set of measure zero.

page 82 in 2nd measure

page 82 in 2nd measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Every y-measurable set is a disjoint union of a Borel set

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let M(y) denote the collection of all y-measurable subsets of |R^p.

Every y-measurable set is a disjoint union of a Borel set

Every y-measurable set is a disjoint union of a Borel set and
of a set of measure zero.

page 82 in 2nd measure

page 82 in 2nd measure

Let F be a s-algebra in X.

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : \/A:-|R* lim(n->oo) f[n](x) = A}

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : \/A:-|R* lim(n->oo) f[n](x) = A}

it belongs to F

page 83 in 2nd measure

page 83 in 2nd measure

Let F be a s-algebra in X.

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : lim(n->oo) f[n](x) = oo}

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : lim(n->oo) f[n](x) = oo}

it belongs to F

page 83 in 2nd measure

page 83 in 2nd measure

Let F be a s-algebra in X.

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : lim(n->oo) f[n](x) = -oo}

Let f[n] : X -> |R* be a sequence of measurable functions.

What can we say about this set:

{x:-X : lim(n->oo) f[n](x) = -oo}

it belongs to F

page 83 in 2nd measure

page 83 in 2nd measure

Let F be a s-algebra in X.

Let f[n] : X -> |R* be a sequence of measurable functions.

Prove that

/\g:-|R {x:-X : lim(n->oo) f[n](x) = g} ???

Let f[n] : X -> |R* be a sequence of measurable functions.

Prove that

/\g:-|R {x:-X : lim(n->oo) f[n](x) = g} ???

/\g:-|R {x:-X : lim(n->oo) f[n](x) = g} :- F

page 83 in 2nd measure

page 83 in 2nd measure

Let F be a s-algebra in X.

Let f,g : X -> |R* be measurable.

What can we conclude about function f*g ?

Let f,g : X -> |R* be measurable.

What can we conclude about function f*g ?

f*g is also measurable

Notice that f,g are allowed to assume the infinite values: oo, - oo.

In the proof, we rely on the theorem that if f,g are measurable and finite,

then their product is also measurable.

Notice that f,g are allowed to assume the infinite values: oo, - oo.

In the proof, we rely on the theorem that if f,g are measurable and finite,

then their product is also measurable.

Let f : X -> Y. Let W c P(Y).

Let K = { f-1(A) : A:-W }.

Suppose that W is a ring.

What can we conclude about K?

Let K = { f-1(A) : A:-W }.

Suppose that W is a ring.

What can we conclude about K?

K is also a ring

page 85 in 2nd measure

page 85 in 2nd measure

Let f : X -> Y. Let W c P(Y).

Let K = { f-1(A) : A:-W }.

Suppose that W is a s-ring.

What can we conclude about K?

Let K = { f-1(A) : A:-W }.

Suppose that W is a s-ring.

What can we conclude about K?

K is also a s-ring

page 85 in 2nd measure

page 85 in 2nd measure

Let f : X -> Y. Let W c P(Y).

Let K = { f-1(A) : A:-W }.

Suppose that W is an algebra.

What can we conclude about K?

Let K = { f-1(A) : A:-W }.

Suppose that W is an algebra.

What can we conclude about K?

K is also an algebra

page 85 in 2nd measure

page 85 in 2nd measure

Let f : X -> Y. Let W c P(Y).

Let K = { f-1(A) : A:-W }.

Suppose that W is a s-algebra in Y.

What can we conclude about K?

Let K = { f-1(A) : A:-W }.

Suppose that W is a s-algebra in Y.

What can we conclude about K?

K is a s-algebra in X

page 85 in 2nd measure

page 85 in 2nd measure

Let f : X -> Y. Let A,B c Y.

Suppose that A c f(X) and f-1(A) c f-1(B).

Does this imply that A c B ?

Suppose that A c f(X) and f-1(A) c f-1(B).

Does this imply that A c B ?

YES

page 87 in 2nd measure

page 87 in 2nd measure

Let f : X -> Y.

Prove that

/\BcY f-1(B) = f-1(B n ???)

Prove that

/\BcY f-1(B) = f-1(B n ???)

/\BcY f-1(B) = f-1(B n f(X))

page 97 in 2nd measure

page 97 in 2nd measure

Let f : X -> Y. Let W c P(Y).

Let K = { f-1(B) : B:-W }.

Suppose that W is monotone.

Does K have to be monotone, too?

Let K = { f-1(B) : B:-W }.

Suppose that W is monotone.

Does K have to be monotone, too?

NO

page 88 in 2nd measure

f : {1,2,3,4,...} -> {0,1,-1,2,-2,3,-3,4,-4,...}; f(n)=n

W = { {1,-1}, {1,2,-2}, {1,2,3,-3}, {1,2,3,4,-4}, ... }

K = { f-1(B) : B:-W } = { {1}, {1,2}, {1,2,3}, {1,2,3,4}, ... }

W is monotone but K is not monotone

page 88 in 2nd measure

f : {1,2,3,4,...} -> {0,1,-1,2,-2,3,-3,4,-4,...}; f(n)=n

W = { {1,-1}, {1,2,-2}, {1,2,3,-3}, {1,2,3,4,-4}, ... }

K = { f-1(B) : B:-W } = { {1}, {1,2}, {1,2,3}, {1,2,3,4}, ... }

W is monotone but K is not monotone

Let F a s-algebra in X.

Let f : X -> |R* be measurable.

Prove that

/\Gc|R [ G is open => ??? ]

Let f : X -> |R* be measurable.

Prove that

/\Gc|R [ G is open => ??? ]

/\Gc|R [ G is open => f-1(G) :- F ]

page 89 in 2nd measure

page 89 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

Prove that

/\Gc|R [ ??? => f-1(G) :- F ]

Let f : X -> |R* be measurable.

Prove that

/\Gc|R [ ??? => f-1(G) :- F ]

(1) /\Gc|R [ G is open => f-1(G) :- F ]

(2) /\Gc|R [ G is Borel => f-1(G) :- F ]

page 89,90 in 2nd measure

In order to prove (2) we use (1).

To prove (1), we recall that every open set

(2) /\Gc|R [ G is Borel => f-1(G) :- F ]

page 89,90 in 2nd measure

In order to prove (2) we use (1).

To prove (1), we recall that every open set

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a ring.

What can we conclude about W ?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a ring.

What can we conclude about W ?

W is also a ring

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-ring.

What can we conclude about W ?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-ring.

What can we conclude about W ?

W is also a s-ring

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is an algebra in X.

What can we conclude about W ?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is an algebra in X.

What can we conclude about W ?

W is an algebra in Y

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-algebra.

What can we conclude about W ?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-algebra.

What can we conclude about W ?

W is a s-algebra in Y

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is an ideal.

Does W have to be an ideal, too?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is an ideal.

Does W have to be an ideal, too?

YES.

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-ideal.

Does W have to be a s-ideal, too?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a s-ideal.

Does W have to be a s-ideal, too?

YES

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is monotone.

What can we conclude about W ?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is monotone.

What can we conclude about W ?

W is also monotone

page 89 in 2nd measure

page 89 in 2nd measure

Let f : X -> Y. Let M c P(X).

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a topology.

Does W also have to be a topology?

Let W = { A c Y : f-1(A) :- M }.

Suppose that M is a topology.

Does W also have to be a topology?

YES

page 89 in 2nd measure

page 89 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> |R\{-oo,oo} be measurable.

Let g:|R->|R be such that /\a:-|R {x:-|R : g(x)>a} is Borel.

What can we conclude?

Let f : X -> |R\{-oo,oo} be measurable.

Let g:|R->|R be such that /\a:-|R {x:-|R : g(x)>a} is Borel.

What can we conclude?

gof is measurable

/\a:-|R {x:-X : g(f(x))>a} :- F

page 90 in 2nd measure

/\a:-|R {x:-X : g(f(x))>a} :- F

page 90 in 2nd measure

Every bounded sequence of rational numbers contains a Cauchy
subsequence.

How can that be proven without using the Dedekind axiom?

How can that be proven without using the Dedekind axiom?

Without using the Dedekind axiom, we can prove that every
bounded subset of |Q is totally bounded. And every sequence
contained in a totally bounded set has a Cauchy subsequence.

page 141 in golden gate

page 141 in golden gate

Let f : |R^2 -> |R have both partial derivatives: f1,f2.

(1) /\a,b:-|R |f1(a,b)| <= 2*|b-a|

(2) /\a,b:-|R |f2(a,b)| <= 2*|b-a|

(3) f(0,0) = 0

Prove that |f(5,4)| <= 1.

(1) /\a,b:-|R |f1(a,b)| <= 2*|b-a|

(2) /\a,b:-|R |f2(a,b)| <= 2*|b-a|

(3) f(0,0) = 0

Prove that |f(5,4)| <= 1.

page 77 in OLDTIMER

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Does MF(y) have to be a s-ring?

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

Does MF(y) have to be a s-ring?

NO.

We know that if A:-MF(y), then y(A)<oo.

A countable union of elementary sets can be unbounded and have infinite measure.

We know that if A:-MF(y), then y(A)<oo.

A countable union of elementary sets can be unbounded and have infinite measure.

Let M be a s-algebra in X. Let f : X -> |R* be constant.

Does f have to be M-measurable?

Does f have to be M-measurable?

[ \/K:-|R* /\x:-X f(x)=K ] ==> f is measurable

page 136 in 1st measure

page 136 in 1st measure

Let M be a s-algebra in X.

Let f:X->|R* be M-measurable. Let b:-|R.

What can we conclude about function g:X->|R* defined by g(x)=b*f(x) ?

Let f:X->|R* be M-measurable. Let b:-|R.

What can we conclude about function g:X->|R* defined by g(x)=b*f(x) ?

Function g:X->|R* defined by g(x)=b*f(x) is M-measurable.

page 136 in 1st measure

page 136 in 1st measure

Let A[n] c X for all natural n.

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

Prove that for all x:-X

/\x:-X lim_sup 1(A[n])(x) = ???

If AcX, then let 1(A) denote the characteristic function of the set A defined on X.

Prove that for all x:-X

/\x:-X lim_sup 1(A[n])(x) = ???

/\x:-X lim_sup 1(A[n])(x) = 1(lim_sup A[n])(x)

page 11 in 1st measure

page 11 in 1st measure

Let A,B c X.

(X\A) \ (X\B) = ???

(X\A) \ (X\B) = ???

(X\A) \ (X\B) = B\A

(B\C) \ (A\D) c ???

(B\C) \ (A\D) c (B\A) u (D\C)

Let F be a s-algebra in X.

Let f : X -> |R*.

Suppose that |f| is measurable.

Does f have to be measurable?

Let f : X -> |R*.

Suppose that |f| is measurable.

Does f have to be measurable?

NO.

Let E c X such that E does not belong to F.

Define f : X -> |R* as follows: f = 1 on E and f = -1 on X\E

{x:-X : f(x)>0} = E does not belong to F

Hence f is not measurable.

But |f| is constant hence measurable.

Let E c X such that E does not belong to F.

Define f : X -> |R* as follows: f = 1 on E and f = -1 on X\E

{x:-X : f(x)>0} = E does not belong to F

Hence f is not measurable.

But |f| is constant hence measurable.

Let R be a ring of sets in X.

Describe K(R) - the algebra generated by R.

Describe K(R) - the algebra generated by R.

K(R) = {EcX : E:-R or X\E:-R}

page 67 in 2nd measure

page 67 in 2nd measure

Let A[1],A[2],A[3],... be a decreasing sequence of sets, having
empty intersection.

What can we conclude from this?

What can we conclude from this?

A[1] = \\*//(n=1 to oo) A[n] \ A[n+1]

/\m:-|N A[m] = \\*//(n=m to oo) A[n] \ A[n+1]

page 97 in 2nd measure

/\m:-|N A[m] = \\*//(n=m to oo) A[n] \ A[n+1]

page 97 in 2nd measure

Let A[1],A[2],A[3],... be a decreasing sequence of sets, having
empty intersection.

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

YES

page 97 in 2nd measure

page 97 in 2nd measure

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

NO

1) Let A[1] = {-7}.

For n>=2, let A[n] = {n,n+1,n+2,n+3,...}.

2) Let A[1]=O, A[2]={7}, A[n] = O, for n>=3

1) Let A[1] = {-7}.

For n>=2, let A[n] = {n,n+1,n+2,n+3,...}.

2) Let A[1]=O, A[2]={7}, A[n] = O, for n>=3

Let A[1],A[2],A[3],... be a decreasing sequence of sets.

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

NO

Let A[n] = {1} for all n:-|N.

Let A[n] = {1} for all n:-|N.

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

YES

The right side is easily seen to be contained in A[1].

To prove the inverse inclusion one argues by contradiction.

Suppose that x:-A[1] and /\n:-|N ( x:-A[n] => x:-A[n+1] ).

Then it follows that x :- /\n:-|N A[n].

This is a contradiction because the intersection is assumed to be empty.

The right side is easily seen to be contained in A[1].

To prove the inverse inclusion one argues by contradiction.

Suppose that x:-A[1] and /\n:-|N ( x:-A[n] => x:-A[n+1] ).

Then it follows that x :- /\n:-|N A[n].

This is a contradiction because the intersection is assumed to be empty.

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

NO.

Although we can conclude that A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1],

we cannot conclude that the union is disjoint.

Let A[1] = A[3] = {1}. Let A[2] = A[4] = O. Let A[n] = O for n>4.

Then (A[1]\A[2]) n (A[3]\A[4]) is not empty.

Although we can conclude that A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1],

we cannot conclude that the union is disjoint.

Let A[1] = A[3] = {1}. Let A[2] = A[4] = O. Let A[n] = O for n>4.

Then (A[1]\A[2]) n (A[3]\A[4]) is not empty.

Let A[1],A[2],A[3],... be a decreasing sequence of sets.

Can we conclude that {A[n]\A[n+1]}(n:-|N) is a disjoint collection?

Can we conclude that {A[n]\A[n+1]}(n:-|N) is a disjoint collection?

YES

page 97 in 2nd measure

page 97 in 2nd measure

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Decide if we can conclude that

\/k:-|N {A[n]\A[n+1]}_(n=k to n=oo) is disjoint.

Decide if we can conclude that

\/k:-|N {A[n]\A[n+1]}_(n=k to n=oo) is disjoint.

NO

(1) Let A[2n]={1}, A[2n+1]=O.

//Let A[2n] = {2n, 2n+1, 2n+2, 2n+3, ...}. Let A[2n+1] = {2n}.

//Then //\\(n:-|N) A[n] = O. And for every even n:-|N,

//(A[n]\A[n+1]) n (A[n+2]\A[n+3]) = {n+1,n+2,n+3,n+4,...} n {n+3,n+4,...} != O

(1) Let A[2n]={1}, A[2n+1]=O.

//Let A[2n] = {2n, 2n+1, 2n+2, 2n+3, ...}. Let A[2n+1] = {2n}.

//Then //\\(n:-|N) A[n] = O. And for every even n:-|N,

//(A[n]\A[n+1]) n (A[n+2]\A[n+3]) = {n+1,n+2,n+3,n+4,...} n {n+3,n+4,...} != O

Let B[1],B[2],B[3],... be an increasing sequence of sets.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

YES

page 98 in 2nd measure

page 98 in 2nd measure

Let B[1],B[2],B[3],... be a sequence of sets.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

YES

The right-hand set is easily seen to be contained in the left- hand set.

Take any x belonging to the left-hand set. We have (1) \/n:-|N x:-B[n].

Let's argue by contradiction. Suppose that (2) /\n:-|N ( x:- B[n] => x:-B[n-1] ).

The right-hand set is easily seen to be contained in the left- hand set.

Take any x belonging to the left-hand set. We have (1) \/n:-|N x:-B[n].

Let's argue by contradiction. Suppose that (2) /\n:-|N ( x:- B[n] => x:-B[n-1] ).

Let B[1],B[2],B[3],... be a sequence of sets.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

NO

We can indeed conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1]

but the right-hand union does not have to be disjoint.

We can indeed conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1]

but the right-hand union does not have to be disjoint.

Let B[1],B[2],B[3],... be an increasing sequence of sets.

Decide if we can conclude that

/\n:-|N B[n] = \\*//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

Decide if we can conclude that

/\n:-|N B[n] = \\*//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

YES

page 42 in 1st measure

page 42 in 1st measure

Let B[1],B[2],B[3],... be a sequence of sets.

Decide if we can conclude that

/\n:-|N B[n] = \\//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

Decide if we can conclude that

/\n:-|N B[n] = \\//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

NO

Let B[1] = {7}, B[2] = O.

Then B[2] != B[1] u B[2]\B[1].

We can only conclude:

/\n:-|N B[n] c \\//(k=1 to k=n) B[k]\B[k-1],

Let B[1] = {7}, B[2] = O.

Then B[2] != B[1] u B[2]\B[1].

We can only conclude:

/\n:-|N B[n] c \\//(k=1 to k=n) B[k]\B[k-1],

Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.

Suppose that both sequences a[n] and b[n] are increasing.

What can we conclude?

Suppose that both sequences a[n] and b[n] are increasing.

What can we conclude?

There exist a,b:-[0,oo] such that

(1) lim a[n] = a

(2) lim b[n] = b

(3) lim a[n]*b[n] = a*b

page 93 in 2nd measure

(1) lim a[n] = a

(2) lim b[n] = b

(3) lim a[n]*b[n] = a*b

page 93 in 2nd measure

Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.

Let a,b:-[0,oo].

(1) lim a[n] = a,

(2) lim b[n] = b.

Decide if we can conclude that

(3) lim a[n]*b[n] = a*b.

Let a,b:-[0,oo].

(1) lim a[n] = a,

(2) lim b[n] = b.

Decide if we can conclude that

(3) lim a[n]*b[n] = a*b.

NO

1) a[n] = 1/n; b[n] = n

2) a[n] = 1/n; b[n] = n^2

1) a[n] = 1/n; b[n] = n

2) a[n] = 1/n; b[n] = n^2

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

What can we conclude about the measurability of 1/f ?

Let f : X -> |R* be measurable.

What can we conclude about the measurability of 1/f ?

It is measurable, no matter how we define: 1/0, 1/oo, 1/-oo.

page 93 in 2nd measure

page 93 in 2nd measure

Let W c P(X) and let W contain the empty set.

Let y : W -> [0,oo] and y(O)=0.

In this broad setting, define an outer measure on X.

Prove that it is indeed an outer measure.

Let y : W -> [0,oo] and y(O)=0.

In this broad setting, define an outer measure on X.

Prove that it is indeed an outer measure.

y*(A) = inf { +(k=1 to oo) y(E[k]) : E[k]:-W and A c \\//(k:-
|N)_E[k] }

inf O = oo

page 95 in 2nd measure

inf O = oo

page 95 in 2nd measure

Let f : X -> Y.

Let J be an ideal in Y.

Let K = {f-1(A) : A :- J}.

Can we conclude that K is also an ideal?

Let J be an ideal in Y.

Let K = {f-1(A) : A :- J}.

Can we conclude that K is also an ideal?

NO

f : {1,2} -> {3}

J = { O, {3} }

K = { O, {1,2} }

J is an ideal but K is not

f : {1,2} -> {3}

J = { O, {3} }

K = { O, {1,2} }

J is an ideal but K is not

(1) f : X -> Y

(2) D,E c Y

(3) f-1(D) c f-1(E)

Can we conclude that D c E ?

(2) D,E c Y

(3) f-1(D) c f-1(E)

Can we conclude that D c E ?

NO

f : {1,2} -> {3,4}, f(1)=f(2)=3

D={3,4}, E={3}

f-1(D) = f-1(E) = {1,2}

----------------------------

f : {1} -> {2,3}, f(1)=3

D={2}, E={3}

f : {1,2} -> {3,4}, f(1)=f(2)=3

D={3,4}, E={3}

f-1(D) = f-1(E) = {1,2}

----------------------------

f : {1} -> {2,3}, f(1)=3

D={2}, E={3}

f : X -> Y; g : Y -> Z; E c Z

f-1(g-1(E)) = { x:-X : ????? }

f-1(g-1(E)) = { x:-X : ????? }

f-1(g-1(E)) = { x:-X : g(f(x)):-E }

f : X -> Y; g : Y -> Z; E c Z

{ x:-X : g(f(x)):-E } = ???

{ x:-X : g(f(x)):-E } = ???

{ x:-X : g(f(x)):-E } = f-1(g-1(E))

Let f : |C\{0} -> |C be f(z)=1/z.

Derive the formula for the nth derivative of f.

Derive the formula for the nth derivative of f.

f|n| (z) = (-1)^n * n! / z^(n+1)

Give the power series expansion of exp(z)*sin(z) about 0 and
find its radius of convergence.

exp(z)*sin(z) = +(n=0 to n=oo) a[n]*z^n

a[n] = ( (1+i)^n - (1-i)^n ) / ( 2i * n! )

radius = oo

hint: sin(z) = ( exp (iz) - exp(-iz) ) / 2i

a[n] = ( (1+i)^n - (1-i)^n ) / ( 2i * n! )

radius = oo

hint: sin(z) = ( exp (iz) - exp(-iz) ) / 2i

Let G be an open subset of |C.

Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

What does it mean that z0 is a zero of function f of multiplicity m ?

Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

What does it mean that z0 is a zero of function f of multiplicity m ?

(1) f(z0) = 0

there exists g:G->|C diffable such that

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

page 87 in OLDTIMER

there exists g:G->|C diffable such that

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

page 87 in OLDTIMER

Let G be an open subset of |C. Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

There exists g:G->|C diffable such that

(1) f(z0) = 0

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

Let z0 :- G and let m:-|N, m>=1.

There exists g:G->|C diffable such that

(1) f(z0) = 0

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

z0 is a zero of function f of multiplicity m

page 87 in OLDTIMER

multiplicity [m^l t(i) 'pli s(i) ti(:)]

page 87 in OLDTIMER

multiplicity [m^l t(i) 'pli s(i) ti(:)]

f(z) = z^2 * ( exp(z^2) - 1 )

Find the multiplicity of 0.

Find the multiplicity of 0.

4

page 87 in OLDTIMER

page 87 in OLDTIMER

Give the power series expansion of f(z) = z^2 * exp(z-1) about
1 and find its radius of convergence.

f(z) = +(n=0 to n=oo) a[n] * (z-1)^n

a[n] = 1/(n-2)! + 2/(n-1)! + 1/(n!)

hint: z^2 * exp(z-1) = ((z-1) + 1)^2 * exp(z-1)

a[n] = 1/(n-2)! + 2/(n-1)! + 1/(n!)

hint: z^2 * exp(z-1) = ((z-1) + 1)^2 * exp(z-1)

f(z) = 6*sin(z^3) + z^3*(z^6 - 6)

Find the multiplicity of 0.

Find the multiplicity of 0.

15

Determine the multiplicity of all zeros of cos.

1

hint: cos(z) = -sin(z - pi/2)

hint: cos(z) = -sin(z - pi/2)

Determine the multiplicity of all zeros of sin.

1

page 87 in OLDTIMER

page 87 in OLDTIMER

Give the power series expansion of cosine about pi/2.

+(k=0 to k=oo) [ (-1)^(k+1) * (z-pi/2)^(2k+1) / (2k+1)! ]

hint: cos(z) = -sin(z - pi/2)

hint: cos(z) = -sin(z - pi/2)

Let G be an open connected subset of |C.

Let f,g : G -> |C be diffable.

What theorem can be used to prove that:

f*g=0 on G ==> f=0 on G or g=0 on G

Let f,g : G -> |C be diffable.

What theorem can be used to prove that:

f*g=0 on G ==> f=0 on G or g=0 on G

Let G be an open connected subset of |C.

Let f : G -> |C be diffable.

(1) {z:-G : f(z)=0} has a limit point in G

(2) /\z:-G f(z)=0

thesis: (1)=>(2)

REMARK: MRW hasn't done this proof yet. 2001.02.23; 2001.05.15

Let f : G -> |C be diffable.

(1) {z:-G : f(z)=0} has a limit point in G

(2) /\z:-G f(z)=0

thesis: (1)=>(2)

REMARK: MRW hasn't done this proof yet. 2001.02.23; 2001.05.15

Let W be a non-empty collection of sets such that

(1) A,B:-W => AnB:-W

(2) A,B:-W => A\B is a finite disjoint union of sets in W

Let y : W -> |R* be additive.

Can we conclude that y is finitely additive?

(1) A,B:-W => AnB:-W

(2) A,B:-W => A\B is a finite disjoint union of sets in W

Let y : W -> |R* be additive.

Can we conclude that y is finitely additive?

NO

W = { O, {1}, {2}, {3}, {1,2,3} }

y({1,2,3}) = 1

y(O) = y({1}) = y({2}) = y({3}) = 0

y is additive but not finitely additive

W = { O, {1}, {2}, {3}, {1,2,3} }

y({1,2,3}) = 1

y(O) = y({1}) = y({2}) = y({3}) = 0

y is additive but not finitely additive

G c |C; f : G -> |C; a:-|C

What does it mean that function f has an isolated singularity at a?

What does it mean that function f has an isolated singularity at a?

(1) a doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

G c |C; f : G -> |C; a:-|C

(1) a doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

What is the verbal description of this situation?

(1) a doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

What is the verbal description of this situation?

function f has an isolated singularity at a

Let G be a subset of |C. Let f : G -> |C. Let a:-|C.

What does it mean that function f has a removable isolated singularity at a?

What does it mean that function f has a removable isolated singularity at a?

There exists a number r > 0 and function g : B(a;r) -> |C such
that:

(1) a doesn't belong to G

(2) B(a;r) \ {a} c G

(3) g is diffable

(4) g=f on B(a;r)\{a}

(1) a doesn't belong to G

(2) B(a;r) \ {a} c G

(3) g is diffable

(4) g=f on B(a;r)\{a}

Let G be a subset of |C. Let f : G -> |C. Let a:-|C.

What is the verbal description of this situation:

There exists a number r > 0 and function g : B(a;r) -> |C such that:

(1) a doesn't belong to G

(2) B(a;r) \ {a} c G

(3) g is diffable

What is the verbal description of this situation:

There exists a number r > 0 and function g : B(a;r) -> |C such that:

(1) a doesn't belong to G

(2) B(a;r) \ {a} c G

(3) g is diffable

function f has a removable isolated singularity at a

Let G be an open subset of a metric space X.

Let A c X. Suppose that G n Clo(A) is non-empty.

What can we conclude from this?

Let A c X. Suppose that G n Clo(A) is non-empty.

What can we conclude from this?

G n A is non-empty

Let G be an open subset of a metric space X.

Let A c X. Suppose that G n A is empty.

What can we conclude from this?

Let A c X. Suppose that G n A is empty.

What can we conclude from this?

G n Clo(A) is empty

Let G,A be subsets of a metric space X.

Suppose that G n A is empty and G n Clo(A) is non-empty.

What can we conclude from this?

Suppose that G n A is empty and G n Clo(A) is non-empty.

What can we conclude from this?

G is not open

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a,b:-|R [a] + [b] <= ???.

Prove that /\a,b:-|R [a] + [b] <= ???.

/\a,b:-|R [a] + [b] <= [a + b]

page 120 in OLDTIMER

page 120 in OLDTIMER

(1) 0 <= a[n] <= oo

(2) 0 <= b[n] <= oo,

(3) lim a[n] = a, 0 < a <= oo

(4) lim b[n] = b, 0 < b <= oo

What can we conclude from this?

(2) 0 <= b[n] <= oo,

(3) lim a[n] = a, 0 < a <= oo

(4) lim b[n] = b, 0 < b <= oo

What can we conclude from this?

lim a[n]*b[n] = a*b

G c |C; f : G -> |C; a:-|C

Suppose that f has an isolated singularity at a.

(1) f has a removable singularity at a

(2) ??????

Recall an equivalent condition without proof.

Suppose that f has an isolated singularity at a.

(1) f has a removable singularity at a

(2) ??????

Recall an equivalent condition without proof.

(2) lim(z->a) (z-a)*f(z) = 0

MRW hasn't done this proof yet. 2001.02.27; 2001.05.26

(page 103, Conway, "Functions of One Complex Variable", chapter V)

MRW hasn't done this proof yet. 2001.02.27; 2001.05.26

(page 103, Conway, "Functions of One Complex Variable", chapter V)

G c |C; f : G -> |C; b:-|C

What does it mean that f has a pole at b?

What does it mean that f has a pole at b?

(1) b doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-b| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

or shorter:

(1) f has an isolated singularity at b

(2) there exists B = {z:-|C : 0 < |z-b| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

or shorter:

(1) f has an isolated singularity at b

G c |C; f : G -> |C; b:-|C

(1) b doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-b| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

What is the verbal description of this situation?

(1) b doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-b| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

What is the verbal description of this situation?

function f has a pole at point b

G c |C; f : G -> |C; b:-|C

Suppose that function f has a removable isolated singularity at point b.

Can it also have a pole at b?

Suppose that function f has a removable isolated singularity at point b.

Can it also have a pole at b?

NO.

By virtue of our supposition there exists a number r > 0

and a function g : B(b;r) -> |C such that:

(1) b doesn't belong to G

(2) B(b;r) \ {b} c G

(3) g is diffable

(4) g=f on B(b;r)\{b}

By virtue of our supposition there exists a number r > 0

and a function g : B(b;r) -> |C such that:

(1) b doesn't belong to G

(2) B(b;r) \ {b} c G

(3) g is diffable

(4) g=f on B(b;r)\{b}

G c |C; f : G -> |C; b:-|C

Suppose that function f has a pole at point b.

Can it also have a removable singularity at b?

Suppose that function f has a pole at point b.

Can it also have a removable singularity at b?

NO.

By virtue of our supposition:

(1) a doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

By virtue of our supposition:

(1) a doesn't belong to G

(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B

(3) lim(z->b) |f(z)| = oo

G c |C; f : G -> |C; b:-|C

What does it mean that function f has an essential isolated singularity at point b?

What does it mean that function f has an essential isolated singularity at point b?

(1) f has an isolated singularity at b

(2) the singularity is not removable

(3) the singularity is not a pole

(2) the singularity is not removable

(3) the singularity is not a pole

Consider the complex function f(z)=exp(1/z).

What kind of isolated singularity does it have at 0?

What kind of isolated singularity does it have at 0?

essential

(1) lim(n->oo) |f(1/n)| = oo

(2) lim(n->oo) |f(i/n)| = 1 < oo

(1) not removable

(2) not a pole

(1) lim(n->oo) |f(1/n)| = oo

(2) lim(n->oo) |f(i/n)| = 1 < oo

(1) not removable

(2) not a pole

Let m be a positive integer.

Consider the complex function f(z)=z^m * sin(1/z).

What kind of isolated singularity does it have at 0?

Consider the complex function f(z)=z^m * sin(1/z).

What kind of isolated singularity does it have at 0?

essential

(1) lim(n->oo) |f(i/n)| = oo

(2) lim(n->oo) |f(1/n)| = 0

(1) not removable

(2) not a pole

(1) lim(n->oo) |f(i/n)| = oo

(2) lim(n->oo) |f(1/n)| = 0

(1) not removable

(2) not a pole

Consider the complex function f(z) = sin(z)/z.

What kind of isolated singularity does it have at 0?

What kind of isolated singularity does it have at 0?

removable

====================

You can prove your answer by referring to the definition only.

This is also an opportunity to test the unproven theorem. (MRW 2001.02.27)

THEOREM: G c |C; f : G -> |C; b:-|C

thesis: If function f has an isolated singularity at poin

====================

You can prove your answer by referring to the definition only.

This is also an opportunity to test the unproven theorem. (MRW 2001.02.27)

THEOREM: G c |C; f : G -> |C; b:-|C

thesis: If function f has an isolated singularity at poin

Consider the complex function f(z) = ( cos(z) - 1 ) / z.

What kind of isolated singularity does it have at 0?

What kind of isolated singularity does it have at 0?

removable

====================

You can prove your answer by referring to the definition only.

This is an opportunity to test the unproven theorem. (MRW 2001.02.04)

THEOREM: G c |C; f : G -> |C; b:-|C

thesis: If function f has an isolated singularity at poin

====================

You can prove your answer by referring to the definition only.

This is an opportunity to test the unproven theorem. (MRW 2001.02.04)

THEOREM: G c |C; f : G -> |C; b:-|C

thesis: If function f has an isolated singularity at poin

lim(x->0,y->0) (x-sin(y))*y^2 / (y^2 + sin(x-y)*sin(x-y) ) = ???

= 0

hint: y^2 <= y^2 + sin(x-y)*sin(x-y)

hint: y^2 <= y^2 + sin(x-y)*sin(x-y)

lim(x->0) x*sin(x) / ( x^2 + sin(x)*sin(x) ) = ???

= 1/2

hint: use the squeeze theorem

hint: use the squeeze theorem

Consider the complex function f(z) = (z+1)/z.

Does it have a removable isolated singularity at 0 ?

Justify your answer in two different ways.

Does it have a removable isolated singularity at 0 ?

Justify your answer in two different ways.

NO. It actually has a pole. Notice that lim(z->0) |f(z)| = oo.

---------------------------------

(1) lim(z->0) z*f(z) = 1

hence not (2) lim(z->0) (z-0)*f(z) = 0.

And (2) is a necessary condition for a removable isolated singularity.

(page 103, Conway, "Functions of One Complex Variable", cha

---------------------------------

(1) lim(z->0) z*f(z) = 1

hence not (2) lim(z->0) (z-0)*f(z) = 0.

And (2) is a necessary condition for a removable isolated singularity.

(page 103, Conway, "Functions of One Complex Variable", cha

w:-|C

2*sin^2(w) = ???

2*sin^2(w) = ???

2*sin^2(w) = 1 - cos(2*w)

Integral(0,pi,cos(2x),dx) = ???

Integral(0,pi,cos(2x),dx) = 0

hint: [ sin(2x)/2 ]' = cos(2x)

hint: [ sin(2x)/2 ]' = cos(2x)

Integral(0,pi,sin(x)sin(x),dx) = ???

Integral(0,pi,sin(x)sin(x),dx) = pi/2

There are at least two ways:

(1) integrate by parts: page 87 in OLDTIMER

(2) use: 2sin(x)sin(x) = 1-cos(2x), and then Integral(0,pi,cos(2x),dx)=0

There are at least two ways:

(1) integrate by parts: page 87 in OLDTIMER

(2) use: 2sin(x)sin(x) = 1-cos(2x), and then Integral(0,pi,cos(2x),dx)=0

Integral(-1,1,sqrt(1-x^2),dx) = ???

Integral(-1,1,sqrt(1-x^2),dx) = pi/2

hint: x=cos(t), later use: Integral(0,pi,sin(x)sin(x),dx) = pi/2

Notice the significance of this integral.

It calculates half the area of the unit circle, which is pi.

hint: x=cos(t), later use: Integral(0,pi,sin(x)sin(x),dx) = pi/2

Notice the significance of this integral.

It calculates half the area of the unit circle, which is pi.

Integral(0,pi,cos(x)cos(x),dx) = ???

Integral(0,pi,cos(x)cos(x),dx) = pi/2

Use:

(1) cos(x)cos(x) = 1 - sin(x)sin(x)

(2) Integral(0,pi,sin(x)sin(x),dx) = pi/2

Or you can integrate by parts: page 87 in OLDTIMER

Use:

(1) cos(x)cos(x) = 1 - sin(x)sin(x)

(2) Integral(0,pi,sin(x)sin(x),dx) = pi/2

Or you can integrate by parts: page 87 in OLDTIMER

Let F be an algerba in X.

Let f : X -> |R\{oo,-oo} be F-measurable.

What can we conclude about function [f] ?

([x] denotes the integer part of real number x)

Let f : X -> |R\{oo,-oo} be F-measurable.

What can we conclude about function [f] ?

([x] denotes the integer part of real number x)

g:|R->|R, g(x)=[x] is an increasing function,

hence gof:X->|R is measurable.

page 91,90 in 2nd measure

hence gof:X->|R is measurable.

page 91,90 in 2nd measure

Let (X,d) be a metric space.

Let A be an uncountable subset of X satisfying:

(1) \/(a>0) /\(x,y:-A) x!=y => d(x,y)>=a

What can we conclude?

Let A be an uncountable subset of X satisfying:

(1) \/(a>0) /\(x,y:-A) x!=y => d(x,y)>=a

What can we conclude?

X does not contain a countable dense subset.

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

(1) f : |R x |R -> |R

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R f(x,y) <= f(a,b)

Can we conclude that a=b ?

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R f(x,y) <= f(a,b)

Can we conclude that a=b ?

NO.

Let f(1,0)=f(0,1)=7 and otherwise: f(x,y)=0.

Let a=1, b=0.

Let f(1,0)=f(0,1)=7 and otherwise: f(x,y)=0.

Let a=1, b=0.

(1) f : |R x |R -> |R

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R (x,y)!=(a,b) ==> f(x,y) < f(a,b)

Can we conclude that a=b ?

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R (x,y)!=(a,b) ==> f(x,y) < f(a,b)

Can we conclude that a=b ?

YES.

Transform (4) into (5) /\x,y:-|R f(x,y) >= f(a,b) ==> (x,y)=(a,b).

Then by (2), f(b,a)=f(a,b), and f(b,a)>=f(a,b).

By (5), (b,a)=(a,b). Hence a=b.

Transform (4) into (5) /\x,y:-|R f(x,y) >= f(a,b) ==> (x,y)=(a,b).

Then by (2), f(b,a)=f(a,b), and f(b,a)>=f(a,b).

By (5), (b,a)=(a,b). Hence a=b.

Draw the set { (x,y):-|R^2 : x^2 + y^2 <= x and x^2 + y^2 <=
y }.

It is the intersection of two closed disks.

(1) center (0 , 1/2) radius 1/2

(2) center (1/2 , 0) radius 1/2

(1) center (0 , 1/2) radius 1/2

(2) center (1/2 , 0) radius 1/2

Let (x,y):-|R^2 satisfy x^2 + y^2 = y and x>=0.

How can we express the distance of point between (x,y) and (0,0)?

How can we express the distance of point between (x,y) and (0,0)?

There are two ways:

1) sqrt(y)

2) sin(a), where a = arctan(y/x) = sin( arg(x+iy) )

page 91 in OLDTIMER

1) sqrt(y)

2) sin(a), where a = arctan(y/x) = sin( arg(x+iy) )

page 91 in OLDTIMER

Integral( 0, pi/2, sin^2(x)*cos(x), dx ) = ???

Integral( 0, pi/2, sin(x)sin(x)cos(x), dx ) = 1/3

Notice that there are two easy ways to calculate this integral:

1) F(x) = 1/3 * sin3(x); F'(x) = sin2(x)cos(x)

2) sin2(x)cos(x) = sin2(x)sin'(x); change variable cos(x)=t

Notice that there are two easy ways to calculate this integral:

1) F(x) = 1/3 * sin3(x); F'(x) = sin2(x)cos(x)

2) sin2(x)cos(x) = sin2(x)sin'(x); change variable cos(x)=t

Integral( pi/6, pi/2, ctg(x), dx ) = ???

Integral( pi/6, pi/2, ctg(x), dx ) = log(2)

hint: [ log(sin(x)) ]' = ctg(x)

sin(pi/6) = 1/2

hint: [ log(sin(x)) ]' = ctg(x)

sin(pi/6) = 1/2

Integral( 0, pi/2, sin^2(2x), dx ) = ???

Integral( 0, pi/2, sin^2(2x), dx ) = pi/4

Use: Integral( 0, pi, sin^2(x), dx ) = pi/2

Use: Integral( 0, pi, sin^2(x), dx ) = pi/2

Integral( 0, pi/2, sin^2(x)*cos^2(x), dx ) = ???

Integral( 0, pi/2, sin^2(x)*cos^2(x), dx ) = pi/16

Use: Integral( 0, pi/2, sin^2(2x), dx ) = pi/4

hint: sin(2x)=2cos(x)sin(x)

Use: Integral( 0, pi/2, sin^2(2x), dx ) = pi/4

hint: sin(2x)=2cos(x)sin(x)

Integer( 0, 1, x^2 * sqrt(1-x^2), dx ) = ???

Integer( 0, 1, x^2 * sqrt(1-x^2), dx ) = pi/16

Use: Integral( 0, pi/2, sin2(x)cos2(x), dx ) = pi/16

hint: x=sin(t)

Use: Integral( 0, pi/2, sin2(x)cos2(x), dx ) = pi/16

hint: x=sin(t)

Integral( 0, 2, x^2 * sqrt(1 - 1/4*x^2), dx ) = ???

Integral( 0, 2, x^2 * sqrt(1 - 1/4*x^2), dx ) = pi/2

hint: x=2sin(t)

Integral( 0, pi/2, sin^2(x)*cos^2(x), dx ) = pi/16

Integral( 0, pi, sin^2(x), dx ) = pi/2 (page 87 in OLDTIMER)

hint: x=2sin(t)

Integral( 0, pi/2, sin^2(x)*cos^2(x), dx ) = pi/16

Integral( 0, pi, sin^2(x), dx ) = pi/2 (page 87 in OLDTIMER)

/\n:-|N lim(x->0) ((1+x)^n - 1)/x = ???

/\n:-|N lim(x->0) ((1+x)^n - 1)/x = n

If n:-|N, let f(x) = (1+x)^n.

Notice that this limit calculates the derivative of f at 0.

If n:-|N, let f(x) = (1+x)^n.

Notice that this limit calculates the derivative of f at 0.

a,b :- |C; a!=1, b!=1, b!=a

1 / (1-a)(1-b) = ???

1 / (1-a)(1-b) = ???

1 / (1-a)(1-b) = (1/(1-a) - 1/(1-b)) / (a-b)

Let x:-|C\{0,1}.

1 / x(x-1) = ???

1 / x(x-1) = ???

1 / x(x-1) = 1/(x-1) - 1/x

Let x:-|C\{0,1}.

1/(x-1) - 1/x = ???

1/(x-1) - 1/x = ???

1/(x-1) - 1/x = 1 / x(x-1)

a,b :- |C; a!=1, b!=1

(a-b) / (1-a)(1-b) = ???

(a-b) / (1-a)(1-b) = ???

(a-b) / (1-a)(1-b) = 1/(1-a) - 1/(1-b)

a,b :- |C; a!=1, b!=1

1/(1-a) - 1/(1-b) = ???

1/(1-a) - 1/(1-b) = ???

(1/(1-a) - 1/(1-b)) = (a-b) / (1-a)(1-b)

For z:-|C, |z|<1, let f(z) = 1 / (z+1)(z-2).

Give the power series expansion of function f about 0.

Find its radius of convergence.

Give the power series expansion of function f about 0.

Find its radius of convergence.

f(z) = +(n=0 to oo) a[n] * z^n

a[n] = -1/(6 * 2^n) - ((-1)^n / 3)

R=1

hint: (1/(z-2) - 1/(z+1))/3

a[n] = -1/(6 * 2^n) - ((-1)^n / 3)

R=1

hint: (1/(z-2) - 1/(z+1))/3

x :- |R

Im( i / (1 + ix) ) = ???

Im( i / (1 + ix) ) = ???

Im( i / (1 + ix) ) = 1 / (1 + x^2)

For z:-|C, |z|<1, let f(z) = 1 / (1+z^3)^2.

Give the power series expansion of function f about point 0.

Find the radius of convergence.

Give the power series expansion of function f about point 0.

Find the radius of convergence.

f(z) = +(n=0 to n=oo) (-1)^n * (n+1) * z^(3n)

R=1

hint: for |w|<1, 1 / (1+w) = +(n= .... ; then differentiate the series.

R=1

hint: for |w|<1, 1 / (1+w) = +(n= .... ; then differentiate the series.

For z:-|C, |z|<1, let f(z) = +(n=0 to n=oo) (-1)^n * (n+1) * z^n.

Find a direct formula for f(z).

Find a direct formula for f(z).

f(z) = 1 / (1+z)^2

hint: f(z) = +(n=1 to n=oo) (-1)^(n+1) * n * z^(n-1)

g(z) = +(n=0 to n=oo) (-1)^(n+1) * z^n

g'(z) = f(z)

g(z) = -1/(1+z)

hint: f(z) = +(n=1 to n=oo) (-1)^(n+1) * n * z^(n-1)

g(z) = +(n=0 to n=oo) (-1)^(n+1) * z^n

g'(z) = f(z)

g(z) = -1/(1+z)

Let w be a complex number.

1 + cos(2w) = 2* ???

1 + cos(2w) = 2* ???

1 + cos(2w) = 2*cos^2(w)

hint: cos(2w) = cos^2(w) - sin^2(w)

hint: cos(2w) = cos^2(w) - sin^2(w)

Let f : |R -> |R be defined f(x) = x * sqrt(1+x^2).

Find a differentiable function F : |R -> |R such that

/\x:-|R F'(x) = f(x).

Find a differentiable function F : |R -> |R such that

/\x:-|R F'(x) = f(x).

F(x) = sqrt(1+x^2) * (1+x^2) * 1/3

hint: Integral( 0, A, f(x), dx ) = ???

u(x) = sqrt(1+x^2)

hint: Integral( 0, A, f(x), dx ) = ???

u(x) = sqrt(1+x^2)

/\a:-|R Integral( 0, a, sqrt(a^2 - x^2), dx ) = ???

/\a:-|R Integral( 0, a, sqrt(a^2 - x^2), dx ) = a*|a|*pi/4

Notice that this integral calculates one fourth of the area of a circle of radius a.

Notice that this integral calculates one fourth of the area of a circle of radius a.

a:-|C

lim(z->0) sin(az)/z = ???

lim(z->0) sin(az)/z = ???

/\a:-|C lim(z->0) sin(az)/z = a

lim(x->0+) log(x^x) = ???

lim(x->0+) log(x^x) = 0

hint: Use L'Hospital's Rule for ln(x)/(1/x)

hint: Use L'Hospital's Rule for ln(x)/(1/x)

lim(x->0+) x^x = ???

lim(x->0+) x^x = 1

hint: lim(x->0+) log(x^x) = 0, and then exp is continuous at 0

hint: lim(x->0+) log(x^x) = 0, and then exp is continuous at 0

lim((x,y)->(0,0)) x*y*log(x^2 + y^2) = ???

lim((x,y)->(0,0)) x*y*log(x^2 + y^2) = 0

hint: x=r*cos(a), y=r*sin(a)

hint: x=r*cos(a), y=r*sin(a)

+(n=1 to n=oo) 1/(n^2 + n) = ???

+(n=1 to n=oo) 1/(n^2 + n) = 1

hint: 1/(n^2 + n) = 1/n - 1/(n+1)

hint: 1/(n^2 + n) = 1/n - 1/(n+1)

inf { M:-|R : /\x:-|R |sin(x)cos(x)|<M } = ???

inf { M:-|R : /\x:-|R |sin(x)cos(x)|<M } = 1/2

A c |R; f,g : A -> |R

Suppose that functions f and g are increasing.

Can we conclude that function f*g is increasing?

Suppose that functions f and g are increasing.

Can we conclude that function f*g is increasing?

NO

A = {1,2}

f(1) = g(1) = -7

f(2) = g(2) = 0

A = {1,2}

f(1) = g(1) = -7

f(2) = g(2) = 0

A c |R; f,g : A -> |R

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0.

Can we conclude that function f*g is increasing?

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0.

Can we conclude that function f*g is increasing?

NO

A = {1,2}

f(1)=10, f(2)=20

g(1)=g(2)= (-1)

f(1) = 1; f(2) = 2; g(1) = -7; g(2) = -5

1<2, f(1)<f(2), g(1)<g(2), but f(1)g(1) > f(2)g(2)

A = {1,2}

f(1)=10, f(2)=20

g(1)=g(2)= (-1)

f(1) = 1; f(2) = 2; g(1) = -7; g(2) = -5

1<2, f(1)<f(2), g(1)<g(2), but f(1)g(1) > f(2)g(2)

A c |R; f,g : A -> |R

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.

Can we conclude that function f*g is increasing?

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.

Can we conclude that function f*g is increasing?

YES

A c |R; f,g : A -> |R

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.

What can we conclude?

Suppose that functions f and g are increasing.

Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.

What can we conclude?

We can conclude that function f*g is increasing.

Let A = { (x,y) : 0<=x, 0<=y, x^2+y^2 <= 1 }.

Let f : A -> |R, be f(0,0)=0 and f(x,y) = x*y*log(x^2+y^2).

Investigate max_f(A) and min_f(A).

Let f : A -> |R, be f(0,0)=0 and f(x,y) = x*y*log(x^2+y^2).

Investigate max_f(A) and min_f(A).

First, check that f is continuous on compact A.

Hence both max and min exist.

max_f(A) = 0

min_f(A) = -1/(2e) = f( 1/sqrt(2e) , 1/sqrt(2e) )

hint: x=r*cos(a), y=r*sin(a)

g(a,r) = r^2*cos(a)*sin(a)*log(r^2)

Hence both max and min exist.

max_f(A) = 0

min_f(A) = -1/(2e) = f( 1/sqrt(2e) , 1/sqrt(2e) )

hint: x=r*cos(a), y=r*sin(a)

g(a,r) = r^2*cos(a)*sin(a)*log(r^2)

(1) lim(n->oo) x[n] = 0

(2) \/M>0 /\n:-|N M <= |y[n]|

What can we conclude from this?

(2) \/M>0 /\n:-|N M <= |y[n]|

What can we conclude from this?

lim(n->oo) x[n]/y[n] = 0

Let a > 0.

Integeral( 0, 1, 1/(a+x), dx ) = ???

Integeral( 0, 1, 1/(a+x), dx ) = ???

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Let a > 1.

Integral( 0, 1, 1/(a-x), dx ) = ???

Integral( 0, 1, 1/(a-x), dx ) = ???

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Let a > 1.

Integral( 0, 1, 1/(a^2 - x^2), dx ) = ???

Integral( 0, 1, 1/(a^2 - x^2), dx ) = ???

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

Use:

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Use:

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Integral( 0, pi/2, sin^2(x)*cos(x) / (1+cos^2(x)), dx ) = ???

hint

hint

Integral( 0, pi/2, sin2(x)cos(x) / (1+cos2(x)), dx )

= 1/sqrt(2) * log( (sqrt(2)+1) / (sqrt(2)-1) ) - 1

hint: sinx = u

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

= 1/sqrt(2) * log( (sqrt(2)+1) / (sqrt(2)-1) ) - 1

hint: sinx = u

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

(1) a : |N -> |C

(2) 0 < R <= oo

(3) f : {z:-|C : |z| < R} -> |C

(4) f(z) = +(n=0 to n=oo) a(n) * z^(n+1) / (n+1)

What can we conclude from this?

(2) 0 < R <= oo

(3) f : {z:-|C : |z| < R} -> |C

(4) f(z) = +(n=0 to n=oo) a(n) * z^(n+1) / (n+1)

What can we conclude from this?

function f is diffable and

/\|z|<R f'(z) = +(n=0 to n=oo) a(n) * z^n

page 81 in OLDTIMER

/\|z|<R f'(z) = +(n=0 to n=oo) a(n) * z^n

page 81 in OLDTIMER

A = {z:-|C : |z|<1}

Let k be an integer.

f : A -> |C

/\z:-A f(z) = log(|1+z|) + i*Arg(1+z) + k*2*pi*i

Give the power series expansion of function f about 0.

Let k be an integer.

f : A -> |C

/\z:-A f(z) = log(|1+z|) + i*Arg(1+z) + k*2*pi*i

Give the power series expansion of function f about 0.

/\z:-A f(z) = +(n=0 to n=oo) a[n] * z^n

a[0] = k*2*pi*i

a[n] = (-1)^(n+1) / n

page 82 in OLDTIMER

a[0] = k*2*pi*i

a[n] = (-1)^(n+1) / n

page 82 in OLDTIMER

Express log(2) as the sum of a series.

log(2) = +(n=1 to oo) (-1)^(n+1) / n = 1 - 1/2 + 1/3 - 1/4 +
1/5 - 1/6 + ...

page 83 in OLDTIMER

page 83 in OLDTIMER

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... = ???

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... = log(2)

page 83 in OLDTIMER

page 83 in OLDTIMER

D = {z:-|C : z:-|R => z > 0}

Let k be an integer.

log : D -> |C, log(z) = log(|z|) + i*Arg(z) + k*2*pi*i

Let a:-|C.

f : D -> |C, f(z) = exp(a*log(z))

What can we conclude about the differentiability of function f?

Let k be an integer.

log : D -> |C, log(z) = log(|z|) + i*Arg(z) + k*2*pi*i

Let a:-|C.

f : D -> |C, f(z) = exp(a*log(z))

What can we conclude about the differentiability of function f?

Notation: if b:-|C, z:-D, let z^b = exp(b*log(z)).

We can conclude that f is diffable and

/\z:-D f'(z) = a * z^(a-1)

In other words: /\b:-|C (z^b)' = b * z^(b-1).

We can conclude that f is diffable and

/\z:-D f'(z) = a * z^(a-1)

In other words: /\b:-|C (z^b)' = b * z^(b-1).

/\n:-|N 2^(2n-1) = +( k=1 to k=n ) ???

2^(2n-1) = +(k=1 to n) Newton(2n,2k-1)

page 85 in OLDTIMER

page 85 in OLDTIMER

Give the power series expansion of sin(z)sin(z) about 0.

sin(z)sin(z) = +(n=1 to n=oo) (-1)^(n+1) * 2^(2n-1) / (2n)! *
z^(2n)

page 85 in OLDTIMER

hint1: 2^(2n-1), multiplication of series

hint2: 2sin(x)cos(x)=sin(2x), differentiation of power series

page 85 in OLDTIMER

hint1: 2^(2n-1), multiplication of series

hint2: 2sin(x)cos(x)=sin(2x), differentiation of power series

Integral( 0, pi, sin^4(x), dx ) = ???

You can use: Integral( 0, pi, sin2(x), dx ) = pi/2.

You can use: Integral( 0, pi, sin2(x), dx ) = pi/2.

Integral( 0, pi, sin4(x), dx ) = 3/8 * pi

page 88 in OLDTIMER

page 88 in OLDTIMER

Integral( 0, pi, cos^4(x), dx ) = ???

Integral( 0, pi, cos^4(x), dx ) = 3/8 * pi

page 89 in OLDTIMER

page 89 in OLDTIMER

Let z be a complex number other than zero.

sin(Arg(z)) * |z| = ???

sin(Arg(z)) * |z| = ???

sin(Arg(z)) * |z| = Im(z)

hint: /\x:-|R sin(x) = Im(exp(ix))

page 90 in OLDTIMER

hint: /\x:-|R sin(x) = Im(exp(ix))

page 90 in OLDTIMER

Let z be a complex number other than zero.

cos(Arg(z)) * |z| = ???

cos(Arg(z)) * |z| = ???

cos(Arg(z)) * |z| = Re(z)

hint: /\x:-|R cos(x) = Re(exp(ix))

page 90 in OLDTIMER

hint: /\x:-|R cos(x) = Re(exp(ix))

page 90 in OLDTIMER

/\z:-|C\{0} Im(z) = |z| * sin( ??? )

/\z:-|C Im(z) = |z| * sin( Arg(z) )

page 90 in OLDTIMER

page 90 in OLDTIMER

/\z:-|C\{0} Re(z) = |z| * cos( ??? )

/\z:-|C\{0} Re(z) = |z| * cos( Arg(z) )

page 90 in OLDTIMER

page 90 in OLDTIMER

Let x>0, y>0, z = x + iy.

Prove that arctan(y/x) = Arg(z).

Prove that arctan(y/x) = Arg(z).

page 91 in OLDTIMER

D = {(x,y):-|C : x^2 + y^2 = y and x > 0}

Prove that /\z:-D |z| = sqrt(Im(z)) = ???

Prove that /\z:-D |z| = sqrt(Im(z)) = ???

/\z:-D |z| = sqrt(Im(z)) = sin(Arg(z)) = sin(arctan(y/x))

page 91 in OLDTIMER

page 91 in OLDTIMER

Let f:|R->|R be continuous.

Does there exist a diffable function F:|R->|R

such that /\x:-|R F'(x) = f(x) ?

Does there exist a diffable function F:|R->|R

such that /\x:-|R F'(x) = f(x) ?

YES.

F(x) = Integral( 0, x, f(t), dt )

page 92 in OLDTIMER

F(x) = Integral( 0, x, f(t), dt )

page 92 in OLDTIMER

Let f : |R -> |R be continuous. Let a:-|R.

Let g : |R -> |R be defined g(x) = Integral( a, x, f(t), dt ).

What can we conclude from this?

Let g : |R -> |R be defined g(x) = Integral( a, x, f(t), dt ).

What can we conclude from this?

/\x:-|R g'(x) = f(x)

page 92 in OLDTIMER

page 92 in OLDTIMER

Let a,b:-|R and a<b.

Let f : ]a,b[ -> |R be continuous and bounded.

Does there exist a diffable function g : ]a,b[ -> |R

such that g'(x) = f(x) for a<x<b ?

Let f : ]a,b[ -> |R be continuous and bounded.

Does there exist a diffable function g : ]a,b[ -> |R

such that g'(x) = f(x) for a<x<b ?

YES

page 92 in OLDTIMER

page 92 in OLDTIMER

(1) a,b:-|R, a!=b

(2) A,B :- |R, A<B

(3) g : *[a,b]* -> |R continuously diffable

(4) f : [A,B] -> continuous

(5) g(*[a,b]*) c [A,B]

What can we conclude from this?

(2) A,B :- |R, A<B

(3) g : *[a,b]* -> |R continuously diffable

(4) f : [A,B] -> continuous

(5) g(*[a,b]*) c [A,B]

What can we conclude from this?

Integral( g(a), g(b), f(x), dx ) = Integral( a, b,
f(g(x))*g'(x), dx )

page 93 in OLDTIMER

page 93 in OLDTIMER

(1) a,b :- |R, a < b

(2) g : [a,b] -> |R continuously diffable

(3) f : |R -> |R continuous

(4) g(a) = g(b)

Integral( a, b, f(g(x))*g'(x), dx ) = ???

(2) g : [a,b] -> |R continuously diffable

(3) f : |R -> |R continuous

(4) g(a) = g(b)

Integral( a, b, f(g(x))*g'(x), dx ) = ???

Integral( a, b, f(g(x))*g'(x), dx ) = 0

page 94 in OLDTIMER

page 94 in OLDTIMER

Let a[1], a[2], a[3], ..., a[n], a[n+1], ..., a[2n] be numbers.

+(k=1 to k=2n) a[k] = +(k=1 to k=n) ???

+(k=1 to k=2n) a[k] = +(k=1 to k=n) ???

+(k=1 to k=2n) a[k] = +(k=1 to k=n) a[2k-1] + a[2k]

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/F:-W F is closed and ??????????????

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/F:-W F is closed and ??????????????

/\A:-W /\E>0 \/F:-W F is closed and FcA and m(A) <= m(F) + E.

page 106 in 1st measure

page 106 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/H:-W ???????????? and m(A) <= m(H) + E.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/H:-W ???????????? and m(A) <= m(H) + E.

/\A:-W /\E>0 \/H:-W H is closed and HcA and m(A) <= m(H) + E.

page 106 in 1st measure

page 106 in 1st measure

Let K be a s-algebra in X. Let f,g : X -> |R* be K-measurable.

Can we conclude that function max(f,g) is K-measurable?

Can we conclude that function max(f,g) is K-measurable?

YES: max(f,g) is K-measurable

Take any a:-|R. Since f,g are K-measurable, {x : f(x)>a}:-K and {x : g(x)>a}:-K. Since K is an algebra, {x : f(x)>a or g(x)>a}:- K. Hence {x : max(f(x),g(x))>a}:-K. We showed that max(f,g) is K-measurable.

Take any a:-|R. Since f,g are K-measurable, {x : f(x)>a}:-K and {x : g(x)>a}:-K. Since K is an algebra, {x : f(x)>a or g(x)>a}:- K. Hence {x : max(f(x),g(x))>a}:-K. We showed that max(f,g) is K-measurable.

Let K be a s-algebra in X. Let f,g : X -> |R* be K-measurable.

Can we conclude that function min(f,g) is K-measurable?

Can we conclude that function min(f,g) is K-measurable?

YES: min(f,g) is K-measurable

Take any a:-|R. Since f,g are K-measurable, {x : f(x)<a}:-K and {x : g(x)<a}:-K. Since K is an algebra, {x : f(x)<a or g(x)<a}:- K. Hence {x : min(f(x),g(x))<a}:-K. We showed that min(f,g) is K-measurable.

Take any a:-|R. Since f,g are K-measurable, {x : f(x)<a}:-K and {x : g(x)<a}:-K. Since K is an algebra, {x : f(x)<a or g(x)<a}:- K. Hence {x : min(f(x),g(x))<a}:-K. We showed that min(f,g) is K-measurable.

(1) W is a s-algebra in X

(2) f,g : X -> |R \ {-oo,oo} are W-measurable

(3) F : |R^2 -> |R is continuous

(4) h : X -> |R , h(x) = F(f(x),g(x))

What can we conclude from this?

(2) f,g : X -> |R \ {-oo,oo} are W-measurable

(3) F : |R^2 -> |R is continuous

(4) h : X -> |R , h(x) = F(f(x),g(x))

What can we conclude from this?

h is W-measurable

page 137 in 1st measure

page 137 in 1st measure

(1) D is an open connected subset of |C

(2) g,h : D -> |C are diffable

(3) /\z:-D g'(z) = h'(z)

What can we conclude from this?

(2) g,h : D -> |C are diffable

(3) /\z:-D g'(z) = h'(z)

What can we conclude from this?

\/A:-|C /\z:-D g(z) = h(z) + A

hint:

(g-h)' = g' - h' = 0

g-h is constant

hint:

(g-h)' = g' - h' = 0

g-h is constant

Let F be a s-algebra in X.

A:-F <=> ???

A:-F <=> ???

A:-F <=> 1(A) is F-measurable

1(A) is the characteristic function of A

1(A) : X -> |R*

1(A) is the characteristic function of A

1(A) : X -> |R*

Let F be a s-algebra in X. Let A c X.

1(A) is the characteristic function of A.

1(A) is F-measurable <=> ???

1(A) is the characteristic function of A.

1(A) is F-measurable <=> ???

1(A) is F-measurable <=> A:-F

f : |R -> (-1;1) , f(x) = x / ( |x|+1 )

What is interesting about this function?

What is interesting about this function?

It shows that |R is homeomorphic with (-1;1).

It is even continuously differentiable.

It can be used to define a metric on [-oo,oo].

Then [-oo,oo] is homeorphic with [-1,1].

It is even continuously differentiable.

It can be used to define a metric on [-oo,oo].

Then [-oo,oo] is homeorphic with [-1,1].

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable.

Let A,B:-F and AnB=O.

Prove that Leb*(AuB,f,dy) = ???.

Let f:X->[0,oo] be measurable.

Let A,B:-F and AnB=O.

Prove that Leb*(AuB,f,dy) = ???.

Leb*(AuB,f,dy) = Leb*(A,f,dy) + Leb*(B,f,dy)

page 170 in 1st measure

page 170 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)>g(x)} ?

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)>g(x)} ?

{x:-X : f(x)>g(x)} :- F

hint:

{x: f(x)>g(x)} = \\//a:-|Q {x: f(x)>a} n {x: a>g(x)}

page 139 in 1st measure

hint:

{x: f(x)>g(x)} = \\//a:-|Q {x: f(x)>a} n {x: a>g(x)}

page 139 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)>=g(x)} ?

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)>=g(x)} ?

{x:-X : f(x)>=g(x)} :- F

hint:

{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}

page 139 in 1st measure

hint:

{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}

page 139 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)=g(x)} ?

Let f,g : X -> |R* be F-measurable.

What can we conclude about the set {x:-X : f(x)=g(x)} ?

{x:-X : f(x)=g(x)} :- F

hint:

{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}

page 139 in 1st measure

hint:

{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}

page 139 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are two two ways of proving that f+g is F-measurable?

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are two two ways of proving that f+g is F-measurable?

WAY 1:

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x+y (page 138 in 1st measure)

WAY 2:

(-1)*g is measurable,

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x+y (page 138 in 1st measure)

WAY 2:

(-1)*g is measurable,

Let F be a s-algebra in X.

Let f : X -> |R \ {-oo,oo} be F-measurable.

Prove that function f*f is F-measurable.

Let f : X -> |R \ {-oo,oo} be F-measurable.

Prove that function f*f is F-measurable.

Take any a:-|R. If a<0, then {x: f^2(x) >= a} :- F.

If a>=0, then {x: f^2(x) > a} = {x: |f(x)| > sqrt(a)} =

= {x: f(x) > sqrt(a) or f(x) < -sqrt(a)} =

= {x: f(x) > sqrt(a)} u {x: f(x) < -sqrt(a)} :- F

If a>=0, then {x: f^2(x) > a} = {x: |f(x)| > sqrt(a)} =

= {x: f(x) > sqrt(a) or f(x) < -sqrt(a)} =

= {x: f(x) > sqrt(a)} u {x: f(x) < -sqrt(a)} :- F

Let F be a s-algebra in X.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

Prove that f*g is F-measurable by using a trick.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

Prove that f*g is F-measurable by using a trick.

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

Use the facts that f+g, f-g are F-measurable,

and that if h:X->|R is measurable then so is h^2.

page 138 in 1st measure

Use the facts that f+g, f-g are F-measurable,

and that if h:X->|R is measurable then so is h^2.

page 138 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are the two ways of proving that f*g is F-measurable?

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are the two ways of proving that f*g is F-measurable?

WAY 1:

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x*y (page 138 in 1st measure)

WAY 2:

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x*y (page 138 in 1st measure)

WAY 2:

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

Let W c P(X). Let U c X.

Suppose that /\(a:-U) \/(B:-W) a:-B and B c U.

Can we conclude that U is W-open?

Suppose that /\(a:-U) \/(B:-W) a:-B and B c U.

Can we conclude that U is W-open?

YES.

page 78,79 in gen top

page 78,79 in gen top

f : [0,oo[ -> |R , f(x) = 1/(1+sqrt(x))

Find a diffable function F : [0,oo[ -> |R

such that /\x>=0 F'(x) = f(x).

Find a diffable function F : [0,oo[ -> |R

such that /\x>=0 F'(x) = f(x).

F(x) = 2 * ( sqrt(x) - log(1+sqrt(x) )

page 116 in OLDTIMER

page 116 in OLDTIMER

(1) 0 < a <= oo

(2) F,f : [0,a) -> |R

(3) /\0<x<a F'(x) = f(x)

(4) F and f are both continuous at 0

(5) F(0) = 0

What can we conclude from this?

(2) F,f : [0,a) -> |R

(3) /\0<x<a F'(x) = f(x)

(4) F and f are both continuous at 0

(5) F(0) = 0

What can we conclude from this?

F'(0) = f(0)

page 117 in OLDTIMER

page 117 in OLDTIMER

Prove that |R is homeomorphic with the bounded open interval (-
1;1).

Do not use trigonometric functions.

Do not use trigonometric functions.

f : |R -> (-1;1) , f(x) = x / ( |x|+1 )

f is 1-1, onto, continuously diffable

f is 1-1, onto, continuously diffable

Let H be a s-algebra in X.

Let f : X -> |R \ {-oo,oo} be H-measurable.

Let F : |R -> |R be a Borel function.

Let g : X -> |R , g(x) = F(f(x)).

What can we conclude?

Let f : X -> |R \ {-oo,oo} be H-measurable.

Let F : |R -> |R be a Borel function.

Let g : X -> |R , g(x) = F(f(x)).

What can we conclude?

function g is H-measurable

Take any a:-|R. Let G={x:-|R : F(x)>a}. G is Borel in |R.

Function f is H-measurable, hence f-1(G) :- H.

{x:-X : g(x)>a} = {x:-X : F(f(x))>a} = {x:-X : f(x) :- G} = f- 1(G) :- H

Take any a:-|R. Let G={x:-|R : F(x)>a}. G is Borel in |R.

Function f is H-measurable, hence f-1(G) :- H.

{x:-X : g(x)>a} = {x:-X : F(f(x))>a} = {x:-X : f(x) :- G} = f- 1(G) :- H

In the context of measure theory, what does it mean that a
function X->|R* is simple?

(0) f : X -> |R* is simple

iff

(1) /\x:-X f(x):-|R

(2) f(X) is finite

===================

f(X) = { a[1], a[2], ..., a[n] } c |R

iff

(1) /\x:-X f(x):-|R

(2) f(X) is finite

===================

f(X) = { a[1], a[2], ..., a[n] } c |R

(1) f : X -> |R*.

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] c X , none of them is empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] c X , none of them is empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

f(X) = { a[1] , ... , a[n] } c |R

f is a simple function

page 105 in 2nd measure

page 140 in 1st measure

f is a simple function

page 105 in 2nd measure

page 140 in 1st measure

Let F be a s-algebra in X.

(1) f : X -> |R*

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(1) f : X -> |R*

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

f is simple and F-measurable

page 106 in 2nd measure

page 141 in 1st measure

page 106 in 2nd measure

page 141 in 1st measure

Let F be a s-algebra in X.

Suppose that f : X -> |R is simple and F-measurable.

How can function f be conveniently expressed?

Suppose that f : X -> |R is simple and F-measurable.

How can function f be conveniently expressed?

There exist:

(1) n:-|N

(2) a[1], ... , a[n] :- |R ; all different ; n numbers altogether

(3) A[1], ... , A[n] :- F ; disjoint and non-empty

such that

(4) X = \\*//(k=1 to n) A[k]

(5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(1) n:-|N

(2) a[1], ... , a[n] :- |R ; all different ; n numbers altogether

(3) A[1], ... , A[n] :- F ; disjoint and non-empty

such that

(4) X = \\*//(k=1 to n) A[k]

(5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that A[n] is increasing?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that A[n] is increasing?

YES

page 99 in 2nd measure

page 99 in 2nd measure

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

What can we conclude from this?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

What can we conclude from this?

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

page 99 in 2nd measure

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

page 99 in 2nd measure

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that { A[k]\A[k-1] }(k:-|N) is a disjoint sequence?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that { A[k]\A[k-1] }(k:-|N) is a disjoint sequence?

YES. page 99 in 2nd measure

We can even conclude much more:

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

We can even conclude much more:

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

{A[k]\A[k-1]}(k:-|N) is a disjoint sequence.

Can we conclude that A[n] is increasing?

{A[k]\A[k-1]}(k:-|N) is a disjoint sequence.

Can we conclude that A[n] is increasing?

NO

A[n] = {n}

A[n] = {n}

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

\\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Can we conclude that A[n] is increasing?

\\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Can we conclude that A[n] is increasing?

NO

A[n] = {n}

A[n] = {n}

If a :- [-oo,oo], what does (a+) mean?

(a+) = max(0 , a)

page 103 in 2nd measure

page 103 in 2nd measure

If a :- [-oo,oo], what does (a-) mean?

(a-) = max(0 , -a)

page 103

page 103

In the context of measure theory,

what does it mean: f~g ?

what does it mean: f~g ?

In this context:

(1) F is a s-algebra in X

(2) y : F -> [0,oo] is countably additive

(3) f,g : X -> |R* are F-measurable

f~g <=> y( {x:-X : f(x)!=g(x)} ) = 0

(1) F is a s-algebra in X

(2) y : F -> [0,oo] is countably additive

(3) f,g : X -> |R* are F-measurable

f~g <=> y( {x:-X : f(x)!=g(x)} ) = 0

What does it mean that (X,F,y) is a measure space in MRW math
materials?

(1) X is a non-empty set

(2) F is a s-algebra in X

(3) y : F -> [0,oo]

(4) y is countably additive and y(O)=0

page 102 in 2nd measure

(2) F is a s-algebra in X

(3) y : F -> [0,oo]

(4) y is countably additive and y(O)=0

page 102 in 2nd measure

If f : X -> |R*, what does (f+) mean?

(f+) : X -> [0,oo]

/\x:-X (f+)(x) = max( 0, f(x) )

page 168 in 1st measure

page 102 in 2nd measure

/\x:-X (f+)(x) = max( 0, f(x) )

page 168 in 1st measure

page 102 in 2nd measure

If f : X -> |R*, what does (f-) mean?

(f-) : X -> [0,oo]

/\x:-X (f-)(x) = max( 0, -f(x) )

page 168 in 1st measure

page 102 in 2nd measure

/\x:-X (f-)(x) = max( 0, -f(x) )

page 168 in 1st measure

page 102 in 2nd measure

Let a,b :- |R*.

Suppose that (a+) = (b+).

What can we conclude?

Suppose that (a+) = (b+).

What can we conclude?

( a<=0 and b<=0 ) or a=b

page 104 in 2nd measure

page 104 in 2nd measure

Let a,b :- |R*.

Suppose that (a-) = (b-).

What can we conclude?

Suppose that (a-) = (b-).

What can we conclude?

(a>=0 and b>=0) or a=b

page 104 in 2nd measure

page 104 in 2nd measure

Let a,b :- |R*.

Suppose that (a+)=(b+) and (a-)=(b-).

Is this enough to conclude that a=b ?

Suppose that (a+)=(b+) and (a-)=(b-).

Is this enough to conclude that a=b ?

YES

page 104 in 2nd measure

page 104 in 2nd measure

Let a,b :- |R*.

Suppose that (a+)!=(b+) or (a-)!=(b-).

What can we conclude from this?

Suppose that (a+)!=(b+) or (a-)!=(b-).

What can we conclude from this?

a != b

page 104 in 2nd measure

page 104 in 2nd measure

Let (X,F,y) be a measure space.

Let f,g : X -> |R* be measurable.

Suppose that (f+)~(g+) and (f-)~(g-).

Is this enough to conclude that f~g ?

Let f,g : X -> |R* be measurable.

Suppose that (f+)~(g+) and (f-)~(g-).

Is this enough to conclude that f~g ?

YES

page 105 in 2nd measure

page 105 in 2nd measure

Let E be the collection of all elementary subsets of |R^p.

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

(1) Ac|R^p ; /\(n:-|N) A[n]:-MF(y)

(2) lim(n->oo) y*(A[n]+A)=0

Let y : E -> [0,oo) be additive and regular.

Let y* be the corresponding outer measure on |R^p.

Let MF(y) denote the collection of all finitely y-measurable subsets of |R^p.

(1) Ac|R^p ; /\(n:-|N) A[n]:-MF(y)

(2) lim(n->oo) y*(A[n]+A)=0

A:-MF(y)

page 21 in 2nd measure

page 21 in 2nd measure

(1) f : X -> |R*.

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] c X , some of them may be empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] c X , some of them may be empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

f(X) c { a[1] , ... , a[n] } c |R

f is a simple function

page 248 in 2nd measure

f is a simple function

page 248 in 2nd measure

Let F be a s-algebra in X.

(1) f : X -> |R*.

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F , some of them may be empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(1) f : X -> |R*.

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F , some of them may be empty

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

f is simple and F-measurable

page 109 in 2nd measure

page 109 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> [0,oo] be F-measurable.

In this setting, how can we interestingly approximate function f ?

(In this item, prove the existence of this interesting approximation.)

Let f : X -> [0,oo] be F-measurable.

In this setting, how can we interestingly approximate function f ?

(In this item, prove the existence of this interesting approximation.)

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple and measurable

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 113 in 2nd measure

(1) /\n:-|N s[n] : X -> [0,oo) is simple and measurable

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 113 in 2nd measure

f : X -> [0,oo]

In this broad setting, how can we interestingly approximate function f ?

(In this item, do not prove the existence of this interesting approximation.)

In this broad setting, how can we interestingly approximate function f ?

(In this item, do not prove the existence of this interesting approximation.)

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 110 in 2nd measure

(1) /\n:-|N s[n] : X -> [0,oo) is simple

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 110 in 2nd measure

Let f : X -> [0,M] , M < oo.

In this broad setting, how can we interestingly approximate function f ?

In this broad setting, how can we interestingly approximate function f ?

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple and bounded

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

(4) sequence s[n] converges uniformly to function f

page 113 in 2nd measure

======================

(1) /\n:-|N s[n] : X -> [0,oo) is simple and bounded

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

(4) sequence s[n] converges uniformly to function f

page 113 in 2nd measure

======================

lim(x,y)->(0,0) y*(x-sin(x)) / (x^4 + y^2) = ???

lim(x,y)->(0,0) y*(x-sin(x)) / (x^4 + y^2) = 0

hint:

x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6

|xy/(x^2+y^2)|<=1/2

page 100 in OLDTIMER

hint:

x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6

|xy/(x^2+y^2)|<=1/2

page 100 in OLDTIMER

Let a : |N -> |C.

Let z be a complex number.

Suppose that series a[n]*|z|^n converges.

Can we conclude that series a[n]*z^n converges?

Let z be a complex number.

Suppose that series a[n]*|z|^n converges.

Can we conclude that series a[n]*z^n converges?

NO

(1) Let a[n] = 1/n * (-1)^n. Let z = (-1).

(2) Consider series (-1)^n / n * z^(3n). Let z = -1.

(1) Let a[n] = 1/n * (-1)^n. Let z = (-1).

(2) Consider series (-1)^n / n * z^(3n). Let z = -1.

(1) W is a ring of sets in X

(2) y : W -> [0,oo) is additive

(3) A[1] , ... , A[n] :- W (disjoint sets)

(5) B[1] , ... , B[m] :- W (disjoint sets)

(6) X = U(k=1 to n)_A[k] = U(j=1 to m)_B[j]

(7) a[1], ... , a[n] , b[1] , ... , b[m] are positive real numbers

(2) y : W -> [0,oo) is additive

(3) A[1] , ... , A[n] :- W (disjoint sets)

(5) B[1] , ... , B[m] :- W (disjoint sets)

(6) X = U(k=1 to n)_A[k] = U(j=1 to m)_B[j]

(7) a[1], ... , a[n] , b[1] , ... , b[m] are positive real numbers

+(k=1 to n) a[k]*y(A[k]) = +(j=1 to m) b[j]*y(B[j])

page 147 in 1st measure

page 147 in 1st measure

Let W be a s-algebra in X.

Let y : W -> [0,oo] be additive, y(O) = 0.

Let f : X -> [0,oo) be a simple and W-measurable function.

Define the Lebesgue integral of function f with respect to y.

LebS(X,f,dy) = ???

Let y : W -> [0,oo] be additive, y(O) = 0.

Let f : X -> [0,oo) be a simple and W-measurable function.

Define the Lebesgue integral of function f with respect to y.

LebS(X,f,dy) = ???

Since f is simple and W-measurable, there exist:

(1) disjoint sets A[1], A[2], ..., A[n] :- W, with X = U(k=1 to n) A[k]

(2) a[1], a[2], ..., a[n] :- [0,oo)

such that (5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])

Define LebS(X,f,dy) := +(k=1 to n) a[k]*y(A[k]).

(1) disjoint sets A[1], A[2], ..., A[n] :- W, with X = U(k=1 to n) A[k]

(2) a[1], a[2], ..., a[n] :- [0,oo)

such that (5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])

Define LebS(X,f,dy) := +(k=1 to n) a[k]*y(A[k]).

Let W be a s-algebra in X.

Let y : W -> |R* be additive.

Let A :- W.

LebS(X,1(A),dy) = ???

Let y : W -> |R* be additive.

Let A :- W.

LebS(X,1(A),dy) = ???

LebS(X,1(A),dy) = y(A)

page 149 in 1st measure

page 149 in 1st measure

Let W be a s-algebra in X.

Let f,g : X -> |R be simple and W-measurable.

Prove that there exist:

(1) E[1], ... , E[n] :- W (disjoint, having union X)

(2) a[1], ... , a[n], b[1], ... , b[n] :- |R

such that

(3) f = +(k=1 to n) a[k]*1(E[k])

Let f,g : X -> |R be simple and W-measurable.

Prove that there exist:

(1) E[1], ... , E[n] :- W (disjoint, having union X)

(2) a[1], ... , a[n], b[1], ... , b[n] :- |R

such that

(3) f = +(k=1 to n) a[k]*1(E[k])

page 149 in 1st measure

Notice the technical difficulty here.

One day I want to go deeper into this theorem.

MRW 2001.02.24; 2001.06.02

Notice the technical difficulty here.

One day I want to go deeper into this theorem.

MRW 2001.02.24; 2001.06.02

Let f,g : X -> |R be simple functions.

Can we conclude that f+g is also a simple function?

Can we conclude that f+g is also a simple function?

YES

Use the theorem on page 149 in 1st measure.

Use the theorem on page 149 in 1st measure.

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo.

Prove that LebS(X,af,dy) = a*LebS(X,f,dy).

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo.

Prove that LebS(X,af,dy) = a*LebS(X,f,dy).

page 151 in 1st measure

Notice that y,f,a are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

Notice that y,f,a are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let E :- W and let f : X -> [0,oo) be simple and measurable.

Define the Lebesgue integral of function f over set E with respect to y.

LebS(E,f,dy) = ???

Let E :- W and let f : X -> [0,oo) be simple and measurable.

Define the Lebesgue integral of function f over set E with respect to y.

LebS(E,f,dy) = ???

LebS(E,f,dy) = LebS(X,1(E)*f,dy)

Notice that here we have to check that 1(E)*f is simple and measurable.

page 152 in 1st measure

Notice that here we have to check that 1(E)*f is simple and measurable.

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Prove that LebS(X,f,dy) + LebS(X,g,dy) = LebS(X,f+g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Prove that LebS(X,f,dy) + LebS(X,g,dy) = LebS(X,f+g,dy).

page 151 in 1st measure

Notice that y,f,g are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

We also want the formula /\a,b,c:-[0,oo] a*(b+c) = (a*b)+(a*c).

Also notice that here we have to check if f+g is simple.

Notice that y,f,g are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

We also want the formula /\a,b,c:-[0,oo] a*(b+c) = (a*b)+(a*c).

Also notice that here we have to check if f+g is simple.

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Suppose that f <= g.

Prove that LebS(X,f,dy) <= LebS(X,g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Suppose that f <= g.

Prove that LebS(X,f,dy) <= LebS(X,g,dy).

page 151 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Suppose that /\x:-E f(x)<=g(x).

Prove that LebS(E,f,dy) <= LebS(E,g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Suppose that /\x:-E f(x)<=g(x).

Prove that LebS(E,f,dy) <= LebS(E,g,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Prove that LebS(E,f,dy) + LebS(E,g,dy) = LebS(E,f+g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Prove that LebS(E,f,dy) + LebS(E,g,dy) = LebS(E,f+g,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo. Let E :- W.

Prove that LebS(E,af,dy) = a*LebS(E,f,dy).

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo. Let E :- W.

Prove that LebS(E,af,dy) = a*LebS(E,f,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is additive?

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is additive?

Y is additive, Y(O)=0

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 119 in 1st measure

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 119 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be countably
additive.

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is countably additive?

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is countably additive?

Y is countably additive, Y(O)=0

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 155 in 1st measure

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 155 in 1st measure

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a:-|R /\n:-|N n*[a] <= ???

Prove that /\a:-|R /\n:-|N n*[a] <= ???

/\a:-|R /\n:-|N n*[a] <= [n*a]

page 119 in OLDTIMER

page 119 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a:-|R /\n:-|N ??? <= [n*a]/n <= a

Prove that /\a:-|R /\n:-|N ??? <= [n*a]/n <= a

/\a:-|R /\n:-|N [a] <= [n*a]/n <= a

page 119 in OLDTIMER

page 119 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\n:-|N /\a(1),...,a(n):-|R +(k=1 to n) [a(k)] <= ???.

Prove that /\n:-|N /\a(1),...,a(n):-|R +(k=1 to n) [a(k)] <= ???.

/\n:-|N /\a(1),...,a(n):-|R +(k=1 to n) [a(k)] <= [ +(k=1
to n) a(k) ]

page 120 in OLDTIMER

page 120 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a,b:-|R ??? <= [a + b].

Prove that /\a,b:-|R ??? <= [a + b].

/\a,b:-|R [a] + [b] <= [a + b]

page 120 in OLDTIMER

page 120 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\n:-|N /\a(1),...,a(n):-|R ??? <= [ +(k=1 to n) a(k) ].

Prove that /\n:-|N /\a(1),...,a(n):-|R ??? <= [ +(k=1 to n) a(k) ].

/\n:-|N /\a(1),...,a(n):-|R +(k=1 to n) [a(k)] <= [ +(k=1
to n) a(k) ]

page 120 in OLDTIMER

page 120 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a:-|R /\n:-|N [a] <= ??? <= a

Prove that /\a:-|R /\n:-|N [a] <= ??? <= a

/\a:-|R /\n:-|N [a] <= [n*a]/n <= a

page 119 in OLDTIMER

page 119 in OLDTIMER

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Prove that /\a:-|R /\n:-|N 0 <= a - [n*a]/n < ???.

Prove that /\a:-|R /\n:-|N 0 <= a - [n*a]/n < ???.

/\a:-|R /\n:-|N 0 <= a - [n*a]/n < 1/n

page 121 in OLDTIMER

page 121 in OLDTIMER

Prove that every real number is the limit of a sequence of
rational numbers.

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

If a:-|R, let x[n] = [n*a]/n. Clearly, /\n:-|N x[n]:-|Q.

Recall that /\a:-|R /\n:-|N 0 <= a - [n*a]/n < 1/n,

and conclude that lim(n->oo) x[n] = a.

If a:-|R, let x[n] = [n*a]/n. Clearly, /\n:-|N x[n]:-|Q.

Recall that /\a:-|R /\n:-|N 0 <= a - [n*a]/n < 1/n,

and conclude that lim(n->oo) x[n] = a.

Let k and p be integers, k>=1, p>=0.

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

What is the radius of convergence of the power series a[n]*z^(kn+p) ?

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

What is the radius of convergence of the power series a[n]*z^(kn+p) ?

( 1/( lim_sup(n->oo) |a[n]|^(1/n) ) )^(1/k)

page 118 in OLDTIMER

page 118 in OLDTIMER

Let f : ]0,oo[ x ]0,oo[ -> |R be the function satisfying:

/\x>0,y>0 sin(arctan(y/x)) = f(x,y).

Find a formula for f(x,y) that does not involve trig functions.

/\x>0,y>0 sin(arctan(y/x)) = f(x,y).

Find a formula for f(x,y) that does not involve trig functions.

f(x,y) = y / (x^2 + y^2)

page 95 in OLDTIMER

page 95 in OLDTIMER

Let f : ]0,oo[ x ]0,oo[ -> |R be the function satisfying:

/\(x>0,y>0) cos(arctan(y/x)) = f(x,y).

Find a formula for f(x,y) that does not involve trig functions.

/\(x>0,y>0) cos(arctan(y/x)) = f(x,y).

Find a formula for f(x,y) that does not involve trig functions.

f(x,y) = x / (x^2 + y^2)

page 95 in OLDTIMER

page 95 in OLDTIMER

lim(x,y)->(0,0) (x^3 * y) / ((x^2 - y^2)^2 + y^2) = ???

lim(x,y)->(0,0) (x^3 * y) / ((x^2 - y^2)^2 + y^2) = 0

page 99 in OLDTIMER

page 99 in OLDTIMER

lim(x,y)->(0,0) (x-sin(x))*y / ((x^2-y^2)^2 + y^2) = ???

lim(x,y)->(0,0) (x-sin(x))*y / ((x^2 - y^2)^2 + y^2) = 0

hint: x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6.

further hint: |ab/(a^2+b^2)|<=K.

page 100 in OLDTIMER

hint: x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6.

further hint: |ab/(a^2+b^2)|<=K.

page 100 in OLDTIMER

Let Y : P(X) -> |R* satisfy:

(1) Y(O)=0.

(2) AcBcX => Y(A)<=Y(B).

(3) Y(AuB) <= Y(A)+Y(B).

Let M be a ring in X.

Let F = { EcX : /\a>0 \/A:-M Y(E+A) < a }.

In what sense can set F be said to be closed?

(1) Y(O)=0.

(2) AcBcX => Y(A)<=Y(B).

(3) Y(AuB) <= Y(A)+Y(B).

Let M be a ring in X.

Let F = { EcX : /\a>0 \/A:-M Y(E+A) < a }.

In what sense can set F be said to be closed?

Define d:P(X)xP(X)->|R*, d(A,B)=Y(A+B).

Notice that d is a oo-pseudometric.

Now, if (1) /\n:-|N A[n]:-F and (2) lim(n->oo) d(A[n],A) = 0, then (3) A:-F.

See pages 21,64 in 2nd measure for relevant theorems.

Notice that d is a oo-pseudometric.

Now, if (1) /\n:-|N A[n]:-F and (2) lim(n->oo) d(A[n],A) = 0, then (3) A:-F.

See pages 21,64 in 2nd measure for relevant theorems.

Find the power series expansion of cos^2(z) about 0.

cos^2(z) = +(n=0 to oo) a[n]*z^(2n)

a[0] = 1

a[n] = (-1)^n * 2^(2n-1) / (2n)!

page 101 in OLDTIMER

a[0] = 1

a[n] = (-1)^n * 2^(2n-1) / (2n)!

page 101 in OLDTIMER

Let 0 < R <= oo.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Prove that f is infinitely diffable and derive the formula for its derivatives.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Prove that f is infinitely diffable and derive the formula for its derivatives.

/\k:-|N /\|z|<R f|k|(z) = +(n=k to oo) c[n]* n!/(n-k)! * z^(n-k)

page 102 in OLDTIMER

/\k:-|N /\|z|<R f|k|(z) = +(n=0 to oo) c[n+k]* (n+k)!/(n!) * z^n

page 102 in OLDTIMER

/\k:-|N /\|z|<R f|k|(z) = +(n=0 to oo) c[n+k]* (n+k)!/(n!) * z^n

Let 0 < R <= oo.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^n = 0.

What can we conclude from this?

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^n = 0.

What can we conclude from this?

0 = c[0] = c[1] = c[2] = ... = c[n] = c[n+1] = ...

page 103 in OLDTIMER

page 103 in OLDTIMER

Let 0 < R <= oo.

Let a[n],b[n] be sequences of complex numbers, indexed from zero.

Suppose that /\|z|<R +(n=0 to oo) a[n]*z^n = +(n=0 to oo) b[n]*z^n.

In the line above we understand that both series converge.

What can we conclude from this?

Let a[n],b[n] be sequences of complex numbers, indexed from zero.

Suppose that /\|z|<R +(n=0 to oo) a[n]*z^n = +(n=0 to oo) b[n]*z^n.

In the line above we understand that both series converge.

What can we conclude from this?

a[n] = b[n] for all n=0,1,2,3,4,...

page 104 in OLDTIMER

page 104 in OLDTIMER

Prove that /\n:-|N 2^(2n) = +(k=0 to n) Newton(2n+1,2k).

page 106 in OLDTIMER

Interestingly, this formula can be derived by writing the power series expansion of cos(z)sin(z) in two different ways. One way involves the formula for multiplying series.

See page 104 in OLDTIMER.

Interestingly, this formula can be derived by writing the power series expansion of cos(z)sin(z) in two different ways. One way involves the formula for multiplying series.

See page 104 in OLDTIMER.

Prove that /\n:-|N 2^(2n) = +(k=0 to n) Newton(2n+1,2k+1).

page 107 in OLDTIMER

Interestingly, this formula can be derived by writing the power series expansion of cos(z)sin(z) in two different ways. One way involves the formula for multiplying series.

See page 104 in OLDTIMER.

Interestingly, this formula can be derived by writing the power series expansion of cos(z)sin(z) in two different ways. One way involves the formula for multiplying series.

See page 104 in OLDTIMER.

Prove that /\n:-|N 2^(2n-1) = +(k=0 to n) Newton(2n,2k)

page 107 in OLDTIMER

Let 0 < R <= oo.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Suppose that /\|z|<R f(-z)=f(z).

What can we conclude from this?

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Suppose that /\|z|<R f(-z)=f(z).

What can we conclude from this?

/\n:-|N a[2n-1] = 0

page 108 in OLDTIMER

page 108 in OLDTIMER

Let 0 < R <= oo.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Suppose that /\|z|<R f(-z) = -f(z).

What can we conclude from this?

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) c[n]*z^n.

Suppose that /\|z|<R f(-z) = -f(z).

What can we conclude from this?

/\n:-|N c[2n] = 0

page 108 in OLDTIMER

page 108 in OLDTIMER

Let 0 < R <= oo. Let k be an integer, k>=1.

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) a[n]*z^(kn) converges.

Prove that /\|z|<??? +(n=0 to oo) a[n]*z^n converges.

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) a[n]*z^(kn) converges.

Prove that /\|z|<??? +(n=0 to oo) a[n]*z^n converges.

/\|z|<R^k +(n=0 to oo) a[n]*z^n converges

page 111 in OLDTIMER

page 111 in OLDTIMER

Let 0 < R <= oo. Let k be an integer, k>=1.

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) a[n]*z^(kn).

What can we conclude about the differentiability of function f?

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) a[n]*z^(kn).

What can we conclude about the differentiability of function f?

/\|z|<R f'(z) = +(n=1 to oo) a[n]*k*n*z^(n*k-1)

page 112 in OLDTIMER

page 112 in OLDTIMER

Let 0 < R <= oo. Let k,p be integers, k>=1, p>=1.

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) a[n]*z^(kn+p).

What can we conclude about the differentiability of function f?

Let { a[n] }(n=0 to oo) be a sequence of complex numbers.

Let f : {z:-|C : |z|<R} -> |C be defined f(z) = +(n=0 to oo) a[n]*z^(kn+p).

What can we conclude about the differentiability of function f?

/\|z|<R f'(z) = +(n=0 to oo) a[n]*(kn+p)*z^(kn+p-1)

page 113 in OLDTIMER

page 113 in OLDTIMER

Let 0 < R <= oo. Let k be an integer, k>=1.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^(k*n) = 0.

What can we conclude from this?

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^(k*n) = 0.

What can we conclude from this?

0 = c[0] = c[1] = c[2] = ... = c[n] = c[n+1] = ...

page 114 in OLDTIMER

page 114 in OLDTIMER

Let 0 < R <= oo. Let k,p be integers, k>=1, p>=1.

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^(k*n+p) = 0.

What can we conclude from this?

Let { c[n] }(n=0 to oo) be a sequence of complex numbers.

Suppose that /\|z|<R +(n=0 to oo) c[n]*z^(k*n+p) = 0.

What can we conclude from this?

0 = c[0] = c[1] = c[2] = ... = c[n] = c[n+1] = ...

page 115 in OLDTIMER

page 115 in OLDTIMER

(1) A is a connected subset of |R

(2) /\n:-|N f[n] : A -> |R is diffable on A

(3) sequence f[n] converges uniformly to a diffable function f:A->|R

(4) sequence f[n]' converges pointwise to a function g:A->|R

Is it possible that f' != g ?

(2) /\n:-|N f[n] : A -> |R is diffable on A

(3) sequence f[n] converges uniformly to a diffable function f:A->|R

(4) sequence f[n]' converges pointwise to a function g:A->|R

Is it possible that f' != g ?

YES.

A = (0,1]

f[n](x) = (x^n)/n

(see page 145 in OLDTIMER for a relevant theorem)

A = (0,1]

f[n](x) = (x^n)/n

(see page 145 in OLDTIMER for a relevant theorem)

(1) A is an open connected subset of |R

(2) /\n:-|N f[n] : A -> |R is diffable on A

(3) sequence f[n] converges uniformly to a diffable function f:A->|R

(4) sequence f[n]' converges pointwise to a function g:A->|R

Is it possible that f != g ?

(2) /\n:-|N f[n] : A -> |R is diffable on A

(3) sequence f[n] converges uniformly to a diffable function f:A->|R

(4) sequence f[n]' converges pointwise to a function g:A->|R

Is it possible that f != g ?

YES.

A = (-1;1)

f[n](x) = 1/n * ( 1- (1-|x|)^n ) * sgn(x)

A = (-1;1)

f[n](x) = 1/n * ( 1- (1-|x|)^n ) * sgn(x)

Let y* be the outer Lebesgue measure on |R^p.

Let Ac|R^p. Suppose /\R>0 \/K:-M(y) [ AcK and y*(K\A) < R ].

What can we conclude about set A ?

Let Ac|R^p. Suppose /\R>0 \/K:-M(y) [ AcK and y*(K\A) < R ].

What can we conclude about set A ?

A:-M(y)

page 124 in 2nd measure

page 22 in 2nd measure

page 124 in 2nd measure

page 22 in 2nd measure

Let (X,F,y) be a measure space. Let E:-F. Let f:X->[0,oo].

(1) /\n:-|N f[n]:X->[0,oo) simple and measurable

(2) /\x:-E /\n:-|N f[n](x)<=f[n+1](x)

(3) /\x:-E lim(n->oo) f[n](x) = f(x)

Prove that: lim(n->oo) LebS(E,f[n],dy) = ???.

(1) /\n:-|N f[n]:X->[0,oo) simple and measurable

(2) /\x:-E /\n:-|N f[n](x)<=f[n+1](x)

(3) /\x:-E lim(n->oo) f[n](x) = f(x)

Prove that: lim(n->oo) LebS(E,f[n],dy) = ???.

lim(n->oo) Leb(E,f[n],dy) = sup{ LebS(E,s,dy) : s:-SM, 0<=s<=f
on E }

s:-SM <=> s:X->[0,oo) is simple and measurable

page 163 in 1st measure

s:-SM <=> s:X->[0,oo) is simple and measurable

page 163 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Let f,g:X->[0,oo] be simple and measurable.

Suppose that /\x:-E f(x)=g(x).

Can we conclude that LebS(E,f,dy)=LebS(E,g,dy) ?

Let f,g:X->[0,oo] be simple and measurable.

Suppose that /\x:-E f(x)=g(x).

Can we conclude that LebS(E,f,dy)=LebS(E,g,dy) ?

Yes.

page 166 in 1st measure

page 166 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Let f:X->[0,oo] be measurable.

How is the Lebesgue integral of f defined?

Leb*(E,f,dy) = ???

Let f:X->[0,oo] be measurable.

How is the Lebesgue integral of f defined?

Leb*(E,f,dy) = ???

Leb*(E,f,dy) = sup{ LebS(E,s,dy) : s:-SM, 0<=s<=f on E }

s:-SM <=> s:X->[0,oo) is simple and measurable

page 168 in 1st measure

s:-SM <=> s:X->[0,oo) is simple and measurable

page 168 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Let f:X->|R* be measurable.

What does it mean that f is Lebesgue-integrable on set E?

Let f:X->|R* be measurable.

What does it mean that f is Lebesgue-integrable on set E?

(1) Leb*(E,(f+),dy) < oo

(2) Leb*(E,(f-),dy) < oo

(f+):X->[0,oo], /\x:-X (f+)(x) = max(0,f(x))

(f-):X->[0,oo], /\x:-X (f-)(x) = max(0,-f(x))

page 168 in 1st measure

(2) Leb*(E,(f-),dy) < oo

(f+):X->[0,oo], /\x:-X (f+)(x) = max(0,f(x))

(f-):X->[0,oo], /\x:-X (f-)(x) = max(0,-f(x))

page 168 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Prove that LebS(E,0,dy) = 0.

Prove that LebS(E,0,dy) = 0.

Notice that 0*oo=0 is essential here.

page 168 in 1st measure

page 168 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Let f:X->|R* be measurable.

Suppose that f is Lebesgue-integrable on set E.

How do we define the Lebesgue integral of function f on set E?

Let f:X->|R* be measurable.

Suppose that f is Lebesgue-integrable on set E.

How do we define the Lebesgue integral of function f on set E?

Since f is Lebesgue-integrable on E, we have:

(1) Leb*(E,(f+),dy) < oo,

(2) Leb*(E,(f-),dy) < oo.

Define Leb(E,f,dy) := Leb*(E,(f+),dy) - Leb*(E,(f-),dy).

page 169 in 1st measure

(1) Leb*(E,(f+),dy) < oo,

(2) Leb*(E,(f-),dy) < oo.

Define Leb(E,f,dy) := Leb*(E,(f+),dy) - Leb*(E,(f-),dy).

page 169 in 1st measure

Let (X,F,y) be a measure space. Let E:-F.

Let f:X->|R* be measurable.

Consider this symbol: Leb(E,f,dy).

What does it mean, when does it make sense?

Let f:X->|R* be measurable.

Consider this symbol: Leb(E,f,dy).

What does it mean, when does it make sense?

The symbol makes sense if one of these two conditions is
satisfied:

(1) Leb*(E,(f+),dy) < oo,

(2) Leb*(E,(f-),dy) < oo,

Leb(E,f,dy) = Leb*(E,(f+),dy) - Leb*(E,(f-),dy).

It could be +oo, or -oo, or a real number.

(1) Leb*(E,(f+),dy) < oo,

(2) Leb*(E,(f-),dy) < oo,

Leb(E,f,dy) = Leb*(E,(f+),dy) - Leb*(E,(f-),dy).

It could be +oo, or -oo, or a real number.

Let (X,F,y) be a measure space. Let E:-F.

Let f:X->[0,oo] be measurable. Suppose that /\x:-E f(x)=0.

Prove that Leb*(E,f,dy) = 0.

Let f:X->[0,oo] be measurable. Suppose that /\x:-E f(x)=0.

Prove that Leb*(E,f,dy) = 0.

page 169 in 1st measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable.

Let E:-F and M:-|R.

(1) y(E) < oo

(2) /\x:-E f(x)<=M

What can we conclude from this?

Let f:X->[0,oo] be measurable.

Let E:-F and M:-|R.

(1) y(E) < oo

(2) /\x:-E f(x)<=M

What can we conclude from this?

Leb*(E,f,dy) <= M*y(E)

page 171 in 1st measure

page 171 in 1st measure

Let (X,F,y) be a measure space.

Let f:X->|R* be measurable. Let E:-F.

(1) f is bounded on E

(2) y(E) < oo

What can we conclude from this?

Let f:X->|R* be measurable. Let E:-F.

(1) f is bounded on E

(2) y(E) < oo

What can we conclude from this?

f is integrable on E

page 172 in 1st measure

page 172 in 1st measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable. Let E:-F. Let a:-|R.

(1) /\(x:-E) 0 <= a <= f(x)

(2) y(E) < oo

What can we conclude from this?

Let f:X->[0,oo] be measurable. Let E:-F. Let a:-|R.

(1) /\(x:-E) 0 <= a <= f(x)

(2) y(E) < oo

What can we conclude from this?

a*y(E) <= Leb*(E,f,dy)

page 173 in 1st measure

page 173 in 1st measure

Let (X,F,y) be a measure space.

Let f:X->|R* be measurable. Let E:-F. Let a,b:-|R.

(1) y(E) < oo

(2) /\x:-E a <= f(x) <= b

What can we conclude from this?

(Do not prove the asnwer, but state it very clearly.)

Let f:X->|R* be measurable. Let E:-F. Let a,b:-|R.

(1) y(E) < oo

(2) /\x:-E a <= f(x) <= b

What can we conclude from this?

(Do not prove the asnwer, but state it very clearly.)

We can conclude that f is integrable on E and

a*y(E) <= Leb(E,f,dy) <= b*y(E).

page 174 in 1st measure

a*y(E) <= Leb(E,f,dy) <= b*y(E).

page 174 in 1st measure

Let (X,M,u) be a measure space.

Let E:-M. Let f,g:X->[0,oo] be measurable.

Suppose that /\x:-E f(x)<=g(x).

What can we conclude from this?

Let E:-M. Let f,g:X->[0,oo] be measurable.

Suppose that /\x:-E f(x)<=g(x).

What can we conclude from this?

Leb*(E,f,u) <= Leb*(E,g,u)

page 175 in 1st measure

page 175 in 1st measure

Let (X,M,u) be a measure space.

Let E:-M. Let f,g:X->|R* be integrable on E.

Suppose that /\x:-E f(x)<=g(x).

What can we conclude from this?

Let E:-M. Let f,g:X->|R* be integrable on E.

Suppose that /\x:-E f(x)<=g(x).

What can we conclude from this?

Leb(E,f,du) <= Leb(E,g,du)

page 175 in 1st measure

page 175 in 1st measure

Let (X,M,u) be a measure space.

Let E:-M. Let f:X->[0,oo] be integrable on E.

Let a:-[0,oo).

Prove that a*Leb*(E,f,du) = ???.

Let E:-M. Let f:X->[0,oo] be integrable on E.

Let a:-[0,oo).

Prove that a*Leb*(E,f,du) = ???.

a*Leb*(E,f,du) = Leb*(E,a*f,du)

page 177 in 1st meaure

If a=oo, then the thesis holds too.

See page 115 in 2nd measure.

page 177 in 1st meaure

If a=oo, then the thesis holds too.

See page 115 in 2nd measure.

Let a:-[-oo,0]. Let x:-|R*.

((a*x)+) = ???

((a*x)+) = ???

((a*x)+) = (-a)*(x-)

page 122 in 2nd measure

page 122 in 2nd measure

Let a:-[-oo,0]. Let x:-|R*.

(-a)*(x-) = ???

(-a)*(x-) = ???

(-a)*(x-) = ((a*x)+) = ((-a*x)-)

page 122 in 2nd measure

page 122 in 2nd measure

Let a:-[-oo,0]. Let x:-|R*.

((a*x)-) = ???

((a*x)-) = ???

((a*x)-) = (-a)*(x+)

page 122 in 2nd measure

page 122 in 2nd measure

Let a:-[-oo,0]. Let x:-|R*.

(-a)*(x-) = ???

(-a)*(x-) = ???

(-a)*(x-) = ((ax)+) = ((-ax)-)

page 122 in 2nd measure

page 122 in 2nd measure

Let (X,F,y) be a measure space.

Let E:-F. Let f:X->|R* be integrable on E. Let a:-|R.

Prove that a*Leb(E,f,dy) = ???.

Let E:-F. Let f:X->|R* be integrable on E. Let a:-|R.

Prove that a*Leb(E,f,dy) = ???.

a*Leb(E,f,dy) = Leb(E,a*f,dy)

page 177 in 1st measure

page 177 in 1st measure

Let (X,F,y) be a measure space.

Let E:-F. Let f:X->|R* be measurable.

Suppose that y(E)=0.

What can we conclude from this?

Let E:-F. Let f:X->|R* be measurable.

Suppose that y(E)=0.

What can we conclude from this?

function f is integrable on E and Leb(E,f,dy) = 0

page 179 in 1st measure

page 179 in 1st measure

Let (X,M,u) be a measure space.

Let A,B :- M. Let f:X->[0,oo] be measurable.

Suppose that A c B. What can we conclude?

Let A,B :- M. Let f:X->[0,oo] be measurable.

Suppose that A c B. What can we conclude?

Leb*(A,f,dy) <= Leb*(B,f,dy)

page 179 in 1st measure

page 179 in 1st measure

Let (X,M,u) be a measure space.

Let A,E :- M. Let f:X->|R* be measurable.

Suppose that AcE and f is integrable on E.

What can we conclude?

Let A,E :- M. Let f:X->|R* be measurable.

Suppose that AcE and f is integrable on E.

What can we conclude?

f is integrable on A

page 180 in 1st measure

page 180 in 1st measure

Let (X,M,u) be a measure space.

Let A,B:-M and AnB=O.

Let f:X->|R* be integrable on A and integrable on B.

What can we conclude?

Let A,B:-M and AnB=O.

Let f:X->|R* be integrable on A and integrable on B.

What can we conclude?

f is integrable on AuB and Leb(AuB,f,du) = Leb(A,f,du) +
Leb(B,f,du)

page 181 in 1st measure

page 181 in 1st measure

Let (X,M,u) be a measure space.

Let A,B:-M and AcB.

Let f:X->|R* be integrable on B.

Suppose that u(B\A)=0.

What can we conclude?

Let A,B:-M and AcB.

Let f:X->|R* be integrable on B.

Suppose that u(B\A)=0.

What can we conclude?

f is integrable on A and Leb(A,f,du) = Leb(B,f,du)

page 181 in 1st measure

page 181 in 1st measure

(1) a : |N x |N -> [0,oo]

(2) /\n:-|N /\k:-|N a(n,k) <= a(n,k+1)

What can we conclude about the limit:

lim(k->oo) +(n=1 to oo) a(n,k)

(2) /\n:-|N /\k:-|N a(n,k) <= a(n,k+1)

What can we conclude about the limit:

lim(k->oo) +(n=1 to oo) a(n,k)

(3) lim(k->oo) +(n=1 to oo) a(n,k) = +(n=1 to oo) lim(k->oo)
a(n,k)

page 182 in 1st measure

page 182 in 1st measure

(1) a : |N x |N -> [0,oo]

(2) /\n:-|N /\k:-|N a(n,k) <= a(n,k+1)

Decide if we can conclude that

(3) +(n=1 to oo) lim(k->oo) a(n,k) = lim(k->oo) +(n=1 to oo) a(n,k)

(2) /\n:-|N /\k:-|N a(n,k) <= a(n,k+1)

Decide if we can conclude that

(3) +(n=1 to oo) lim(k->oo) a(n,k) = lim(k->oo) +(n=1 to oo) a(n,k)

(3) +(n=1 to oo) lim(k->oo) a(n,k) = lim(k->oo) +(n=1 to oo)
a(n,k)

page 182 in 1st measure

page 182 in 1st measure

(1) a : |N x |N -> [0,oo]

Decide if we can conclude that

(2) +(n=1 to oo) lim(k->oo) a(n,k) = lim(k->oo) +(n=1 to oo) a(n,k)

Decide if we can conclude that

(2) +(n=1 to oo) lim(k->oo) a(n,k) = lim(k->oo) +(n=1 to oo) a(n,k)

NO.

Let a(n,k) = n/k

see page 182 in 1st measure

Let a(n,k) = n/k

see page 182 in 1st measure

Let f:[0,1]->|R be f(x)=sqrt(1-x).

Find a diffable function F:[0,1]->|R such that /\0<=x<=1 F'(x)=f(x).

Find a diffable function F:[0,1]->|R such that /\0<=x<=1 F'(x)=f(x).

F(x) = -2/3 * (1-x) * sqrt(1-x)

page 125 in OLDTIMER

page 125 in OLDTIMER

Let f:[0,1]->|R be f(x)=x*sqrt(1-x^2).

Find a diffable function F:[0,1]->|R such that /\0<=x<=1 F'(x)=f(x).

Find a diffable function F:[0,1]->|R such that /\0<=x<=1 F'(x)=f(x).

F(x) = -1/3 * (1-x^2) * sqrt(1-x^2)

page 125 in OLDTIMER

page 125 in OLDTIMER

(1) W is a set of sets

(2) T:W->[0,oo)

(3) T is finitely additive

(4) T is not countably additive

Is this setup possible?

(2) T:W->[0,oo)

(3) T is finitely additive

(4) T is not countably additive

Is this setup possible?

Yes.

W = { {n} : n:-|N } u { |N }

T( {n} ) = 0

T( |N ) = 1

W = { {n} : n:-|N } u { |N }

T( {n} ) = 0

T( |N ) = 1

(1) W is a ring of sets

(2) T:W->[0,oo)

(3) T is finitely additive

(4) T is not countably additive

Is this setup possible?

(2) T:W->[0,oo)

(3) T is finitely additive

(4) T is not countably additive

Is this setup possible?

Yes.

W = { Ec|N : E is finite or |N\E is finite }

If E is finite, then let T(E) = 0.

If |N\E is finite, then let T(E) = 1.

W = { Ec|N : E is finite or |N\E is finite }

If E is finite, then let T(E) = 0.

If |N\E is finite, then let T(E) = 1.

(1) W = { Ec|N : E is finite or |N\E is finite }

(2) T : W -> |R*

(3) If E is finite, then let T(E) = 0.

(4) If |N\E is finite, then let T(E) = 1.

What is interesting about this situation?

(2) T : W -> |R*

(3) If E is finite, then let T(E) = 0.

(4) If |N\E is finite, then let T(E) = 1.

What is interesting about this situation?

T is finitely additive but not countably additive

Let (X,M,u) be a measure space.

Let f:X->[0,oo] be measurable.

Let's define Y:M->[0,oo] by Y(A)=Leb*(A,f,du).

Is Y countably additive?

Let f:X->[0,oo] be measurable.

Let's define Y:M->[0,oo] by Y(A)=Leb*(A,f,du).

Is Y countably additive?

Yes.

page 183 in 1st measure

page 59 in MSW measure

page 183 in 1st measure

page 59 in MSW measure

Let (X,M,u) be a measure space.

Let f:X->|R* be integrable on X.

Let's define Y:M->[0,oo) by Y(A)=Leb(A,f,du).

Is Y countably additive?

Let f:X->|R* be integrable on X.

Let's define Y:M->[0,oo) by Y(A)=Leb(A,f,du).

Is Y countably additive?

Yes.

page 184 in 1st measure

page 184 in 1st measure

Let (X,M,u) be a measure space.

Consider {E:-M : u(E)=0}.

What can we say about this set?

Consider {E:-M : u(E)=0}.

What can we say about this set?

It is a s-ring.

page 185 in 1st measure

page 185 in 1st measure

Let (X,M,u) be a measure space.

Let f,g:X->|R* be measurable.

We write f~g iff u( {x:-X : f(x)!=g(x)} )=0.

What kind of relation is ~?

Let f,g:X->|R* be measurable.

We write f~g iff u( {x:-X : f(x)!=g(x)} )=0.

What kind of relation is ~?

It is an equivalence relation.

page 185 in 1st measure

page 185 in 1st measure

Let A be a set. Let F c AxA.

(1) /\x:-A [ (x,x):-F ]

(2) /\x,y:-A [ (x,y):-F => (y,x):-F ]

(3) /\x,y,z:-A [ (x,y):-F and (y,z):-F |=> (x,z):-F ]

What is the verbal description of F?

(1) /\x:-A [ (x,x):-F ]

(2) /\x,y:-A [ (x,y):-F => (y,x):-F ]

(3) /\x,y,z:-A [ (x,y):-F and (y,z):-F |=> (x,z):-F ]

What is the verbal description of F?

F is an equivalence relation.

Let (X,M,u) be a measure space.

Let f,g:X->|R* be measurable.

Let E:-M.

We write "f~g on E" iff u( {x:-E : f(x)!=g(x)} )=0.

What kind of relation is this ~?

Let f,g:X->|R* be measurable.

Let E:-M.

We write "f~g on E" iff u( {x:-E : f(x)!=g(x)} )=0.

What kind of relation is this ~?

It is an equivalence relation.

page 185 in 1st measure

hint: a!=c => ( a!=b v b!=c )

page 185 in 1st measure

hint: a!=c => ( a!=b v b!=c )

Let (X,M,u) be a measure space.

Let E:-M. Let f,g:X->|R* be integrable on E.

Suppose that f~g on E.

Prove that Leb(E,???????) = Leb(E,???????).

Let E:-M. Let f,g:X->|R* be integrable on E.

Suppose that f~g on E.

Prove that Leb(E,???????) = Leb(E,???????).

Leb(E,f,du) = Leb(E,g,du)

page 186 in 1st measure

page 186 in 1st measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that y( {x:-E : f(x)=oo} ) > 0.

What can we conclude?

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that y( {x:-E : f(x)=oo} ) > 0.

What can we conclude?

Leb*(E,f,dy) = oo

page 187 in 1st measure

page 121 in 2nd measure

page 187 in 1st measure

page 121 in 2nd measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that y( {x:-E : f(x)=oo} ) > 0.

Is it possible that Leb*(E,f,dy) < oo.

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that y( {x:-E : f(x)=oo} ) > 0.

Is it possible that Leb*(E,f,dy) < oo.

No.

page 187 in 1st measure

page 121 in 2nd measure

page 187 in 1st measure

page 121 in 2nd measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that Leb*(E,f,dy) < oo.

Is it possible that y( {x:-E : f(x)=oo} ) > 0.

Let f:X->[0,oo] be measurable. Let E:-F.

Suppose that Leb*(E,f,dy) < oo.

Is it possible that y( {x:-E : f(x)=oo} ) > 0.

No.

page 187 in 1st measure

page 121 in 2nd measure

page 187 in 1st measure

page 121 in 2nd measure

Let a,b :- |R, a<b. Let g : [a,b] -> |R.

(1) g is continuous on [a,b]

(2) g is continuously diffable on ]a,b[

(3) Integral(a, b, g'(x), dx) exists

What can we conclude?

(1) g is continuous on [a,b]

(2) g is continuously diffable on ]a,b[

(3) Integral(a, b, g'(x), dx) exists

What can we conclude?

Integral(a, b, g'(x), dx) = g(b) - g(a)

page 121 in OLDTIMER

page 121 in OLDTIMER

Let a,b:-|R, a!=b, A,B:-|R, A<B.

(1) g : *[a,b]* -> |R continuous

(2) g is continuously diffable on *]a,b[*.

(3) f : [A,B] -> |R continuous

(4) g(*[a,b]*) c [A,B]

(5) Integral(a, b, f(g(x))*g'(x), dx) exists

What can we conclude?

(1) g : *[a,b]* -> |R continuous

(2) g is continuously diffable on *]a,b[*.

(3) f : [A,B] -> |R continuous

(4) g(*[a,b]*) c [A,B]

(5) Integral(a, b, f(g(x))*g'(x), dx) exists

What can we conclude?

Integral(a, b, f(g(x))*g'(x), dx) = Integral( g(a), g(b), f(x),
dx )

page 123 in OLDTIMER

page 123 in OLDTIMER

Integral( 0, 1, x*sqrt(1-x^2), dx ) = ???

Integral( 0, 1, x*sqrt(1-x^2), dx ) = 1/3

page 124 in OLDTIMER

page 124 in OLDTIMER

Integral( 0, 1, sqrt(1-x), dx ) = ???

Integral( 0, 1, sqrt(1-x), dx ) = 2/3

page 125 in OLDTIMER

page 125 in OLDTIMER

Verify if this is true:

Integral( 0, 1, x*sqrt(1-x^2), dx ) = Integral( 0, 1, sqrt(1- x), dx )

Integral( 0, 1, x*sqrt(1-x^2), dx ) = Integral( 0, 1, sqrt(1- x), dx )

It is false. It should be:

Integral( 0, 1, x*sqrt(1-x^2), dx ) = 1/2 * Integral( 0, 1, sqrt(1-x), dx )

page 125 in OLDTIMER

Integral( 0, 1, x*sqrt(1-x^2), dx ) = 1/2 * Integral( 0, 1, sqrt(1-x), dx )

page 125 in OLDTIMER

Let a,b,c,d be real numbers.

(1) 0 < a < b

(2) c < d

(3) a*c > b*d

Is this setup possible?

(1) 0 < a < b

(2) c < d

(3) a*c > b*d

Is this setup possible?

0 < 1 < 20

-10 < -1

-10 > -20

-10 < -1

-10 > -20

lim(x,y)->(0,0) x*(1-cos(y)) / (2*x^2 - xy + y^2) = ???

lim(x,y)->(0,0) x*(1-cos(y)) / (2*x^2 - xy + y^2) = 0

hint: 1-cos(y) = y^2 * w(y), and lim(y->0)_w(y)=1/2

Use the power series expansion of cos.

x=r*cos(a), y=r*sin(a)

hint: 1-cos(y) = y^2 * w(y), and lim(y->0)_w(y)=1/2

Use the power series expansion of cos.

x=r*cos(a), y=r*sin(a)

Let (X,M,u) be a measure space.

Let E:-M. Let f:X->|R* be integrable on E.

Which set can be concluded to have measure zero?

Let E:-M. Let f:X->|R* be integrable on E.

Which set can be concluded to have measure zero?

{ x:-E : f(x) = oo v f(x) = -oo }

page 188 in 1st measure

page 188 in 1st measure

Let (X,M,u) be a measure space.

Let E:-M. Let f:X->|R* be integrable on E.

What inequality involving integrals on E can be written?

Let E:-M. Let f:X->|R* be integrable on E.

What inequality involving integrals on E can be written?

|Leb(E,f,du)| <= Leb*(E,|f|,du) < oo

page 188 in 1st measure

page 188 in 1st measure

Let (X,M,u) be a measure space. Let E:-M.

(1) /\n:-|N f[n]:X->[0,oo] measurable

(2) /\x:-E /\n:-|N f[n](x)<=f[n+1](x)

(3) f:X->[0,oo]

(4) /\x:-E lim(n->oo) f[n](x) = f(x)

What can we conclude?

(1) /\n:-|N f[n]:X->[0,oo] measurable

(2) /\x:-E /\n:-|N f[n](x)<=f[n+1](x)

(3) f:X->[0,oo]

(4) /\x:-E lim(n->oo) f[n](x) = f(x)

What can we conclude?

lim(n->oo) Leb*(E,f[n],du) = Leb*(E,f,du)

page 189 in 1st measure

page 119 in 2nd measure

page 189 in 1st measure

page 119 in 2nd measure

Prove that in a topological space, the union of two closed sets
is again a closed set.

page 7 in gen top

Let W c P(X) satisfy:

(1) /\B,A:-W [ BnA=O => BuA:-W ]

(2) /\B:-W X\B :- W

(3) X :- W

Decide if we can conclude that

(4) /\A,B:-W AuB:-W

(1) /\B,A:-W [ BnA=O => BuA:-W ]

(2) /\B:-W X\B :- W

(3) X :- W

Decide if we can conclude that

(4) /\A,B:-W AuB:-W

NO.

X = {1,2,3,4}

W = {O, X, {1,3}, {1,4}, {2,4}, {2,3} }

X = {1,2,3,4}

W = {O, X, {1,3}, {1,4}, {2,4}, {2,3} }

Suppose that a set on the real line has Lebesgue measure
greater than zero.

Can we conclude that this set contains an open interval?

Can we conclude that this set contains an open interval?

NO.

Let A = [0,1] \ |Q

Let A = [0,1] \ |Q

Integral( 0, pi, exp(sin(x))cos(x), dx ) = ???

Integral( 0, pi, exp(sin(x))cos(x), dx ) = 0

hint: page 94 in OLDTIMER

hint: page 94 in OLDTIMER

Integral( 0, pi, sin(sin(x))cos(x), dx ) = ???

Integral( 0, pi, sin(sin(x))cos(x), dx ) = 0

hint: page 94 in OLDTIMER

hint: page 94 in OLDTIMER

Let I be a connected subset of |R.

Let a,f : I -> |R be continuous functions.

Let t0:-I, x0:-|R. Given these conditions, prove that

there exists a unique function x:I->|R satisfying:

(1) /\t:-I x'(t) + a(t)x(t) = f(t),

(2) x(t0) = x0.

Let a,f : I -> |R be continuous functions.

Let t0:-I, x0:-|R. Given these conditions, prove that

there exists a unique function x:I->|R satisfying:

(1) /\t:-I x'(t) + a(t)x(t) = f(t),

(2) x(t0) = x0.

page 132 in OLDTIMER

Let J = (-pi/2 ; pi/2).

Find all diffable functions g:J->|R satisfying:

/\x:-J g'(x) - g(x)tan(x) = 1 / cos^3(x).

Find all diffable functions g:J->|R satisfying:

/\x:-J g'(x) - g(x)tan(x) = 1 / cos^3(x).

g(x) = sin(x)/cos^2(x) + M/cos(x)

where M is any real number

hint: page 132 in OLDTIMER

where M is any real number

hint: page 132 in OLDTIMER

Let J = (-pi/2 ; pi/2).

Find all diffable functions g:J->|R satisfying:

/\x:-J g'(x)cos(x) - g(x)sin(x) = 2x.

Find all diffable functions g:J->|R satisfying:

/\x:-J g'(x)cos(x) - g(x)sin(x) = 2x.

g(x) = (x^2 + M) / cos(x)

where M is any real number

hint: [g(x)cos(x)]'

hint: page 132 in OLDTIMER

where M is any real number

hint: [g(x)cos(x)]'

hint: page 132 in OLDTIMER

Find all diffable functions h:|R->|R satisfying:

/\x:-|R h'(x) + 2*h(x) = exp(-x).

/\x:-|R h'(x) + 2*h(x) = exp(-x).

h(x) = exp(-x) + M*exp(-2x)

where M is any real number

hint: page 132 in OLDTIMER

where M is any real number

hint: page 132 in OLDTIMER

Find all diffable functions h:|R->|R satisfying:

/\x:-|R h'(x) -2x*h(x) = 2x*exp(x^2).

/\x:-|R h'(x) -2x*h(x) = 2x*exp(x^2).

h(x) = (x^2+M)*exp(x^2)

where M is any real number

hint: page 132 in OLDTIMER

where M is any real number

hint: page 132 in OLDTIMER

Find all diffable functions h:|R->|R satisfying:

(1) /\x:-|R h'(x) - h(x) = 1

(2) h(2)=3

(1) /\x:-|R h'(x) - h(x) = 1

(2) h(2)=3

h(x) = 4*exp(x-2) - 1

hint: page 132 in OLDTIMER

hint: page 132 in OLDTIMER

Let J be a connected subset of |R with 1:-J.

Let h : J -> |R be a diffable function satisfying:

(1) /\x:-J x*h'(x) + h(x) = x+1

(2) h(1)=0.

Find a maximal set J and a function h satisfying the above conditions.

Let h : J -> |R be a diffable function satisfying:

(1) /\x:-J x*h'(x) + h(x) = x+1

(2) h(1)=0.

Find a maximal set J and a function h satisfying the above conditions.

h : (0;oo) -> |R

h(x) = x/2 + 1 - 3/(2x)

hint: [x*h(x)]' = x*h'(x) + h(x)

h(x) = x/2 + 1 - 3/(2x)

hint: [x*h(x)]' = x*h'(x) + h(x)

Let J be a connected subset of |R with 1:-J.

Let g : J -> |R be a diffable function satisfying:

(1) /\x:-J x^2 + x*g'(x) = g(x)

(2) g(1)=0.

Find a maximal set J and a function g satisfying the above conditions.

Let g : J -> |R be a diffable function satisfying:

(1) /\x:-J x^2 + x*g'(x) = g(x)

(2) g(1)=0.

Find a maximal set J and a function g satisfying the above conditions.

g : |R -> |R

g(x) = -x^2 + x

hint: page 134 in OLDTIMER

g(x) = -x^2 + x

hint: page 134 in OLDTIMER

(1) g : |R -> |R is a differentiable function

(2) g(1) = 0

(3) /\x:-|R x^2 + x*g'(x) = g(x)

Prove that there exists a unique object satisfying these conditions.

(2) g(1) = 0

(3) /\x:-|R x^2 + x*g'(x) = g(x)

Prove that there exists a unique object satisfying these conditions.

g(x)= -x^2 + x

page 134 in OLDTIMER

page 134 in OLDTIMER

(1) g : |R -> |R is diffable

(2) g(0) = 1

(3) /\x:-|R g'(x)*cos(x) + g(x)*sin(x) = 1

Does such a function exist?

(2) g(0) = 1

(3) /\x:-|R g'(x)*cos(x) + g(x)*sin(x) = 1

Does such a function exist?

Yes: g(x) = sin(x) + cos(x)

hint: [g(x)/cos(x)]' = ...

hint: [g(x)/cos(x)]' = ...

(1) h : |R -> |R is diffable

(2) h(0) = 0

(3) /\x:-|R h'(x) + h(x)*cos(x) = cos(x)

Find the set of all such functions.

(2) h(0) = 0

(3) /\x:-|R h'(x) + h(x)*cos(x) = cos(x)

Find the set of all such functions.

There is one and only one such function.

/\x:-|R h(x) = 1 - exp(-sin(x))

hint: multiply both sides of (3) by exp(sin(x))

see page 132 in OLDTIMER

/\x:-|R h(x) = 1 - exp(-sin(x))

hint: multiply both sides of (3) by exp(sin(x))

see page 132 in OLDTIMER

(1) h:|R->|R is diffable

(2) h(0)=0

(3) /\x:-|R x*h'(x) = x - h(x)

Find the set of all such functions.

(2) h(0)=0

(3) /\x:-|R x*h'(x) = x - h(x)

Find the set of all such functions.

There is one and only one such function.

/\x:-|R h(x) = x/2

/\x:-|R h(x) = x/2

Find a function whose derivative is 1/sin(x).

log( |tan(x/2)| )

page 181 in OLDTIMER

page 181 in OLDTIMER

(1) h:|R->|R is diffable

(2) /\x:-|R h(x) != 0

(3) /\x:-|R h'(x) + 2x*h(x) = 2x*h(x)*h(x)

Find the set of all such functions.

(2) /\x:-|R h(x) != 0

(3) /\x:-|R h'(x) + 2x*h(x) = 2x*h(x)*h(x)

Find the set of all such functions.

Let E = (-oo;-1) u [0,oo).

Let H = { h:-|R->|R : \/M:-E /\x:-|R h(x) = 1 / (1 + M*exp(x^2)) }.

H is the set of all such functions.

page 135 in OLDTIMER

Let H = { h:-|R->|R : \/M:-E /\x:-|R h(x) = 1 / (1 + M*exp(x^2)) }.

H is the set of all such functions.

page 135 in OLDTIMER

Let {a[n]}(n:-|N) be a sequence of complex numbers.

Suppose that lim(n->oo) sin(a[n]) = 0.

What can we conclude about the sequence cos(2*a[n]) ?

Suppose that lim(n->oo) sin(a[n]) = 0.

What can we conclude about the sequence cos(2*a[n]) ?

lim(n->oo) cos(2*a[n]) = 1

(1) sin^2(a[n])->0

(2) cos^2(a[n])->1

(3) cos(2*a[n])=cos^2(a[n])-sin^2(a[n])->1

(1) sin^2(a[n])->0

(2) cos^2(a[n])->1

(3) cos(2*a[n])=cos^2(a[n])-sin^2(a[n])->1

Let 0 < p < oo.

Let X = { a:|N->|R : +(n=1 to oo)_(|a[n]|^p) < oo }.

Can we conclude that /\a,b:-X a+b:-X ?

Let X = { a:|N->|R : +(n=1 to oo)_(|a[n]|^p) < oo }.

Can we conclude that /\a,b:-X a+b:-X ?

Yes.

hint: |a+b|^p <= 2^p*(|a|^p + |b|^p)

page 137 in OLDTIMER

hint: |a+b|^p <= 2^p*(|a|^p + |b|^p)

page 137 in OLDTIMER

(1) a,b : |N -> |R*

(2) /\n \/k>n a(n)<=b(k)

What can we conclude?

(2) /\n \/k>n a(n)<=b(k)

What can we conclude?

lim_sup(n->oo)_a(n) <= lim_sup(k->oo)_b(k)

(1) a : |N x |N -> |R*

(2) f,p : |N -> |R*

(3) /\n /\k a(n,k) <= a(n,k+1)

(4) /\n lim(k->oo) a(n,k) = f(n)

(5) /\n f(n) <= f(n+1)

(6) /\n p(n) = max(i=1 to n)_a(i,n)

What can we conclude?

(2) f,p : |N -> |R*

(3) /\n /\k a(n,k) <= a(n,k+1)

(4) /\n lim(k->oo) a(n,k) = f(n)

(5) /\n f(n) <= f(n+1)

(6) /\n p(n) = max(i=1 to n)_a(i,n)

What can we conclude?

(1) /\n p(n) <= p(n+1)

(2) lim(n->oo)_p(n) = lim(n->oo)_f(n)

page 118 in 2nd measure

(2) lim(n->oo)_p(n) = lim(n->oo)_f(n)

page 118 in 2nd measure

If x:-|R, then let [x] be the integer, x-1 < [x] <= x.

Let s:-|R. Let a[n] = 1/(2^n) * [2^n * s], for n:-|N.

Can we conclude that sequence a[n] is increasing?

Let s:-|R. Let a[n] = 1/(2^n) * [2^n * s], for n:-|N.

Can we conclude that sequence a[n] is increasing?

Yes.

page 137 in OLDTIMER

page 137 in OLDTIMER

Let (X,F,y) be a measure space.

Let f,g : X -> [0,oo] be measurable.

Let A:-F.

Prove that Leb*(A,f,dy) + Leb*(A,g,dy) = ??? .

Let f,g : X -> [0,oo] be measurable.

Let A:-F.

Prove that Leb*(A,f,dy) + Leb*(A,g,dy) = ??? .

Leb*(A,f,dy) + Leb*(A,g,dy) = Leb*(A,f+g,dy)

page 192 in 1st measure

page 192 in 1st measure

Let (X,M,u) be a measure space.

Let E:-M. Let f:X->[0,oo] be integrable on E.

Decide if we can conclude that:

oo * Leb*(E,f,du) = Leb*(E,oo*f,du).

Let E:-M. Let f:X->[0,oo] be integrable on E.

Decide if we can conclude that:

oo * Leb*(E,f,du) = Leb*(E,oo*f,du).

Yes.

page 115 in 2nd measure

page 115 in 2nd measure

Let f,g : X -> |R be simple functions.

Can we conclude that max(f,g) is also a simple function?

Can we conclude that max(f,g) is also a simple function?

Yes.

page 117 in 2nd measure

page 117 in 2nd measure

Let (X,F,y) be a measure space.

(1) /\n:-|N f[n]:X->[0,oo] measurable

Prove that:

+(n=1 to oo) Leb*(X,f[n],dy) = ???

(1) /\n:-|N f[n]:X->[0,oo] measurable

Prove that:

+(n=1 to oo) Leb*(X,f[n],dy) = ???

+(n=1 to oo) Leb*(X,f[n],dy) = Leb*(X, +(n=1 to oo)_f[n], dy)

page 120 in 2nd measure

page 10 in 2nd measure

page 120 in 2nd measure

page 10 in 2nd measure

Let (X,F,y) be a measure space.

(1) /\n:-|N f[n]:X->[0,oo] measurable

(2) E:-F

Prove that:

Leb*(E, +(n=1 to oo)_f[n], dy) = ???

(1) /\n:-|N f[n]:X->[0,oo] measurable

(2) E:-F

Prove that:

Leb*(E, +(n=1 to oo)_f[n], dy) = ???

Leb*(E, +(n=1 to oo)_f[n], dy) = +(n=1 to oo) Leb*(E,f[n],dy)

page 10 in 2nd measure

page 120 in 2nd measure

page 10 in 2nd measure

page 120 in 2nd measure

Let (X,F,y) be a measure space.

Let f:X->[0,oo] be measurable.

Let 0 < a < oo.

Prove that: y( {x:-X : f(x)>a} ) <= ??? .

Let f:X->[0,oo] be measurable.

Let 0 < a < oo.

Prove that: y( {x:-X : f(x)>a} ) <= ??? .

y( {x:-X : f(x)>a} ) <= 1/a * Leb*(X,f,dy)

page 121 in 2nd measure

page 121 in 2nd measure

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Prove that /\a>=0 /\x:-|R* ((ax)+) = ??? .

Prove that /\a>=0 /\x:-|R* ((ax)+) = ??? .

/\a>=0 /\x:-|R* ((ax)+) = a*(x+)

page 122 in 2nd measure

page 122 in 2nd measure

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Prove that /\a<=0 /\x:-|R* ((ax)+) = ??? .

Prove that /\a<=0 /\x:-|R* ((ax)+) = ??? .

/\a<=0 /\x:-|R* ((ax)+) = (-a)*(x-)

page 122 in 2nd measure

page 122 in 2nd measure

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Prove that /\a>=0 /\x:-|R* ((ax)-) = ??? .

Prove that /\a>=0 /\x:-|R* ((ax)-) = ??? .

/\a>=0 /\x:-|R* ((ax)-) = a*(x-)

page 122 in 2nd measure

page 122 in 2nd measure

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Prove that /\a<=0 /\x:-|R* ((ax)-) = ??? .

Prove that /\a<=0 /\x:-|R* ((ax)-) = ??? .

/\a<=0 /\x:-|R* ((ax)-) = (-a)*(x+)

page 122 in 2nd measure

page 122 in 2nd measure

Let (X,F,y) be a measure space.

Let E:-F. Let f:X->|R* be integrable on E.

Let a:-|R\{-oo,oo}.

What can we conclude about function a*f ?

Let E:-F. Let f:X->|R* be integrable on E.

Let a:-|R\{-oo,oo}.

What can we conclude about function a*f ?

Function a*f is integrable on E.

Leb(E,af,dy) = a*Leb(E,f,dy)

page 122 in 2nd measure

Leb(E,af,dy) = a*Leb(E,f,dy)

page 122 in 2nd measure

(1) f : |R -> |R

(2) /\x,y:-|R f(x+y) = f(x)*f(y)

What can we conclude about f(0) ?

(2) /\x,y:-|R f(x+y) = f(x)*f(y)

What can we conclude about f(0) ?

f(0) = f(0+0) = f(0)*f(0), hence f(0) = 0 or f(0) = 1

(1) g : |R -> |R

(2) /\x:-|R g'(x) = g(x)

Decide if we can conclude that:

(3) /\x:-|R g(x)>=0

(2) /\x:-|R g'(x) = g(x)

Decide if we can conclude that:

(3) /\x:-|R g(x)>=0

No.

f(x) = -exp(x)

f(x) = -exp(x)

Let (X,F,y) be a measure space.

Let E:-F. Let f,g:X->|R* be integrable on E.

What can we conclude about function f+g ?

Let E:-F. Let f,g:X->|R* be integrable on E.

What can we conclude about function f+g ?

Function f+g is also integrable on E.

Leb(E,f+g,dy) = Leb(E,f,dy) + Leb(E,g,dy)

page 128 in 2nd measure

page 123 in 2nd measure

page 9 in 2nd measure

Leb(E,f+g,dy) = Leb(E,f,dy) + Leb(E,g,dy)

page 128 in 2nd measure

page 123 in 2nd measure

page 9 in 2nd measure

Let (X,F,y) be a measure space.

Let f[n]:X->[0,oo] be measurable for all n:-|N.

Prove that:

Leb*(X , lim_inf(n->oo) f[n] , dy) <= ??? .

Let f[n]:X->[0,oo] be measurable for all n:-|N.

Prove that:

Leb*(X , lim_inf(n->oo) f[n] , dy) <= ??? .

Leb*(X , lim_inf(n->oo) f[n] , dy) <= lim_inf(n->oo)
Leb*(X,f[n],dy)

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

What is Fatou's Lemma?

Let (X,F,y) be a measure space.

Let E:-F. Let f[n]:X->[0,oo] be measurable for all n:-|N.

thesis:

Leb*(E , lim_inf(n->oo) f[n] , dy) <= lim_inf(n->oo) Leb*(E,f[n],dy)

page 10 in 2nd measure

Let E:-F. Let f[n]:X->[0,oo] be measurable for all n:-|N.

thesis:

Leb*(E , lim_inf(n->oo) f[n] , dy) <= lim_inf(n->oo) Leb*(E,f[n],dy)

page 10 in 2nd measure

Let (X,F,y) be a measure space.

Let f[n]:X->[0,oo] be a sequence of measurable functions.

Prove that:

lim_inf(n->oo) Leb*(X,f[n],dy) >= ???

Let f[n]:X->[0,oo] be a sequence of measurable functions.

Prove that:

lim_inf(n->oo) Leb*(X,f[n],dy) >= ???

lim_inf(n->oo) Leb*(X,f[n],dy) >= Leb*(X , lim_inf(n->oo) f[n]
, dy)

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

Let (X,F,y) be a measure space.

Let E:-F. Let f[n]:X->[0,oo] be a sequence of measurable functions.

Decide if it is possible that:

Leb*(E , lim_inf(n->oo) f[n] , dy) > lim_inf(n->oo) Leb*(E,f[n],dy)

Let E:-F. Let f[n]:X->[0,oo] be a sequence of measurable functions.

Decide if it is possible that:

Leb*(E , lim_inf(n->oo) f[n] , dy) > lim_inf(n->oo) Leb*(E,f[n],dy)

No.

We can conclude that:

Leb*(E , lim_inf(n->oo) f[n] , dy) <= lim_inf(n->oo) Leb*(E,f[n],dy)

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

We can conclude that:

Leb*(E , lim_inf(n->oo) f[n] , dy) <= lim_inf(n->oo) Leb*(E,f[n],dy)

page 10 in 2nd measure

This theorem is called: Fatou's Lemma.

Let (X,F,y) be a measure space.

Let E:-F. Let f:X->[0,oo] be measurable.

Decide if it is possible that:

Leb*(E , lim_inf(n->oo) f[n] , dy) < lim_inf(n->oo) Leb*(E,f[n],dy)

Let E:-F. Let f:X->[0,oo] be measurable.

Decide if it is possible that:

Leb*(E , lim_inf(n->oo) f[n] , dy) < lim_inf(n->oo) Leb*(E,f[n],dy)

Yes.

see page 19 in 2nd measure for an example

see page 19 in 2nd measure for an example

(1) /\n:-|N { a[n] , b[n] } c |R*

(2) /\n:-|N a[n] >= b[n]

(3) lim(n->oo) b[n] = b

What can we conclude?

(2) /\n:-|N a[n] >= b[n]

(3) lim(n->oo) b[n] = b

What can we conclude?

lim_inf(n->oo) a[n] >= b

This little fact is used in the proof of Fatou's Lemma.

(page 11 in 2nd measure)

This little fact is used in the proof of Fatou's Lemma.

(page 11 in 2nd measure)

Let (X,F,y) be a measure space.

Let E:-F. Let f,g:X->|R* be measurable.

(1) /\x:-E |f(x)| <= g(x)

(2) g is integrable on E

What can we conclude?

Let E:-F. Let f,g:X->|R* be measurable.

(1) /\x:-E |f(x)| <= g(x)

(2) g is integrable on E

What can we conclude?

(3) f is also integrable on E

page 11 in 2nd measure

page 11 in 2nd measure

Let (X,F,y) be a measure space.

Let E:-F. Let f[n]:X->|R* be measurable for all n:-|N.

Let f,g : X->|R*.

(1) /\n:-|N /\x:-E |f[n](x)| <= g(x)

(2) g is integrable on E

(3) /\x:-X lim(n->oo) f[n](x) = f(x)

What can we conclude?

Let E:-F. Let f[n]:X->|R* be measurable for all n:-|N.

Let f,g : X->|R*.

(1) /\n:-|N /\x:-E |f[n](x)| <= g(x)

(2) g is integrable on E

(3) /\x:-X lim(n->oo) f[n](x) = f(x)

What can we conclude?

lim(n->oo) Leb(E,f[n],dy) = Leb(E,f,dy)

This theorem is called: Lebesgue's Dominated Convergence Theorem.

page 13 in 2nd measure

This theorem is called: Lebesgue's Dominated Convergence Theorem.

page 13 in 2nd measure

Let (X,F,y) be a measure space.

Let f[n] : X->|R* be integrable for all n:-|N.

Let f : X->|R* be integrable.

Let /\x:-X lim(n->oo) f[n](x) = f(x).

Decide if we can conclude that

lim(n->oo) Leb(X,f[n],dy) = Leb(X,f,dy).

Let f[n] : X->|R* be integrable for all n:-|N.

Let f : X->|R* be integrable.

Let /\x:-X lim(n->oo) f[n](x) = f(x).

Decide if we can conclude that

lim(n->oo) Leb(X,f[n],dy) = Leb(X,f,dy).

No.

Let X = |R, F = the Lebesgue measurable sets, y = the Lebesgue measure.

Let f[n] = 1([n,2n]) for all n:-|N.

Then /\x:-|R lim(n->oo) f[n](x) = 0 = f(x).

But Leb(X,f[n],dy) = n. And lim(n->oo) n != 0.

Let X = |R, F = the Lebesgue measurable sets, y = the Lebesgue measure.

Let f[n] = 1([n,2n]) for all n:-|N.

Then /\x:-|R lim(n->oo) f[n](x) = 0 = f(x).

But Leb(X,f[n],dy) = n. And lim(n->oo) n != 0.

(1) a,b :- |R, a!=b

(2) g : *[a,b]* -> |R

(3) g is continuous on *[a,b]*

(4) g is continuously diffable on *[a,b)*

lim(t->b) Riemann_Integral(a, t, g'(x), dx) = ???

(2) g : *[a,b]* -> |R

(3) g is continuous on *[a,b]*

(4) g is continuously diffable on *[a,b)*

lim(t->b) Riemann_Integral(a, t, g'(x), dx) = ???

lim(t->b) Riemann_Integral(a, t, g', dx) = g(b) - g(a)

page 126 in OLDTIMER

page 126 in OLDTIMER

Let 0 < p < 1.

Let X = {x:|N->|C : +(n=1 to oo) |x(n)|^p < oo}.

If x,y:-X, let d(x,y) = +(n=1 to oo) |x(n)-y(n)|^p.

Is d a metric on X?

Let X = {x:|N->|C : +(n=1 to oo) |x(n)|^p < oo}.

If x,y:-X, let d(x,y) = +(n=1 to oo) |x(n)-y(n)|^p.

Is d a metric on X?

Yes.

Define g(x) = +(n=1 to oo) |x(n)|^p.

Prove that g(-x)=g(x).

Prove that g(x+y) <= g(x) + g(y), (page 140 in OLDTIMER).

Now d(x,y) = g(x-y) is easily seen to be a metric.

Define g(x) = +(n=1 to oo) |x(n)|^p.

Prove that g(-x)=g(x).

Prove that g(x+y) <= g(x) + g(y), (page 140 in OLDTIMER).

Now d(x,y) = g(x-y) is easily seen to be a metric.

Let 0 < p < 1.

Let X = {x:|N->|C : +(n=1 to oo) |x(n)|^p < oo}.

If x:-X, let g(x) = +(n=1 to oo) |x(n)|^p.

Is g a norm on X?

Let X = {x:|N->|C : +(n=1 to oo) |x(n)|^p < oo}.

If x:-X, let g(x) = +(n=1 to oo) |x(n)|^p.

Is g a norm on X?

No.

g(2x) = 2^p*g(x) != 2*g(x)

g(2x) = 2^p*g(x) != 2*g(x)

Consider two normed vector spaces over the same field.

What does it mean that they are isomorphic?

What does it mean that they are isomorphic?

Let (X,K,g), (Y,K,h) be two normed vector spaces over the field
K.

There exists a bijection U : X -> Y such that:

(1) /\x,y:-X /\a:-K U(ax+y) = a*U(x) + U(y)

(2) For every sequence x[n] contained in X:

lim(n->oo) g(x[n]) = 0 <=> lim(n->oo) h(U(x[n])) = 0.

-------------

There exists a bijection U : X -> Y such that:

(1) /\x,y:-X /\a:-K U(ax+y) = a*U(x) + U(y)

(2) For every sequence x[n] contained in X:

lim(n->oo) g(x[n]) = 0 <=> lim(n->oo) h(U(x[n])) = 0.

-------------

Consider two normed vector spaces over the same field.

What does it mean that they are isometrically isomorphic?

What does it mean that they are isometrically isomorphic?

Let (X,K,g), (Y,K,h) be two normed vector spaces over the field
K.

There exists a bijection U : X -> Y such that:

(1) /\x,y:-X /\a:-K U(ax+y) = a*U(x) + U(y)

(2) For every sequence x[n] contained in X:

lim(n->oo) g(x[n]) = 0 <=> lim(n->oo) h(U(x[n])) = 0

(3) /\x:-X g(x) = h(U(x))

There exists a bijection U : X -> Y such that:

(1) /\x,y:-X /\a:-K U(ax+y) = a*U(x) + U(y)

(2) For every sequence x[n] contained in X:

lim(n->oo) g(x[n]) = 0 <=> lim(n->oo) h(U(x[n])) = 0

(3) /\x:-X g(x) = h(U(x))

Give an example of two normed vector spaces over the same
field, such that they are isomorphic but not isometrically
isomorphic.

Let us consider R^2 over |R.

Let us equip it with two different norms: the taxi norm and the Euclidean norm.

Then the two normed vector spaces are isomorphic.

But they cannot be isometric: see page 142 in OLDTIMER.

Let us equip it with two different norms: the taxi norm and the Euclidean norm.

Then the two normed vector spaces are isomorphic.

But they cannot be isometric: see page 142 in OLDTIMER.

Let a[n] be a sequence of real numbers other than zero.

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Is it generally true that A=B ?

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Is it generally true that A=B ?

No.

==================================

1, -1, 2, -1/2, 3, -1/3, 4, -1/4, 5, -1/5, ...

A = 1/oo = 0; B = -oo

==================================

1, -1, 1/2, -2, 1/3, -3, 1/4, -4, 1/5, -5, ...

A = 1/0 = oo; B = 0

==================================

1, -1, 2, -1/2, 3, -1/3, 4, -1/4, 5, -1/5, ...

A = 1/oo = 0; B = -oo

==================================

1, -1, 1/2, -2, 1/3, -3, 1/4, -4, 1/5, -5, ...

A = 1/0 = oo; B = 0

Let X = { g:[0,1]->|R : g is continuously diffable on [0,1] }.

Let H : X -> |R, H(g) = sup_norm(g) + sup_norm(g').

Is H a norm on X ?

Let H : X -> |R, H(g) = sup_norm(g) + sup_norm(g').

Is H a norm on X ?

Yes.

page 143 in OLDTIMER

page 143 in OLDTIMER

Let -oo < a < b < oo.

Let f : [a,b] -> |R be Riemann-integrable.

Investigate lim(t->b) Integral(a,t,f(x),dx).

Let f : [a,b] -> |R be Riemann-integrable.

Investigate lim(t->b) Integral(a,t,f(x),dx).

lim(t->b) Integral(a,t,f(x),dx) = Integral(a,b,f(x),dx)

page 126 in OLDTIMER

page 126 in OLDTIMER

Let a:-|R.

Integral( 0, a, x*sqrt(a^2-x^2), dx ) = ???

Integral( 0, a, x*sqrt(a^2-x^2), dx ) = ???

Integral( 0, a, x*sqrt(a^2-x^2), dx ) = a^3 / 3

page 42 in redcrest

page 127 in OLDTIMER

page 42 in redcrest

page 127 in OLDTIMER

Let a:-|R\{0}.

Let f : *[0,a]* -> |R be integrable.

Prove that:

Integral( 0, a, f(x), dx ) = ??? Integral( 0, ???, f( ??? ), dx ).

Let f : *[0,a]* -> |R be integrable.

Prove that:

Integral( 0, a, f(x), dx ) = ??? Integral( 0, ???, f( ??? ), dx ).

Integral( 0, a, f(x), dx ) = - Integral( 0, -a, f(-x), dx )

page 128 in OLDTIMER

page 128 in OLDTIMER

Let a:-|R\{0}.

Investigate lim(t->a) Integral( 0, t, x/sqrt(a^2 - x^2), dx).

Investigate lim(t->a) Integral( 0, t, x/sqrt(a^2 - x^2), dx).

lim(t->a) Integral( 0, t, x/sqrt(a^2 - x^2), dx) = |a|

page 128 in OLDTIMER

There is a simpler way than that on page 128 in OLDTIMER.

Hint: F(x)= -sqrt(a^2-x^2)

page 128 in OLDTIMER

There is a simpler way than that on page 128 in OLDTIMER.

Hint: F(x)= -sqrt(a^2-x^2)

(1) g : |R -> |R

(2) /\x:-|R g'(x) = g(x)

Find the set of all such functions.

(2) /\x:-|R g'(x) = g(x)

Find the set of all such functions.

(3) \/A:-|R /\x:-|R g(x) = A*exp(x)

(3) <=> (2)

page 129 in OLDTIMER

(3) <=> (2)

page 129 in OLDTIMER

Let g : |C -> |C.

(1) \/a:-|C /\x:-|C g'(x) = a*g(x)

(2) /\x,y:-|C g(x)*g(y) = g(x+y)*g(0)

Does (2) follow from (1) ?

(1) \/a:-|C /\x:-|C g'(x) = a*g(x)

(2) /\x,y:-|C g(x)*g(y) = g(x+y)*g(0)

Does (2) follow from (1) ?

(1) => (2)

page 130 in OLDTIMER

page 130 in OLDTIMER

Let g : |R -> |R.

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) = 0

What can we conclude?

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) = 0

What can we conclude?

/\x:-|R g(x)=0

page 131 in OLDTIMER

page 131 in OLDTIMER

Let g : |R -> |R.

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) > 0

What can we conclude?

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) > 0

What can we conclude?

(3) /\x:-|R g(x) > 0

page 131 in OLDTIMER

page 131 in OLDTIMER

Let g : |R -> |R.

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) < 0

What can we conclude?

(1) /\x:-|R g'(x) = g(x)

(2) \/x:-|R g(x) < 0

What can we conclude?

(3) /\x:-|R g(x) < 0

page 131 in OLDTIMER

page 131 in OLDTIMER

Let J be a connected subset of |R.

Let a,f : J -> |R be continuous.

Let t0:-J, x0:-|R.

Let x : J -> |R be a diffable function satisfying:

(1) x(t0) = x0

(2) /\t:-J x'(t) + a(t)*x(t) = f(t)

What can we say about the existence of such functions?

Let a,f : J -> |R be continuous.

Let t0:-J, x0:-|R.

Let x : J -> |R be a diffable function satisfying:

(1) x(t0) = x0

(2) /\t:-J x'(t) + a(t)*x(t) = f(t)

What can we say about the existence of such functions?

There exists one and only one such function.

page 132 in OLDTIMER

page 132 in OLDTIMER

Let A and B be non-empty sets.

Let f : AxB -> |R.

sup{ f(x,y) : x:-A and y:-B } = ???

Let f : AxB -> |R.

sup{ f(x,y) : x:-A and y:-B } = ???

sup{ f(x,y) : x:-A and y:-B } = sup{ sup{ f(x,y) : y:-B } : x:-
A }

page 138 in OLDTIMER

page 138 in OLDTIMER

(1) A c |R

(2) a :- Clo( A\{a} )

Can we conclude that sup(A) = sup( A\{a} ) ?

(2) a :- Clo( A\{a} )

Can we conclude that sup(A) = sup( A\{a} ) ?

Yes.

page 139 in OLDTIMER

page 139 in OLDTIMER

Let 0 < p < oo.

sup{ (a+b)^p / (a^p + b^p) : a,b>0 } = ???

sup{ (a+b)^p / (a^p + b^p) : a,b>0 } = ???

sup{ (a+b)^p / (a^p + b^p) : a,b>0 } = max( 1, 2^(p-1) )

page 140 in OLDTIMER

page 140 in OLDTIMER

Let 0 < p < oo.

sup{ |a+b|^p / (|a|^p + |b|^p) : a,b:-|C\{0} } = ???

sup{ |a+b|^p / (|a|^p + |b|^p) : a,b:-|C\{0} } = ???

sup{ |a+b|^p / (|a|^p + |b|^p) : a,b:-|C\{0} } = max( 1, 2^(p-
1) )

page 141 in OLDTIMER

page 141 in OLDTIMER

(1) -oo < a < b < oo

(2) /\n:-|N g[n] : <a,b> -> |R diffable

(3) \/x:-<a,b> { g[n](x) }(n:-|N) is Cauchy

(4) { g'[n] }(n:-|N) converges uniformly on <a,b>

Is this enough to conclude that { g[n] }(n:-|N) converges uniformly on <a,b> ?

(2) /\n:-|N g[n] : <a,b> -> |R diffable

(3) \/x:-<a,b> { g[n](x) }(n:-|N) is Cauchy

(4) { g'[n] }(n:-|N) converges uniformly on <a,b>

Is this enough to conclude that { g[n] }(n:-|N) converges uniformly on <a,b> ?

{ g[n] }(n:-|N) converges uniformly on <a,b>

page 144 in OLDTIMER

page 144 in OLDTIMER

Let A be a connected subset of |R.

(1) /\n:-|N h[n]:A->|R diffable

(2) h,g:A->|R

(3) { h[n] }(n:-|N) converges pointwise to h on A

(4) { h'[n] }(n:-|N) converges uniformly to g on A

What can we conclude?

(1) /\n:-|N h[n]:A->|R diffable

(2) h,g:A->|R

(3) { h[n] }(n:-|N) converges pointwise to h on A

(4) { h'[n] }(n:-|N) converges uniformly to g on A

What can we conclude?

/\x:-A h'(x) = g(x)

page 135 in 2nd measure

page 145 in OLDTIMER

page 135 in 2nd measure

page 145 in OLDTIMER

Let -oo < a < b < oo.

Let C1[a,b] be the set of all continuously diffable real-valued functions on [a,b].

Let's define a norm on C1[a,b], ||h||' = ||h||oo + ||h'||oo.

Is this normed vector space complete?

Let C1[a,b] be the set of all continuously diffable real-valued functions on [a,b].

Let's define a norm on C1[a,b], ||h||' = ||h||oo + ||h'||oo.

Is this normed vector space complete?

Yes.

page 147 in OLDTIMER

page 147 in OLDTIMER

Let X,Y be normed vector spaces.

Let T:X->Y be linear and continuous at the zero vector.

Prove that \/M:-|R /\x:-X ||Tx|| <= M*||x||.

Let T:X->Y be linear and continuous at the zero vector.

Prove that \/M:-|R /\x:-X ||Tx|| <= M*||x||.

page 11 in OLDTIMER

Let X,Y be normed vector spaces.

Let T:X->Y be linear and continuous.

Can we conclude that T satisfies a Lipschitz condition?

Let T:X->Y be linear and continuous.

Can we conclude that T satisfies a Lipschitz condition?

Yes.

page 11 in OLDTIMER

page 11 in OLDTIMER

Let X,Y be normed vector spaces.

Let u,v be distinct vectors in X.

Let T:X->Y be linear and continuous at v.

Can we conclude that T is continuous at u?

Let u,v be distinct vectors in X.

Let T:X->Y be linear and continuous at v.

Can we conclude that T is continuous at u?

Yes.

Let X,Y be normed vector spaces. Let T:X->Y.

Suppose that \/M:-|R /\x:-X ||Tx|| <= M*||x||.

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

What can we conclude about the number ||T|| ?

Suppose that \/M:-|R /\x:-X ||Tx|| <= M*||x||.

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

What can we conclude about the number ||T|| ?

/\x:-X ||Tx|| <= ||T||*||x||

page 148 in OLDTIMER

page 148 in OLDTIMER

Let f:X->|R be bounded above.

Let a = inf{ m:-|R : /\x:-X f(x)<=m }.

Let b = sup{ m:-|R : \/x:-X f(x)=m }.

What can we conclude about numbers a,b ?

Let a = inf{ m:-|R : /\x:-X f(x)<=m }.

Let b = sup{ m:-|R : \/x:-X f(x)=m }.

What can we conclude about numbers a,b ?

a=b

page 149 in OLDTIMER

page 149 in OLDTIMER

Let X,Y be normed vector spaces.

Let f:X->Y, f(0)=0.

Decide if we can conclude that:

inf{ m:-|R : /\x:-X ||f(x)|| <= m*||x|| } = sup{ ||f(x)|| / ||x|| : x:-X , x!=0 }

Let f:X->Y, f(0)=0.

Decide if we can conclude that:

inf{ m:-|R : /\x:-X ||f(x)|| <= m*||x|| } = sup{ ||f(x)|| / ||x|| : x:-X , x!=0 }

Yes.

The assumption f(0)=0 is important.

page 150 in OLDTIMER

The assumption f(0)=0 is important.

page 150 in OLDTIMER

Let X,Y be normed vector spaces.

Let f:X->Y. Decide if we can conclude that:

inf{ m:-|R : /\x:-X ||f(x)|| <= m*||x|| } = sup{ ||f(x)|| / ||x|| : x:-X , x!=0 }

Let f:X->Y. Decide if we can conclude that:

inf{ m:-|R : /\x:-X ||f(x)|| <= m*||x|| } = sup{ ||f(x)|| / ||x|| : x:-X , x!=0 }

No.

Let X,Y be |R.

Let f(0)=1, f(x)=x for x!=0.

page 150 in OLDTIMER

Let X,Y be |R.

Let f(0)=1, f(x)=x for x!=0.

page 150 in OLDTIMER

Let X,Y be normed vector spaces. Let T:X->Y.

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that ||T|| < sup{ ||Tx|| : ||x||<=1 }.

Is this possible?

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that ||T|| < sup{ ||Tx|| : ||x||<=1 }.

Is this possible?

No.

page 151 in OLDTIMER

page 151 in OLDTIMER

Let (X,F,y) be a measure space.

Let E:-F. Let b:-|R.

Let f:X->|R* be measurable, and integrable on E.

(1) y(E) < oo

(2) /\x:-E f(x) <= b

Prove that Leb(E,f,dy) <= b*y(E).

Let E:-F. Let b:-|R.

Let f:X->|R* be measurable, and integrable on E.

(1) y(E) < oo

(2) /\x:-E f(x) <= b

Prove that Leb(E,f,dy) <= b*y(E).

page 174 in 1st measure

Let (X,F,y) be a measure space.

Let E:-F. Let a:-|R.

Let f:X->|R* be measurable and integrable on E.

(1) y(E) < oo

(2) /\x:-E a <= f(x)

Prove that a*y(E) <= Leb(E,f,dy).

Let E:-F. Let a:-|R.

Let f:X->|R* be measurable and integrable on E.

(1) y(E) < oo

(2) /\x:-E a <= f(x)

Prove that a*y(E) <= Leb(E,f,dy).

page 174 in 1st measure

Let X,Y be normed vector spaces. Let T:X->Y be linear.

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that sup{ ||Tx|| : ||x||=1 } < ||T||.

Is this possible?

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that sup{ ||Tx|| : ||x||=1 } < ||T||.

Is this possible?

No.

page 151 in OLDTIMER

page 151 in OLDTIMER

Let X,Y be normed vector spaces.

Let T:X->Y be continuous.

Decide if it is possible that:

sup{ ||Tx|| : ||x||<1 } < sup{ ||Tx|| : ||x||<=1 }

Let T:X->Y be continuous.

Decide if it is possible that:

sup{ ||Tx|| : ||x||<1 } < sup{ ||Tx|| : ||x||<=1 }

No.

page 152 in OLDTIMER

page 152 in OLDTIMER

Let X,Y be normed vector spaces.

Let T:X->Y be linear.

Suppose there exists an open set E c X

such that \/M:-|R /\x:-E ||Tx|| < M.

What can we conclude?

Let T:X->Y be linear.

Suppose there exists an open set E c X

such that \/M:-|R /\x:-E ||Tx|| < M.

What can we conclude?

\/M:-|R /\x:-X ||Tx|| <= M*||x||

page 153 in OLDTIMER

page 153 in OLDTIMER

Let X be a complete normed vector space.

Suppose that a series of vectors in X converges absolutely.

Can we conclude that this series converges to a vector in X?

Suppose that a series of vectors in X converges absolutely.

Can we conclude that this series converges to a vector in X?

Yes.

page 154 in OLDTIMER

page 154 in OLDTIMER

Let X be a normed vector space.

In this space, every absolutely convergent series converges to a vector in X.

What can we conclude about this space?

In this space, every absolutely convergent series converges to a vector in X.

What can we conclude about this space?

It is complete.

page 155 in OLDTIMER

page 155 in OLDTIMER

If k:-|N, and f,g:|R->|R, let d[k](f,g) = sup{|f(x)-g(x)| : -
k<=x<=k}.

Notice that /\k:-|N function d[k] is a pseudometric on |R^|R.

Let W = {U c |R^|R : U is d[k]-open for some k:-|N}.

Let f[n] be a sequence of functions |R->|R, and let f:|R->|R.

Prove that the two conditions are equivalent:

(1) { f[n] } W-converges to f

Notice that /\k:-|N function d[k] is a pseudometric on |R^|R.

Let W = {U c |R^|R : U is d[k]-open for some k:-|N}.

Let f[n] be a sequence of functions |R->|R, and let f:|R->|R.

Prove that the two conditions are equivalent:

(1) { f[n] } W-converges to f

page 72 in gen top

Let f:X->Y.

If ZcX, let B(Z) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let T be a set and /\t:-T Z[t] c X.

Prove that:

B( \\//t:-T Z[t] ) = ??? .

If ZcX, let B(Z) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let T be a set and /\t:-T Z[t] c X.

Prove that:

B( \\//t:-T Z[t] ) = ??? .

B( \\//t:-T Z[t] ) = //\\t:-T B( Z[t] )

page 73 in gen top

page 73 in gen top

Let X,Y be sets.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that:

(1) /\n:-|N U[n]:-W |=> //\\n:-|N_U[n] :- W

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that:

(1) /\n:-|N U[n]:-W |=> //\\n:-|N_U[n] :- W

page 74 in gen top

It is not necessary in this item but you can recall that W is a topology.

It is not necessary in this item but you can recall that W is a topology.

Let X,Y be sets.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that W is a topology.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that W is a topology.

page 74 in gen top

Let X,Y be sets.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that every A c Y^X is sequentially W-closed.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

Prove that every A c Y^X is sequentially W-closed.

page 74 in gen top

It is not necessary in this item but you can recall that W is a topology.

It is not necessary in this item but you can recall that W is a topology.

If Zc|R, f:|R->|R, let B(Z,f) = {g:|R->|R : /\x:-Z f(x)=g(x)}.

Let W = {U c |R^|R : /\f:-U \/Zc|R Z is countable and B(Z,f) c U}.

Prove that:

/\f:|R->|R {f} is not W-open

Let W = {U c |R^|R : /\f:-U \/Zc|R Z is countable and B(Z,f) c U}.

Prove that:

/\f:|R->|R {f} is not W-open

page 74 in gen top

It is not necessary in this item but you can recall that W is a topology.

It is not necessary in this item but you can recall that W is a topology.

Give an example of a topological space in which:

(1) every set is sequentially closed,

(2) uncountably many sets are not closed.

(1) every set is sequentially closed,

(2) uncountably many sets are not closed.

Let X = {0,1}^|R = the set of all functions |R->{0,1}.

Let G = { UcX : /\f:-U \/(Ac|R countable) /\g:-X (g=f on A => g:-U) }.

This (X,G) is such a topological space.

page 74 in gen top

Let G = { UcX : /\f:-U \/(Ac|R countable) /\g:-X (g=f on A => g:-U) }.

This (X,G) is such a topological space.

page 74 in gen top

(1) h : [0,oo) -> |R

(2) h is diffable and strictly increasing

(3) h(0) = 0

(4) /\x>=0 h'(x) = 2*sqrt(h(x))

Find all such functions.

(2) h is diffable and strictly increasing

(3) h(0) = 0

(4) /\x>=0 h'(x) = 2*sqrt(h(x))

Find all such functions.

h(x) = x^2

It is the only one.

page 159 in OLDTIMER

It is the only one.

page 159 in OLDTIMER

(1) h : |R -> |R

(2) h is diffable

(3) h(0) = 0

(4) /\x:-|R h'(x) = 2*sqrt(|h(x)|)

How many such functions are there?

(2) h is diffable

(3) h(0) = 0

(4) /\x:-|R h'(x) = 2*sqrt(|h(x)|)

How many such functions are there?

uncountably many

If b>=0, let's define h relative to b.

h(x) = 0 for x<=b

h(x) = (x-b)^2 for x>b

If b>=0, let's define h relative to b.

h(x) = 0 for x<=b

h(x) = (x-b)^2 for x>b

(1) h : |R -> (-oo,0]

(2) h is diffable

(3) h(0) = 0

(4) /\x:-|R h'(x) = 2*sqrt(-h(x))

How many such functions are there?

(2) h is diffable

(3) h(0) = 0

(4) /\x:-|R h'(x) = 2*sqrt(-h(x))

How many such functions are there?

uncountably many

If a<=0, let's define h relative to a.

h(x) = 0 for x>=a

h(x) = -(x-a)^2 for x<a

If a<=0, let's define h relative to a.

h(x) = 0 for x>=a

h(x) = -(x-a)^2 for x<a

(1) h : |R -> |R diffable

(2) /\x:-|R h'(x) = exp(x-h(x))

Find all such functions.

(2) /\x:-|R h'(x) = exp(x-h(x))

Find all such functions.

\/M>=0 /\x:-|R h(x) = log(exp(x)+M)

page 160 in OLDTIMER

page 160 in OLDTIMER

(1) a,b :- |R, a<b

(2) h : [a,b] -> |R continuous on [a,b]

(3) h is continuously diffable on ]a,b[

(4) h' is Riemann-integrable on [a,b]

Decide if these conditions are sufficient to conclude that:

Integral(a,b,h'(x),dx) = h(b)-h(a)

(2) h : [a,b] -> |R continuous on [a,b]

(3) h is continuously diffable on ]a,b[

(4) h' is Riemann-integrable on [a,b]

Decide if these conditions are sufficient to conclude that:

Integral(a,b,h'(x),dx) = h(b)-h(a)

Yes.

page 156 in OLDTIMER

page 156 in OLDTIMER

Let f:(-1;oo)->|R, f(x) = x/(x+1).

Find function F:(-1;oo)->|R such that F'=f.

Find function F:(-1;oo)->|R such that F'=f.

F(x) = x - log(x+1)

(1) J is a connected subset of |R and 0:-J

(2) h : J -> |R

(3) h(0)=0

(3) /\x:-J h'(x) = 1 + h^2(x)

Find all functions satisfying these conditions.

(2) h : J -> |R

(3) h(0)=0

(3) /\x:-J h'(x) = 1 + h^2(x)

Find all functions satisfying these conditions.

J c ]-pi/2 , pi/2[

h(x) = tan(x)

page 161 in OLDTIMER

h(x) = tan(x)

page 161 in OLDTIMER

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f nearly uniformly;

(2) f[n]->f in measure.

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f nearly uniformly;

(2) f[n]->f in measure.

Is it true that (1)=>(2) ?

Yes.

page 83 in MSW measure

page 86 in MSW measure

page 83 in MSW measure

page 86 in MSW measure

(1) J is a connected subset of |R and 0:-J

(2) h : J -> |R

(3) h(0)=1

(3) /\x:-J h'(x) = - h^2(x)

Find all functions satisfying these conditions.

(2) h : J -> |R

(3) h(0)=1

(3) /\x:-J h'(x) = - h^2(x)

Find all functions satisfying these conditions.

J c ]-1;oo[

h(x) = 1/(1+x)

page 162 in OLDTIMER

h(x) = 1/(1+x)

page 162 in OLDTIMER

If k:-|N, and f,g:|R->|R, let d[k](f,g) = sup{|f(x)-g(x)| : -
k<=x<=k}.

Notice that /\k:-|N function d[k] is a oo-pseudometric on |R^|R.

Let W = {U c |R^|R : U is d[k]-open for some k:-|N}.

Prove that W is a topology.

Notice that /\k:-|N function d[k] is a oo-pseudometric on |R^|R.

Let W = {U c |R^|R : U is d[k]-open for some k:-|N}.

Prove that W is a topology.

page 81 in gen top

Let X and K be non-empty sets.

Let S : X x K -> P(X).

Let W = { U c X : /\a:-U \/p:-K S(a,p) c U }.

What can we conclude about set W ?

Let S : X x K -> P(X).

Let W = { U c X : /\a:-U \/p:-K S(a,p) c U }.

What can we conclude about set W ?

/\t:-T A[t]:-W |=> U(t:-T)_A[t] :- W

page 82 in gen top

page 82 in gen top

Let X,Y be sets.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

If g:X->Y and if ZcX is countable, then B(Z,g) ????.

If ZcX, f:X->Y, let B(Z,f) = {g:X->Y : /\x:-Z f(x)=g(x)}.

Let W = {U c Y^X : /\f:-U \/ZcX Z is countable and B(Z,f) c U}.

If g:X->Y and if ZcX is countable, then B(Z,g) ????.

If g:X->Y and if ZcX is countable, then B(Z,g) :- W.

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Clo( //\\(t:-T)_A[t] ) c ???

(1) /\(t:-T) A[t] c X

Clo( //\\(t:-T)_A[t] ) c ???

Clo( //\\(t:-T) A[t] ) c //\\(t:-T) Clo(A[t])

page 13 in gen top

page 13 in gen top

Let X be the set of all functions |N->{0,1}.

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E be a finite subset of X.

What can we conclude?

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E be a finite subset of X.

What can we conclude?

(1) X\E :- W

(2) if E is not empty, then E #- W

page 79 in gen top

(2) if E is not empty, then E #- W

page 79 in gen top

Let X be the set of all functions |N->{0,1}.

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E be a non-empty finite or infinitely countable subset of X.

Can we conclude that E does not belong to W ?

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E be a non-empty finite or infinitely countable subset of X.

Can we conclude that E does not belong to W ?

E does not belong to W

page 79 in gen top

page 79 in gen top

Let X be the set of all functions |N->{0,1}.

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E = X \ {0, 1({1}), 1({2}), 1({3}), ..., 1({n}), 1({n+1}), ...}.

What can we conclude about set E ?

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Let E = X \ {0, 1({1}), 1({2}), 1({3}), ..., 1({n}), 1({n+1}), ...}.

What can we conclude about set E ?

E :- W

page 79 in gen top

page 79 in gen top

Let X be the set of all functions |N->{0,1}.

Let W = { UcX : /\(f:-U) \/(finite Ac|N) /\(g:-X) (g=f on A => g:-U) }.

Let E be an infinitely countable subset of X.

Can we conclude that X\E does not belong to W ?

Let W = { UcX : /\(f:-U) \/(finite Ac|N) /\(g:-X) (g=f on A => g:-U) }.

Let E be an infinitely countable subset of X.

Can we conclude that X\E does not belong to W ?

No.

Let E = {0, 1({1}), 1({2}), 1({3}), ..., 1({n}), 1({n+1}), ...}.

X\E belongs to W.

page 79 in gen top

Let E = {0, 1({1}), 1({2}), 1({3}), ..., 1({n}), 1({n+1}), ...}.

X\E belongs to W.

page 79 in gen top

Let X be the set of all functions |N->{0,1}.

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Is W a topology?

Let W = { UcX : /\f:-U \/Ac|N A is finite and /\g:-X (g=f on A => g:-U) }.

Is W a topology?

Yes.

page 79 in gen top

page 79 in gen top

Let f:|R->|R be f(x)=x/(1+x^2).

Find a diffable function F:|R->|R such that F'=f.

Find a diffable function F:|R->|R such that F'=f.

F(x) = 1/2 * log(1+x^2)

Let X,Y be normed vector spaces.

Let f:X->Y be linear.

Suppose \/M:-|R /\x:-X ||f(x)|| < M.

What can we conclude?

Let f:X->Y be linear.

Suppose \/M:-|R /\x:-X ||f(x)|| < M.

What can we conclude?

/\x:-X f(x)=0

Let b,a:-|R.

max{a*x^2 + b*y^2 : x,y:-|R, x^2+y^2=1} = ???

max{a*x^2 + b*y^2 : x,y:-|R, x^2+y^2=1} = ???

max{a*x^2 + b*y^2 : x,y:-|R, x^2+y^2=1} = max{a,b}

Let (X,M,u) be a measure space.

Let f:X->[0,oo] be measurable.

Let a>0. Suppose that u( {x:-X : f(x)>a} ) > 0.

What can we conclude?

Let f:X->[0,oo] be measurable.

Let a>0. Suppose that u( {x:-X : f(x)>a} ) > 0.

What can we conclude?

Leb*(X,f,du) > 0

page 15 in 2nd measure

page 15 in 2nd measure

Let (X,M,u) be a measure space.

Let f:X->[0,oo] be measurable.

Suppose that Leb*(X,f,du)=0.

What can we conclude about the set {x:-X : f(x)>a>0} ?

Let f:X->[0,oo] be measurable.

Suppose that Leb*(X,f,du)=0.

What can we conclude about the set {x:-X : f(x)>a>0} ?

It has measure zero.

page 15 in 2nd measure

page 15 in 2nd measure

Let (X,M,u) be a measure space.

Let f:X->[0,oo] be measurable.

Suppose that Leb*(X,f,du)=0.

What can we conclude about the set {x:-X : f(x)>0} ?

Let f:X->[0,oo] be measurable.

Suppose that Leb*(X,f,du)=0.

What can we conclude about the set {x:-X : f(x)>0} ?

It has measure zero.

page 16 in 2nd measure

page 16 in 2nd measure

Let (X,M,u) be a measure space.

Let f:X->[0,oo] be measurable.

Suppose that u( {x:-X : f(x)>0} ) > 0.

What can we conclude?

Let f:X->[0,oo] be measurable.

Suppose that u( {x:-X : f(x)>0} ) > 0.

What can we conclude?

Leb*(X,f,du) > 0

page 16 in 2nd measure

page 16 in 2nd measure

Let (X,M,u) be a measure space.

Let f,g:X->[0,oo] be measurable. Let E:-M.

Suppose that f is integrable on E and g is bounded on E.

What can we conclude?

Let f,g:X->[0,oo] be measurable. Let E:-M.

Suppose that f is integrable on E and g is bounded on E.

What can we conclude?

function f*g is integrable on E

page 17 in 2nd measure

page 17 in 2nd measure

Let (X,M,u) be a measure space.

Let f,g:X->|R* be measurable. Let E:-M.

Suppose that f is integrable on E and g is bounded on E.

What can we conclude?

Let f,g:X->|R* be measurable. Let E:-M.

Suppose that f is integrable on E and g is bounded on E.

What can we conclude?

function f*g is integrable on E

page 17 in 2nd measure

page 17 in 2nd measure

Let y* be the outer Lebesgue measure on |R^p.

Let Ac|R^p, M :- M(y), A c M, M\A :- M(y).

Can we conclude that A :- M(y) ?

Let Ac|R^p, M :- M(y), A c M, M\A :- M(y).

Can we conclude that A :- M(y) ?

Yes.

M\(M\A):-M(y)

hence A:-M(y)

M\(M\A):-M(y)

hence A:-M(y)

Let (X,F,y) be a measure space in which:

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Let f,g:X->|R*. Let f be measurable.

Suppose f~g. Can we conclude that g is also measurable?

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Let f,g:X->|R*. Let f be measurable.

Suppose f~g. Can we conclude that g is also measurable?

Yes. Property (1) guarantees that.

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 25 in 2nd measure

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 25 in 2nd measure

Consider the Lebesgue measure on |R^p.

Let f,g : |R^p -> |R*.

Suppose that f is measurable and f~g.

Can we conclude that g is also measurable?

Let f,g : |R^p -> |R*.

Suppose that f is measurable and f~g.

Can we conclude that g is also measurable?

Yes.

The Lebesgue measure is complete, we have:

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Property (1) guarantees the result.

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 25 in 2nd measure

The Lebesgue measure is complete, we have:

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Property (1) guarantees the result.

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 25 in 2nd measure

Let (X,M,u) be a measure space.

Let A,B:-M, AnB=O. Let f:X->|R*.

Suppose that f is integrable on A, and integrable on B.

Can we conclude that f is integrable on AuB ?

Let A,B:-M, AnB=O. Let f:X->|R*.

Suppose that f is integrable on A, and integrable on B.

Can we conclude that f is integrable on AuB ?

Yes.

page 26 in 2nd measure

page 26 in 2nd measure

Let (X,F,y) be a measure space in which:

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Let f,g:X->|R*. Let f be integrable.

Suppose f~g. Can we conclude that g is also integrable?

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

Let f,g:X->|R*. Let f be integrable.

Suppose f~g. Can we conclude that g is also integrable?

Yes. Property (1) guarantees that.

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 26 in 2nd measure

f~g <=> [ \/(A:-F, y(A)=0) f=g on X\A ]

page 26 in 2nd measure

Consider the Lebesgue measure on |R^p.

Let f,g : |R^p -> |R*.

Suppose that f is integrable and f~g.

Can we conclude that g is also integrable?

Let f,g : |R^p -> |R*.

Suppose that f is integrable and f~g.

Can we conclude that g is also integrable?

Yes.

page 26 in 2nd measure

page 26 in 2nd measure

(1) X is a metric space

(2) f:X->|R

(3) E[n] is a sequence of open subsets of X

(4) x belongs to the intersection of all E[n]

(5) lim(n->oo) sup_f(E[n]) - inf_f(E[n]) = 0

What can we conclude?

(2) f:X->|R

(3) E[n] is a sequence of open subsets of X

(4) x belongs to the intersection of all E[n]

(5) lim(n->oo) sup_f(E[n]) - inf_f(E[n]) = 0

What can we conclude?

function f is continuous at x

page 27 in 2nd measure

page 27 in 2nd measure

(1) X is a metric space

(2) f:X->|R

(3) E[n] is a sequence of subsets of X

(4) lim(n->oo) diam( E[n] ) = 0

(5) x belongs to the intersection of all E[n]

(6) function f is continuous at x

Prove that:

(2) f:X->|R

(3) E[n] is a sequence of subsets of X

(4) lim(n->oo) diam( E[n] ) = 0

(5) x belongs to the intersection of all E[n]

(6) function f is continuous at x

Prove that:

page 27 in 2nd measure

(1) (X,d) is a metric space

(2) f:X->|R

(3) f is continuous at p:-X

(4) f(p)>0

Prove that:

(5) \/M>0 /\x:-X d(x,p)<M => f(x)>M

(2) f:X->|R

(3) f is continuous at p:-X

(4) f(p)>0

Prove that:

(5) \/M>0 /\x:-X d(x,p)<M => f(x)>M

page 37 in 2nd measure

Let f:[a,b]->[0,oo) be Riemann integrable.

Suppose that f is continuous at p:-[a,b] and f(p)>0.

What can we conclude?

Suppose that f is continuous at p:-[a,b] and f(p)>0.

What can we conclude?

Integral(a,b,f(x),dx) > 0

page 37 in 2nd measure

page 37 in 2nd measure

Consider a real-valued continuously diffable function defined
on a bounded closed interval. Does it have to be a Lipschitz
function?

Yes.

When a function defined on a connected subset of |R is diffable and its derivative is bounded, then this function is Lipschitz. In our case, the derivative is continuous on a bounded closed interval, hence the derivative is bounded.

page 159 in golden gate

When a function defined on a connected subset of |R is diffable and its derivative is bounded, then this function is Lipschitz. In our case, the derivative is continuous on a bounded closed interval, hence the derivative is bounded.

page 159 in golden gate

Let f,g:[a,b]->|R be Riemann integrable.

Can we conclude that f*g is also Riemann integrable?

Can we conclude that f*g is also Riemann integrable?

Yes.

page 35 in 2nd measure

page 35 in 2nd measure

Let f,g:[a,b]->|R be Riemann integrable.

Suppose that g>0 on [a,b].

Can we conclude that f/g is also Riemann integrable?

Suppose that g>0 on [a,b].

Can we conclude that f/g is also Riemann integrable?

No.

f=1 on [0,1]

g(0)=7, g(x)=x on (0,1]

f/g is not even bounded on [0,1],

hence it cannot be Riemann integrable

f=1 on [0,1]

g(0)=7, g(x)=x on (0,1]

f/g is not even bounded on [0,1],

hence it cannot be Riemann integrable

Let g:[a,b]->|R be Riemann integrable.

Suppose that /\x:-[a,b] |g(x)|>e>0.

Can we conclude that 1/g is Riemann integrable?

Suppose that /\x:-[a,b] |g(x)|>e>0.

Can we conclude that 1/g is Riemann integrable?

Yes.

page 38 in 2nd measure

page 38 in 2nd measure

Let f,g:[a,b]->|R be Riemann integrable.

Suppose that /\x:-[a,b] |g(x)|>e>0.

Can we conclude that f/g is Riemann integrable?

Suppose that /\x:-[a,b] |g(x)|>e>0.

Can we conclude that f/g is Riemann integrable?

Yes.

pages 38,35 in 2nd measure

pages 38,35 in 2nd measure

Let M be a set whose elements are topologies on X.

Can we conclude that //\\M is also a topology on X?

Can we conclude that //\\M is also a topology on X?

Yes.

Let (X,G) be a topological space.

What does it mean that it is metrizable?

What does it mean that it is metrizable?

There exists a metric d on X such that the d-topology induced
by the metric d is the same as the topology G.

Prove that in a topological space, the intersection of a
collection of closed sets is again a closed set.

page 7 in gen top

Prove that in a topological space X, the empty set and the set
X are both closed.

page 7 in gen top

Let (X,G) be a topological space.

What is the definition of a closed set?

What is the definition of a closed set?

A is closed <=> X\A belongs to G

page 6 in gen top

page 6 in gen top

Let (X,G) be a topological space.

What is the definition of an open set?

What is the definition of an open set?

A is open <=> A belongs to G

page 6 in gen top

page 6 in gen top

What is the definition of a topological space?

(X,G) is a topological space

iff

(1) X is a set and G c P(X)

(2) O:-G and X:-G

(3) A,B:-G => AnB:-G

(4) /\McG \\//M :- G

iff

(1) X is a set and G c P(X)

(2) O:-G and X:-G

(3) A,B:-G => AnB:-G

(4) /\McG \\//M :- G

Let (X,W) be a topological space. Let A c X.

(1) /\x:-A \/G:-W GcA and x:-G

What can we conclude?

(1) /\x:-A \/G:-W GcA and x:-G

What can we conclude?

A is open

page 7 in gen top

page 7 in gen top

Let (X,W) be a topological space. Let AcX.

What is the definition of "Int(A)" ?

What is the definition of "Int(A)" ?

Int(A) = \\//{G:-W : GcA}

Int(A) is the union of all open sets contained in A.

Hence Int(A) is an open set and Int(A) c A.

Int(A) is the union of all open sets contained in A.

Hence Int(A) is an open set and Int(A) c A.

In a topological space, is it always true that Int(A) c A ?

Yes.

page 8 in gen top

page 8 in gen top

Let (X,W) be a topological space. Let AcX.

What is the definition of "Clo(A)" ?

What is the definition of "Clo(A)" ?

Clo(A) = //\\{FcX : AcF and F is closed}

Clo(A) is the intersection of all closed sets containing A.

Hence Clo(A) is a closed set,

Notice that X is closed and AcX, hence A c Clo(A).

page 9 in gen top

Clo(A) is the intersection of all closed sets containing A.

Hence Clo(A) is a closed set,

Notice that X is closed and AcX, hence A c Clo(A).

page 9 in gen top

In a topological space, is it always true that A c Clo(A)?

Yes.

page 9 in gen top

page 9 in gen top

Let (X,W) be a topological space. Let AcX.

Suppose that A c Int(A).

What can we conclude about set A ?

Suppose that A c Int(A).

What can we conclude about set A ?

A is an open set

page 9 in gen top

page 9 in gen top

Let X be a topological space. Let AcX.

Can we conclude that A c Int(A) ?

Can we conclude that A c Int(A) ?

No.

Let A be any non-open set.

Then A is not contained in Int(A).

page 9 in gen top

Let A be any non-open set.

Then A is not contained in Int(A).

page 9 in gen top

Let (X,W) be a topological space. Let AcX.

Suppose that Clo(A) c A.

What can we conclude?

Suppose that Clo(A) c A.

What can we conclude?

A is a closed set

page 9 in gen top

page 9 in gen top

Let (X,W) be a topological space. Let AcX.

Suppose that A is closed.

Can we conclude that Clo(A) c A ?

Suppose that A is closed.

Can we conclude that Clo(A) c A ?

Yes.

page 9 in gen top

page 9 in gen top

Let (X,W) be a topological space. Let AcX.

X \ Int(A) = ???

X \ Int(A) = ???

X \ Int(A) = Clo( X\A )

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

Clo( X\A ) = ???

Clo( X\A ) = ???

Clo( X\A ) = X \ Int(A)

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

X \ Clo(X\A) = ???

X \ Clo(X\A) = ???

X \ Clo(X\A) = Int(A)

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

Int(A) = X \ ???

Int(A) = X \ ???

Int(A) = X \ Clo(X\A)

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

Int(X\A) = ???

Int(X\A) = ???

Int(X\A) = X \ Clo(A)

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

X \ Int(X\A) = ???

X \ Int(X\A) = ???

X \ Int(X\A) = Clo(A)

page 10 in gen top

page 10 in gen top

Let (X,W) be a topological space. Let AcX.

Clo(A) = X \ ???

Clo(A) = X \ ???

Clo(A) = X \ Int(X\A)

page 10 in gen top

page 10 in gen top

Let X be a topological space. Let A,B c X.

Suppose that B is open and Int(A) c B c A.

What can we conclude?

Suppose that B is open and Int(A) c B c A.

What can we conclude?

B = Int(A)

page 11 in gen top

page 11 in gen top

Let X be a topological space. Let A,B c X.

Suppose that B is closed and A c B c Clo(A).

What can we conclude?

Suppose that B is closed and A c B c Clo(A).

What can we conclude?

B = Clo(A)

page 11 in gen top

page 11 in gen top

Let X be a topological space. Let A,B c X.

Suppose that B c A and B is open.

What can we conclude?

Suppose that B c A and B is open.

What can we conclude?

B c Int(A)

page 11 in gen top

page 11 in gen top

Let X be a topological space. Let A,B c X.

Suppose that A c B and B is closed.

What can we conclude?

Suppose that A c B and B is closed.

What can we conclude?

Clo(A) c B

page 12 in gen top

page 12 in gen top

Let X be a topological space. Let A,B c X.

Suppose that A c B.

We can immediately conclude that Int(A) ??? .

Suppose that A c B.

We can immediately conclude that Int(A) ??? .

Int(A) c Int(B)

page 12 in gen top

Int(A) c A c B, hence Int(A) c Int(B)

page 12 in gen top

Int(A) c A c B, hence Int(A) c Int(B)

Let X be a topological space. Let A,B c X.

Suppose that A c B.

We can immediately conclude that Int(B) ??? .

Suppose that A c B.

We can immediately conclude that Int(B) ??? .

Int(A) c Int(B)

page 12 in gen top

Int(A) c A c B, hence Int(A) c Int(B)

page 12 in gen top

Int(A) c A c B, hence Int(A) c Int(B)

Let X be a topological space. Let A,B c X.

Suppose that Int(A) c Int(B).

Can we conclude that A c B ?

Suppose that Int(A) c Int(B).

Can we conclude that A c B ?

No.

A = [0,1].

B = [0,1).

A = [0,1].

B = [0,1).

Let X be a topological space. Let A,B c X.

Suppose that Int(A) c Int(B).

Is it possible that B is properly contained in A ?

Suppose that Int(A) c Int(B).

Is it possible that B is properly contained in A ?

Yes.

A = [0,1]

B = (0,1]

A = [0,1]

B = (0,1]

Let X be a topological space. Let A,B c X.

Suppose that A c B.

We can immediately conclude

that Clo(A) ??? .

Suppose that A c B.

We can immediately conclude

that Clo(A) ??? .

Clo(A) c Clo(B)

page 12 in gen top

A c B c Clo(B), hence Clo(A) c Clo(B)

page 12 in gen top

A c B c Clo(B), hence Clo(A) c Clo(B)

Let X be a topological space. Let A,B c X.

Suppose that A c B.

We can immediately conclude

that Clo(B) ??? .

Suppose that A c B.

We can immediately conclude

that Clo(B) ??? .

Clo(A) c Clo(B)

page 12 in gen top

A c B c Clo(B), hence Clo(A) c Clo(B)

page 12 in gen top

A c B c Clo(B), hence Clo(A) c Clo(B)

Let X be a topological space. Let A,B c X.

Suppose that Clo(B) c Clo(A).

Can we conclude that B c A ?

Suppose that Clo(B) c Clo(A).

Can we conclude that B c A ?

No.

B = [0,1]

A = [0,1)

B = [0,1]

A = [0,1)

Let X be a topological space. Let A,B c X.

Suppose that Clo(B) c Clo(A).

Is it possible that A is properly contained in B?

Suppose that Clo(B) c Clo(A).

Is it possible that A is properly contained in B?

Yes.

B = [0,1]

A = [0,1)

B = [0,1]

A = [0,1)

Let X be a topological space. Let A,B c X.

Suppose that Clo(B) c Clo(A) and A,B are not empty.

Is it possible that B c X\A ?

Suppose that Clo(B) c Clo(A) and A,B are not empty.

Is it possible that B c X\A ?

Yes.

B = |Q

A = |R \ |Q

B = |Q

A = |R \ |Q

Let X be a topological space. Let A,B c X.

Suppose that Int(A) c Int(B) and A,B are not empty.

Is it possible that A c X\B ?

Suppose that Int(A) c Int(B) and A,B are not empty.

Is it possible that A c X\B ?

Yes.

A = |Q

B = |R \ |Q

=========

A = {1}

B = {2}

A = |Q

B = |R \ |Q

=========

A = {1}

B = {2}

Let X be a topological space. Let A,B c X.

Suppose that Int(A) c Int(B) and Int(A) is not empty.

Is it possible that A c X\B ?

Suppose that Int(A) c Int(B) and Int(A) is not empty.

Is it possible that A c X\B ?

No.

We have {x} c Int(A) c Int(B).

Hence {x} c A and {x} c B.

x :- AnB

We have {x} c Int(A) c Int(B).

Hence {x} c A and {x} c B.

x :- AnB

Let X be a topological space. Let A c X.

Int(Int(A)) = ???

Int(Int(A)) = ???

Int(Int(A)) = Int(A)

page 9 in gen top

page 9 in gen top

Let X be a topological space. Let A c X.

Clo(Clo(A)) = ???

Clo(Clo(A)) = ???

Clo(Clo(A)) = Clo(A)

page 9 in gen top

page 9 in gen top

Let (X,M,u) be a measure space.

Let f:X->|R* be integrable.

Can we conclude that |f| is also integrable?

Let f:X->|R* be integrable.

Can we conclude that |f| is also integrable?

Yes.

page 188 in 1st measure

page 188 in 1st measure

Let (X,M,u) be a measure space.

Let f:X->|R* be measurable.

Suppose that |f| is integrable.

Can we conclude that f is also integrable?

Let f:X->|R* be measurable.

Suppose that |f| is integrable.

Can we conclude that f is also integrable?

Yes.

hint:

(f+) <= |f|

(f-) <= |f|

hint:

(f+) <= |f|

(f-) <= |f|

Let (X,M,u) be a measure space.

Let f,g:X->|R* be measurable.

Suppose that |f|<=g and g is integrable.

Can we conclude that f is also integrable?

Let f,g:X->|R* be measurable.

Suppose that |f|<=g and g is integrable.

Can we conclude that f is also integrable?

Yes.

hint:

(f+) <= |f| <= g

(f-) <= |f| <= g

hint:

(f+) <= |f| <= g

(f-) <= |f| <= g

Let (X,M,u) be a measure space.

Let f,g:X->|R* be measurable.

Suppose that f is integrable and f~g.

What can we conclude about function g?

Let f,g:X->|R* be measurable.

Suppose that f is integrable and f~g.

What can we conclude about function g?

g is also integrable and Leb(X,g,du) = Leb(X,f,du)

Let (X,M,u) be a measure space.

Let f:X->|R* be integrable.

Can we conclude that

there exists a function g:X->|R\{-oo,oo}

that is integrable and g~f ?

Let f:X->|R* be integrable.

Can we conclude that

there exists a function g:X->|R\{-oo,oo}

that is integrable and g~f ?

Yes.

Use:

(1) Theorem 151 page 188 in 1st measure

(2) Theorem 170 page 26 in 2nd measure

Use:

(1) Theorem 151 page 188 in 1st measure

(2) Theorem 170 page 26 in 2nd measure

Consider the Lebesgue measure on |R^p: (|R^p, M(y), y).

Let Ac|R^p be open and let f:A->|R be integrable on A and continuous at x:-A.

Prove that:

/\e>0 \/G open, x:-G and /\(open BcG) |Leb(B,f,dy) / y(B) - f(x)| < e.

Let Ac|R^p be open and let f:A->|R be integrable on A and continuous at x:-A.

Prove that:

/\e>0 \/G open, x:-G and /\(open BcG) |Leb(B,f,dy) / y(B) - f(x)| < e.

page 65 in MSW measure

Consider the Lebesgue measure on |R^p: (|R^p, M(y), y).

Let Ac|R^p be open and let f:A->|R be integrable. Let x:-A.

Suppose that:

/\e>0 \/G open, x:-G and /\(open BcG) |Leb(B,f,dy) / y(B) - f(x)| < e.

Can we conclude that f is continuous at x:-A ?

Let Ac|R^p be open and let f:A->|R be integrable. Let x:-A.

Suppose that:

/\e>0 \/G open, x:-G and /\(open BcG) |Leb(B,f,dy) / y(B) - f(x)| < e.

Can we conclude that f is continuous at x:-A ?

No.

If x is irrational, let f(x) = 0.

If x is rational, let f(x) = 7.

Then f satisfies the condition at every irrational point,

and it is discontinuous at every point.

If x is irrational, let f(x) = 0.

If x is rational, let f(x) = 7.

Then f satisfies the condition at every irrational point,

and it is discontinuous at every point.

Let (X,F,y) be a measure space in which:

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

What is the adjective describing this property?

(1) /\EcX (\/A:-F EcA, y(A)=0) => E:-F.

What is the adjective describing this property?

complete

measure y is complete

page 68 in MSW measure

measure y is complete

page 68 in MSW measure

Let f:[a,b)->|R be Lebesgue integrable.

Suppose that f is continuous at a.

lim(h->0+) Leb([a,a+h],f,dy) / h = ???

Suppose that f is continuous at a.

lim(h->0+) Leb([a,a+h],f,dy) / h = ???

lim(h->0+) Leb([a,a+h],f,dy) / h = f(a)

Notice that it is important that y([a,b]) = b-a.

page 67 in MSW measure

page 146 in golden gate

Notice that it is important that y([a,b]) = b-a.

page 67 in MSW measure

page 146 in golden gate

Let X be a topological space. Let AcX.

Is it always true that Clo(Int(A)) = Clo(A) ?

Is it always true that Clo(Int(A)) = Clo(A) ?

No.

Int(|Q) = O

Clo(|Q) = |R

Int(|Q) = O

Clo(|Q) = |R

Let X be a topological space. Let AcX.

Is it always true that Int(Clo(A)) = Int(A) ?

Is it always true that Int(Clo(A)) = Int(A) ?

No.

Int(|Q) = O

Clo(|Q) = |R

Int(|Q) = O

Clo(|Q) = |R

Let X be a topological space. Let AcX.

Suppose that Int(A) is not empty.

Can we conclude that Clo(Int(A)) = Clo(A) ?

Suppose that Int(A) is not empty.

Can we conclude that Clo(Int(A)) = Clo(A) ?

No.

A = |Q u [0,1]

Int(A) = (0,1)

Clo(Int(A)) = [0,1]

Clo(A) = |R

A = |Q u [0,1]

Int(A) = (0,1)

Clo(Int(A)) = [0,1]

Clo(A) = |R

Let X be a topological space. Let AcX.

Suppose that Int(A) is not empty.

Can we conclude that Int(Clo(A)) = Int(A) ?

Suppose that Int(A) is not empty.

Can we conclude that Int(Clo(A)) = Int(A) ?

No.

A = |Q u [0,1]

Int(A) = (0,1)

Clo(A) = |R

Int(Clo(A)) = |R

A = |Q u [0,1]

Int(A) = (0,1)

Clo(A) = |R

Int(Clo(A)) = |R

(1) /\(n:-|N) A[n] c X

(2) \/(k:-|N) A[k] is non-empty

(3) //\\(n:-|N)_A[n] is empty

Decide if we can conclude that

(4) U(n:-|N) A[n]\A[n+1] is non-empty

(2) \/(k:-|N) A[k] is non-empty

(3) //\\(n:-|N)_A[n] is empty

Decide if we can conclude that

(4) U(n:-|N) A[n]\A[n+1] is non-empty

No.

O,{1},{1},{1},{1},...

page 135 in 2nd measure

O,{1},{1},{1},{1},...

page 135 in 2nd measure

Let X be a topological space. Let AcX.

Suppose Int(Clo(A)) c A.

Can we conclude that Clo(A)\A is empty?

Suppose Int(Clo(A)) c A.

Can we conclude that Clo(A)\A is empty?

No.

A=[0,1)

page 84 in gen top

A=[0,1)

page 84 in gen top

Let X be a topological space. Let AcX.

Suppose Int(Clo(A)) c A.

Is it possible that Int(Clo(A)\A) is not empty?

Suppose Int(Clo(A)) c A.

Is it possible that Int(Clo(A)\A) is not empty?

No.

Int(Clo(A)\A) is always empty, no matter what.

page 84 in gen top

Int(Clo(A)\A) is always empty, no matter what.

page 84 in gen top

(1) J is a connected subset of |R

(2) h:J->|R diffable

(3) /\x:-J x*h'(x) = x - h(x)

Find all such functions.

(2) h:J->|R diffable

(3) /\x:-J x*h'(x) = x - h(x)

Find all such functions.

If 0:-J, then h(x)=x/2.

If 0#-J, then h(x) = x/2 + M/x, where M is any real number.

If 0#-J, then h(x) = x/2 + M/x, where M is any real number.

Let W be a monotone class of sets.

Let E be a set. Let K = {BnE : B:-W}.

Can we conclude that K is also monotone?

Let E be a set. Let K = {BnE : B:-W}.

Can we conclude that K is also monotone?

NO.

page 126 in 2nd measure

E = |N

W = { {1,-1}, {1,2,-2}, {1,2,3,-3}, {1,2,3,4,-4}, ... }

K = { B n |N : B:-W } = { {1}, {1,2}, {1,2,3}, {1,2,3,4}, ... }

W is monotone but K is not monotone

page 126 in 2nd measure

E = |N

W = { {1,-1}, {1,2,-2}, {1,2,3,-3}, {1,2,3,4,-4}, ... }

K = { B n |N : B:-W } = { {1}, {1,2}, {1,2,3}, {1,2,3,4}, ... }

W is monotone but K is not monotone

(1) J is a connected subset of |R

(2) h:J->|R is diffable

(3) /\x:-J 1 + h^2(x) - x*h(x)*h'(x) = 0

Find all such functions.

(2) h:J->|R is diffable

(3) /\x:-J 1 + h^2(x) - x*h(x)*h'(x) = 0

Find all such functions.

There exists M>0 such that

(1) J c (1/sqrt(M) ; oo) or J c (-oo ; -1/sqrt(M))

(2) /\x:-J h(x) = sqrt(M*x^2 - 1) OR /\x:-J h(x) = (- 1)*sqrt(M*x^2 - 1).

page 164 in OLDTIMER

(1) J c (M ; oo) or J c (-oo ; -M)

(1) J c (1/sqrt(M) ; oo) or J c (-oo ; -1/sqrt(M))

(2) /\x:-J h(x) = sqrt(M*x^2 - 1) OR /\x:-J h(x) = (- 1)*sqrt(M*x^2 - 1).

page 164 in OLDTIMER

(1) J c (M ; oo) or J c (-oo ; -M)

(1) a,b are distinct real numbers

(2) f : *[a,b]* -> |R is continuous

Check if:

(3) Integral(a,b,f(x),dx) = Integral(-b,-a,f(-x),dx).

(2) f : *[a,b]* -> |R is continuous

Check if:

(3) Integral(a,b,f(x),dx) = Integral(-b,-a,f(-x),dx).

This is true.

page 167 in OLDTIMER

page 167 in OLDTIMER

(1) J is a connected subset of |R

(2) y:J->|R is diffable

(3) /\t:-J t*y'(t) = t + y(t)

Find all such functions.

(2) y:J->|R is diffable

(3) /\t:-J t*y'(t) = t + y(t)

Find all such functions.

(1) J c |R\{0}

(2) \/M:-|R /\t:-J y(t) = t*log(|t|) + M*t

page 167 in OLDTIMER

(2) \/M:-|R /\t:-J y(t) = t*log(|t|) + M*t

page 167 in OLDTIMER

Consider the intersection of two connected sets.

Does it have to be a connected set?

Does it have to be a connected set?

No.

Let y=0 and y=x^2-1 be viewed as connected subsets of |R^2.

Their intersection { (-1,0) , (1,0) } is disconnected.

Let y=0 and y=x^2-1 be viewed as connected subsets of |R^2.

Their intersection { (-1,0) , (1,0) } is disconnected.

(1) g : |R\{0} -> |R

(2) g(x) = log(|x|)

Find the derivative of g wherever it exists.

(2) g(x) = log(|x|)

Find the derivative of g wherever it exists.

/\x:-|R\{0} g'(x) = 1/x

Let f:[1,7)->|R be f(x)=x*sqrt(log(x)).

Is f diffable?

Is f diffable?

No.

f is not diffable at 1

hint: Hospital

f is not diffable at 1

hint: Hospital

(1) J is a connected subset of [0,oo)

(2) y:J->|R is diffable

(3) /\(t:-J) t*y(t)*y'(t) = y(t)*y(t) + t*t

Find all such functions.

(2) y:J->|R is diffable

(3) /\(t:-J) t*y(t)*y'(t) = y(t)*y(t) + t*t

Find all such functions.

There exists a positive real number M such that:

(1) J c (M ; oo)

AND:

(2) /\(t:-J) y(t) = t*sqrt(2*log(t/M))

OR

(3) /\(t:-J) y(t) = - t*sqrt(2*log(t/M))

page 169 in OLDTIMER

(1) J c (M ; oo)

AND:

(2) /\(t:-J) y(t) = t*sqrt(2*log(t/M))

OR

(3) /\(t:-J) y(t) = - t*sqrt(2*log(t/M))

page 169 in OLDTIMER

Let (X,F,y) be a measure space. Let B:-F.

(1) \/A:-F y(A)=0 and X\A c B

(2) y(X\B) = 0

Are these conditions equivalent?

(1) \/A:-F y(A)=0 and X\A c B

(2) y(X\B) = 0

Are these conditions equivalent?

Yes.

Let (X,F,y) be a measure space.

What does it mean that a property holds almost everywhere?

What does it mean that a property holds almost everywhere?

The set of points having this property is contained in a
measurable set of measure zero.

Let (X,F,y) be a measure space.

Let f,g:X->|R* be arbitrary functions, not necessarily measurable.

What does it mean that f~g ?

Let f,g:X->|R* be arbitrary functions, not necessarily measurable.

What does it mean that f~g ?

\/A:-F y(A)=0 and {x:-X : f(x)!=g(x)} c A

there is A:-F such that y(A)=0 and f=g on X\A

page 72 in MSW measure

there is A:-F such that y(A)=0 and f=g on X\A

page 72 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) sequence f[n] converges pointwise to f almost everywhere

(2) /\n:-|N ( f[n] <= g almost everywhere )

Can we conclude that f <= g almost everywhere?

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) sequence f[n] converges pointwise to f almost everywhere

(2) /\n:-|N ( f[n] <= g almost everywhere )

Can we conclude that f <= g almost everywhere?

Yes.

page 73 in MSW measure

page 73 in MSW measure

Let (X,F,y) be a measure space.

What does it mean that y is a complete measure?

What does it mean that y is a complete measure?

/\EcX (\/A:-F EcA, y(A)=0) => E:-F

Every subset of a set of measure zero is measurable.

Every set contained in a set of measure zero is measurable.

N(y) c F, see page 72 in 2nd measure

page 68 in MSW measure

Every subset of a set of measure zero is measurable.

Every set contained in a set of measure zero is measurable.

N(y) c F, see page 72 in 2nd measure

page 68 in MSW measure

Let (X,F,y) be a complete measure space.

Let f[n]:X->|R* be a sequence of measurable functions.

Let f:X->|R* be an arbitrary function.

(1) f[n]->f pointwise almost everywhere

Can we conclude that f is measurable?

Let f[n]:X->|R* be a sequence of measurable functions.

Let f:X->|R* be an arbitrary function.

(1) f[n]->f pointwise almost everywhere

Can we conclude that f is measurable?

Yes.

Completeness guarantees this.

page 73 in MSW measure

Completeness guarantees this.

page 73 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) f[n]->f pointwise almost everywhere

(2) g~f

What can we conclude?

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) f[n]->f pointwise almost everywhere

(2) g~f

What can we conclude?

f[n]->g almost everywhere

page 73 in MSW measure

page 73 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) f[n]->f pointwise almost everywhere

(2) f[n]->g pointwise almost everywhere

What can we conclude?

Let f[n]:X->|R* be a sequence of arbitrary functions.

Let f,g:X->|R* be arbitrary functions.

(1) f[n]->f pointwise almost everywhere

(2) f[n]->g pointwise almost everywhere

What can we conclude?

f~g

page 73 in MSW measure

page 73 in MSW measure

Let (X,F,y) be a measure space.

Let f[n],g[n]:X->|R* be two sequences of arbitrary functions.

Let h:X->|R* be an arbitrary function.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->h pointwise almost everywhere

What can we conclude?

Let f[n],g[n]:X->|R* be two sequences of arbitrary functions.

Let h:X->|R* be an arbitrary function.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->h pointwise almost everywhere

What can we conclude?

g[n]->h pointwise almost everywhere

page 73 in MSW measure

page 73 in MSW measure

(1) f:|R->|R

(2) g:|R->|R, g(x)=exp(f(x))

(3) g is diffable

Can we conclude that f is diffable?

(2) g:|R->|R, g(x)=exp(f(x))

(3) g is diffable

Can we conclude that f is diffable?

Yes.

Since g is diffable, log(g(x)) is diffable.

Hence f is diffable.

Since g is diffable, log(g(x)) is diffable.

Hence f is diffable.

(1) f:|R->|R

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

Can we conclude that f is diffable?

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

Can we conclude that f is diffable?

No.

If x is rational, let f(x) = -x.

If x is irrational, let f(x) = x.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is nowhere diffable.

If x is rational, let f(x) = -x.

If x is irrational, let f(x) = x.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is nowhere diffable.

(1) f:|R->|R is continuous

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

Can we conclude that f is diffable?

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

Can we conclude that f is diffable?

No.

Let f(x) = |x|.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is not diffable at 0.

Let f(x) = |x|.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is not diffable at 0.

(1) f:|R->|R is continuous

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is not diffable

Is this setup possible?

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is not diffable

Is this setup possible?

Yes.

f(x) = |x|

h(x) = x*x

f(x) = |x|

h(x) = x*x

(1) f:|R->|R is continuous

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is not diffable

What can we conclude?

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is not diffable

What can we conclude?

\/x:-|R h'(x)=0

Proof.

Suppose /\x:-|R h'(x)>0. Then h-1 exists and is diffable.

Hence h-1 o h o f is diffable, and f is diffable.

Proof.

Suppose /\x:-|R h'(x)>0. Then h-1 exists and is diffable.

Hence h-1 o h o f is diffable, and f is diffable.

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) /\w>0 lim(n->oo) y( {x: |f[n](x) - f(x)| > w} ) = 0

What is the verbal description of (1) ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) /\w>0 lim(n->oo) y( {x: |f[n](x) - f(x)| > w} ) = 0

What is the verbal description of (1) ?

sequence f[n] converges in measure to f

f[n]->f in measure

page 91 in "Measure Theory" Paul Halmos

f[n]->f in measure

page 91 in "Measure Theory" Paul Halmos

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone and
countably subadditive.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

What does it mean: "sequence f[n] converges in measure to f" ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

What does it mean: "sequence f[n] converges in measure to f" ?

(1) /\w>0 lim(n->oo) y( {x: |f[n](x) - f(x)| > w} ) = 0

(2) /\a>0 \/k:-|N /\n>k y{x: |f[n](x)-f(x)|>a} < a

f[n]->f in measure

page 91 in "Measure Theory" Paul Halmos

page 81 in MSW measure

(2) /\a>0 \/k:-|N /\n>k y{x: |f[n](x)-f(x)|>a} < a

f[n]->f in measure

page 91 in "Measure Theory" Paul Halmos

page 81 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n],g[n]:X->|R be two sequences of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) /\n:-|N f[n]~g[n]

(3) f~g

Let f[n],g[n]:X->|R be two sequences of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) /\n:-|N f[n]~g[n]

(3) f~g

(4) g[n]->g in measure

page 75 in MSW measure

page 75 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) f[n]->g in measure

What can we conclude?

Let f[n]:X->|R be a sequence of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) f[n]->g in measure

What can we conclude?

f~g

page 77 in MSW measure

page 77 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) f[n]->g in measure

Can we conclude that f=g ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure

(2) f[n]->g in measure

Can we conclude that f=g ?

No.

f[n]=g[n]=f=0 on X

g=0 on X\{a}

We can only conclude that f~g.

page 77 in MSW measure

f[n]=g[n]=f=0 on X

g=0 on X\{a}

We can only conclude that f~g.

page 77 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure

Prove that there exists k:|N->|N strictly increasing

such that f[k(n)]->f pointwise almost everywhere as n->oo.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure

Prove that there exists k:|N->|N strictly increasing

such that f[k(n)]->f pointwise almost everywhere as n->oo.

page 137 in 2nd measure

page 77 in MSW measure

page 77 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise almost everywhere

(2) y(X) < oo

Can we conclude that f[n]->f in measure ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise almost everywhere

(2) y(X) < oo

Can we conclude that f[n]->f in measure ?

Yes.

page 77 in MSW measure

page 77 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise almost everywhere

Can we conclude that f[n]->f in measure ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise almost everywhere

Can we conclude that f[n]->f in measure ?

No.

Let X = |R. Let f[n](x) = x/n.

Notice y(X)=oo.

If y(X)<oo, then we could conclude that f[n]->f in measure.

see page 77 in MSW measure

Let X = |R. Let f[n](x) = x/n.

Notice y(X)=oo.

If y(X)<oo, then we could conclude that f[n]->f in measure.

see page 77 in MSW measure

(1) J c |R is connected, 1:-J

(2) y:J->|R is diffable

(3) y(1)=1

(4) /\t:-J t*t*y'(t) + y(t)*y(t) = 0

Find all such functions.

(2) y:J->|R is diffable

(3) y(1)=1

(4) /\t:-J t*t*y'(t) + y(t)*y(t) = 0

Find all such functions.

J c (1/2 ; oo)

/\t:-J y(t) = t/(2t-1)

page 171 in OLDTIMER

/\t:-J y(t) = t/(2t-1)

page 171 in OLDTIMER

(1) y:|R->|R is diffable

(2) y(0) = -1

(3) /\t:-|R y'(t) = - exp(y(t)+t+1)

Find all such functions.

(2) y(0) = -1

(3) /\t:-|R y'(t) = - exp(y(t)+t+1)

Find all such functions.

There is one and only one such function.

y(t) = -(t+1)

===================

-y'(t)*exp(-y(t)) = exp(t+1)

[ exp(-y(t)) ]' = [ exp(t+1) ]'

y(t) = -(t+1)

===================

-y'(t)*exp(-y(t)) = exp(t+1)

[ exp(-y(t)) ]' = [ exp(t+1) ]'

Let X be a topological space. Let AcX.

Suppose that Clo(Int(A)) c A.

Can we conclude that A is closed?

Suppose that Clo(Int(A)) c A.

Can we conclude that A is closed?

No.

A = |Q u [0,1]

A = |Q

Int(A)=O

A = |Q u [0,1]

A = |Q

Int(A)=O

Let A c |R be bounded above.

Let x :- Clo(A) \ Int(A).

Can we conclude that sup(A) = sup(A\{x}) ?

Let x :- Clo(A) \ Int(A).

Can we conclude that sup(A) = sup(A\{x}) ?

No.

A = {1,2}

x = 2

see page 139 in OLDTIMER

A = {1,2}

x = 2

see page 139 in OLDTIMER

Let X be a topological space. Let AcX.

Suppose that A c Clo(Int(A)).

Can we conclude that A is open?

Suppose that A c Clo(Int(A)).

Can we conclude that A is open?

No.

A = [0,1)

A = [0,1)

(1) a,b : |N -> |R*

(2) /\n \/k>n a(n)<=b(k)

Decide if we can conclude that

(3) lim_inf(n->oo)_a(n) <= lim_inf(k->oo)_b(k).

(2) /\n \/k>n a(n)<=b(k)

Decide if we can conclude that

(3) lim_inf(n->oo)_a(n) <= lim_inf(k->oo)_b(k).

No.

a(n)=0 for all n

b(2n)=1

b(2n+1)=-n

a(n)=0 for all n

b(2n)=1

b(2n+1)=-n

Let 0 < x < pi/2.

tan(x - pi/2) = ???

tan(x - pi/2) = ???

tan(x - pi/2) = (-1)/tan(x)

page 173 in OLDTIMER

page 173 in OLDTIMER

(1) J c |R is connected, 1:-J

(2) y:J->|R is diffable

(3) y(1) = -1

(4) /\t:-J y'(t) = ( 1 + y(t)*y(t) ) / (1+t*t)

Find all such functions.

(2) y:J->|R is diffable

(3) y(1) = -1

(4) /\t:-J y'(t) = ( 1 + y(t)*y(t) ) / (1+t*t)

Find all such functions.

J c (0,oo)

y(t) = (-1)/t

page 174 in OLDTIMER

y(t) = (-1)/t

page 174 in OLDTIMER

(1) J c |R is connected, 0:-J

(2) y:J->|R is diffable

(3) y(0) = 0

(4) /\t:-J exp(y(t)) * (y'(t)-1) = 1

Find all such functions.

(2) y:J->|R is diffable

(3) y(0) = 0

(4) /\t:-J exp(y(t)) * (y'(t)-1) = 1

Find all such functions.

J c (-log(2) ; oo)

y(t) = log(2*exp(t) - 1)

page 175 in OLDTIMER

y(t) = log(2*exp(t) - 1)

page 175 in OLDTIMER

Let f:|R->|R be f(x) = exp(x) / (1+exp(x)).

Find a diffable function F:|R->|R such that F'=f.

Find a diffable function F:|R->|R such that F'=f.

F(x) = log( (1+exp(x) )

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Let a,b be real numbers.

((a+b)+) <= ???

Let a,b be real numbers.

((a+b)+) <= ???

((a+b)+) <= (a+) + (b+)

page 127 in 2nd measure

page 127 in 2nd measure

If h:-|R*, then let (h+) = max(0,h), and let (h-) = max(0,-h).

Let a,b be real numbers.

((a+b)-) <= ???

Let a,b be real numbers.

((a+b)-) <= ???

((a+b)-) <= (a-) + (b-)

page 128 in 2nd measure

page 128 in 2nd measure

(1) y:|R->|R diffable

(2) /\t:-|R y'(t) = 2t*y(t)

Find all such functions.

(2) /\t:-|R y'(t) = 2t*y(t)

Find all such functions.

y(t) = M*exp(t^2), where M is any real number

page 176 in OLDTIMER

page 176 in OLDTIMER

(1) y:|R->|R diffable

(2) /\t:-|R t*y'(t) - y(t) = t^2*cos(t)

Find all such functions.

(2) /\t:-|R t*y'(t) - y(t) = t^2*cos(t)

Find all such functions.

y(t) = t*sin(t) + M*t, where M is any real number

hint: [y(t)/t]'

hint: [y(t)/t]'

Suppose that cos(x/2)=0.

Can we conclude that sin(x)=0 ?

Can we conclude that sin(x)=0 ?

Yes.

0 = cos(x/2) = 2*sin(x/2)*cos(x/2) = sin(x)

0 = cos(x/2) = 2*sin(x/2)*cos(x/2) = sin(x)

Suppose that sin(x)=0.

Can we conclude that sin(2x)=0 ?

Can we conclude that sin(2x)=0 ?

Yes.

0 = sin(x) = 2*sin(x)*cos(x) = sin(2x)

0 = sin(x) = 2*sin(x)*cos(x) = sin(2x)

Suppose that sin(x)=0.

Can we conclude that sin(x/2)=0 ?

Can we conclude that sin(x/2)=0 ?

No.

sin(pi) = 0

sin(pi/2) = 1

sin(pi) = 0

sin(pi/2) = 1

Suppose that sin(x)=0.

Can we conclude that sin(x/2) != 0 ?

Can we conclude that sin(x/2) != 0 ?

No.

sin(2*pi) = 0

sin(pi) = 0

===========

sin(0)=0 and sin(0/2)=0

sin(2*pi) = 0

sin(pi) = 0

===========

sin(0)=0 and sin(0/2)=0

Suppose that sin(x)=0.

Can we conclude that cos(x/2)=0 ?

Can we conclude that cos(x/2)=0 ?

No.

1) sin(2*pi) = 0 and cos(pi) = -1

2) sin(0) = 0 and cos(0/2) = 1

1) sin(2*pi) = 0 and cos(pi) = -1

2) sin(0) = 0 and cos(0/2) = 1

Decide if it is true:

cos(x/2)=0 <=> cos(x) = -1

cos(x/2)=0 <=> cos(x) = -1

Yes.

(1) cos(x) = cos^2(x/2) - sin^2(x/2)

(2) cos(x) = -1 + cos^2(x/2)

cos(x)=(-1) <=> cos^2(x/2)=0

(1) cos(x) = cos^2(x/2) - sin^2(x/2)

(2) cos(x) = -1 + cos^2(x/2)

cos(x)=(-1) <=> cos^2(x/2)=0

Decide if it is true:

tan^2(x/2) = (1-cos(x)) / (1+cos(x))

tan^2(x/2) = (1-cos(x)) / (1+cos(x))

Yes.

hint: cos(x) = cos^2(x/2) - sin^2(x/2)

hint: cos(x) = cos^2(x/2) - sin^2(x/2)

(1) J = (-pi ; pi)

(2) y:J->|R diffable

(3) y(pi/2)=0

(4) /\x:-J y'(x)*sin(x) - y(x) = 1

Find all such functions.

(2) y:J->|R diffable

(3) y(pi/2)=0

(4) /\x:-J y'(x)*sin(x) - y(x) = 1

Find all such functions.

y(x) = tan(x/2) - 1

page 178 in OLDTIMER

page 178 in OLDTIMER

Investigate the limit: lim(n->oo) (n!)^(1/n).

lim(n->oo) (n!)^(1/n) = oo

page 182 in OLDTIMER

page 182 in OLDTIMER

Find a diffable function F:(0,oo)->|R such that F'=log(x).

F(x) = x*(log(x)-1)

hint: log(x) = x*log(x)*(1/x) = exp(log(x))*log(x)*log'(x);

[x*exp(x)-exp(x)]' = x*exp(x)

hint: log(x) = x*log(x)*(1/x) = exp(log(x))*log(x)*log'(x);

[x*exp(x)-exp(x)]' = x*exp(x)

Let f:(0,e)u(e,oo)->|R be f(x)=1/( x * (log(x)-1) ).

Find a diffable function F:(0,e)u(e,oo)->|R such that F'=f.

Find a diffable function F:(0,e)u(e,oo)->|R such that F'=f.

F(x) = log( |log(x)-1| )

hint: (1/x)/(lnx-1), 1/x=(lnx-1)'

hint: (1/x)/(lnx-1), 1/x=(lnx-1)'

Let K > 0.

Is it true that [ log(K*t) ]' = 1/t ?

Is it true that [ log(K*t) ]' = 1/t ?

Yes.

(1) y:|R->|R diffable

(2) /\x:-|R x*y'(x) = y(x)

Find all such functions.

(2) /\x:-|R x*y'(x) = y(x)

Find all such functions.

y(x)=M*x, where M is any real number

hint: (xy'-y)/(x^2)

hint: (xy'-y)/(x^2)

(1) y:(0,oo)->|R diffable

(2) y(1)=1

(3) /\t>0 t*y'(t) = y(t)*log(y(t)/t)

Find all such functions.

(2) y(1)=1

(3) /\t>0 t*y'(t) = y(t)*log(y(t)/t)

Find all such functions.

y(t) = t*exp(1-t)

page 182 in OLDTIMER

page 182 in OLDTIMER

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure;

(2) /\a>0 \/k:-|N /\n>k y{x: |f[n](x)-f(x)|>a} < a.

Are (1) and (2) equivalent?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure;

(2) /\a>0 \/k:-|N /\n>k y{x: |f[n](x)-f(x)|>a} < a.

Are (1) and (2) equivalent?

Yes.

page 81 in MSW measure

page 81 in MSW measure

Let (X,F,y) be a measure space.

Let A:-F. Let f:X->|R* be integrable on A.

Suppose that Leb(A,f,dy) is not zero.

What can we conclude?

Let A:-F. Let f:X->|R* be integrable on A.

Suppose that Leb(A,f,dy) is not zero.

What can we conclude?

We can conclude that y(A) > 0.

---------

If y(A)=0, then Leb(A,f,dy)=0, no matter what f is.

---------

If y(A)=0, then Leb(A,f,dy)=0, no matter what f is.

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

What does it mean that sequence f[n] is fundamental in measure?

Let f[n]:X->|R be a sequence of measurable functions.

What does it mean that sequence f[n] is fundamental in measure?

(1) /\(a>0) \/(k:-|N) /\(n,m>k) y{x: |f[n](x)-f[m](x)|>a} < a

(2) /\(a>0) /\(b>0) \/(k:-|N) /\(n,m>k) y{x: |f[n](x)- f[m](x)|>a} < b

(3) /\a>0 lim(n,m->oo) y{x: |f[n](x)-f[m](x)|>a} = 0

page 82 in MSW measure

(2) /\(a>0) /\(b>0) \/(k:-|N) /\(n,m>k) y{x: |f[n](x)- f[m](x)|>a} < b

(3) /\a>0 lim(n,m->oo) y{x: |f[n](x)-f[m](x)|>a} = 0

page 82 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

(1) /\a>0 \/k:-|N /\(n,m>k) y{x: |f[n](x)-f[m](x)|>a} < a;

(2) /\a>0 /\b>0 \/k:-|N /\(n,m>k) y{x: |f[n](x)-f[m](x)|>a} < b.

Are (1) and (2) equivalent?

Let f[n]:X->|R be a sequence of measurable functions.

(1) /\a>0 \/k:-|N /\(n,m>k) y{x: |f[n](x)-f[m](x)|>a} < a;

(2) /\a>0 /\b>0 \/k:-|N /\(n,m>k) y{x: |f[n](x)-f[m](x)|>a} < b.

Are (1) and (2) equivalent?

page 82 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Suppose that f[n] converges in measure.

Can we conclude that f[n] is fundamental in measure?

Let f[n]:X->|R be a sequence of measurable functions.

Suppose that f[n] converges in measure.

Can we conclude that f[n] is fundamental in measure?

Yes.

page 82 in MSW measure

page 82 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of functions. Let f:X->|R* be a function.

What does it mean that f[n]->f nearly uniformly?

Let f[n]:X->|R be a sequence of functions. Let f:X->|R* be a function.

What does it mean that f[n]->f nearly uniformly?

/\a>0 \/E:-F [ y(E)<a and f[n]->f uniformly on X\E ]

Notice that if a sequence converges uniformly, then it converges nearly uniformly.

But not the other way around. Put f[n](x)=x^n, for 0<x<1.

This sequence converges nearly uniformly, but not uniformly.

Notice that if a sequence converges uniformly, then it converges nearly uniformly.

But not the other way around. Put f[n](x)=x^n, for 0<x<1.

This sequence converges nearly uniformly, but not uniformly.

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of functions. Let f:X->|R be a function.

(1) f[n]->f nearly uniformly;

(2) f[n]->f pointwise almost everywhere.

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of functions. Let f:X->|R be a function.

(1) f[n]->f nearly uniformly;

(2) f[n]->f pointwise almost everywhere.

Is it true that (1)=>(2) ?

Yes.

page 83 in MSW measure

page 83 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n],g[n]:X->|R be sequences of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure and g[n]->g in measure

(2) f[n]+g[n]->f+g in measure

Is it true that (1)=>(2) ?

Let f[n],g[n]:X->|R be sequences of measurable functions.

Let f,g:X->|R be measurable functions.

(1) f[n]->f in measure and g[n]->g in measure

(2) f[n]+g[n]->f+g in measure

Is it true that (1)=>(2) ?

Yes.

page 84 in MSW measure

page 84 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function. Let K be a real number.

(1) f[n]->f in measure;

(2) K*f[n]->K*f in measure.

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function. Let K be a real number.

(1) f[n]->f in measure;

(2) K*f[n]->K*f in measure.

Is it true that (1)=>(2) ?

Yes.

page 85 in MSW measure

page 85 in MSW measure

Let 0 < a < oo.

What can we say about the sequence a^(1/n) ?

What can we say about the sequence a^(1/n) ?

lim(n->oo) a^(1/n) = 1

page 133 in 2nd analysis

page 133 in 2nd analysis

Let a[n] > 0 for all n:-|N.

lim_sup a[n]^(1/n) < lim_sup a[n+1]/a[n] < oo

Is this possible?

lim_sup a[n]^(1/n) < lim_sup a[n+1]/a[n] < oo

Is this possible?

Yes.

1,2,1,2,1,2,1,2, ...

1,2,1,2,1,2,1,2, ...

Let a[n] > 0 for all n:-|N.

(1) lim_sup a[n+1]/a[n] < oo

(2) lim_sup a[n+1]/a[n] <= lim_sup a[n]^(1/n)

Is it true that (1)=>(2) ?

(1) lim_sup a[n+1]/a[n] < oo

(2) lim_sup a[n+1]/a[n] <= lim_sup a[n]^(1/n)

Is it true that (1)=>(2) ?

No.

1,2,1,2,1,2,1,2, ...

page 100 in 2nd analysis

1,2,1,2,1,2,1,2, ...

page 100 in 2nd analysis

Let a[n] > 0 for all n:-|N.

(1) lim_sup a[n]^(1/n) <= lim_sup a[n+1]/a[n].

Is (1) generally true?

(1) lim_sup a[n]^(1/n) <= lim_sup a[n+1]/a[n].

Is (1) generally true?

Yes.

page 97 in 2nd analysis

page 97 in 2nd analysis

Let a[n] > 0 for all n:-|N.

(1) lim_sup a[n+1]/a[n] <= lim_sup a[n]^(1/n)

Is (1) generally true?

(1) lim_sup a[n+1]/a[n] <= lim_sup a[n]^(1/n)

Is (1) generally true?

No.

1,2,1,2,1,2,1,2, ...

page 97 in 2nd analysis

page 100 in 2nd analysis

1,2,1,2,1,2,1,2, ...

page 97 in 2nd analysis

page 100 in 2nd analysis

Let a[n] > 0 for all n:-|N.

lim_sup a[n]^(1/n) < lim_sup a[n+1]/a[n] = oo

Is this possible?

lim_sup a[n]^(1/n) < lim_sup a[n+1]/a[n] = oo

Is this possible?

1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,1,11,1,12, ...

Let A c |R.

Is it generally true that inf(Clo(A)) = inf(A) ?

Is it generally true that inf(Clo(A)) = inf(A) ?

Yes.

(1) Jc|R connected; y:J->|R diffable

(2) y(0)=1

(3) /\t:-|R y'(t) - 2ty(t) = 2*(t^3)*y^2(t)

Find all such functions.

HINT

(2) y(0)=1

(3) /\t:-|R y'(t) - 2ty(t) = 2*(t^3)*y^2(t)

Find all such functions.

HINT

J c (-1 ; 1)

y(t) = 1 / (1 - t^2)

page 97 in rownania rozniczkowe zwyczajne

y(t) = 1 / (1 - t^2)

page 97 in rownania rozniczkowe zwyczajne

Let f(x) = -2*(x^3)*exp(x^2).

Find an F such that F' = f.

Find an F such that F' = f.

F(x) = (1-x^2)*exp(x^2)

Let X,Y be normed vector spaces. Let T:X->Y.

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that sup{ ||Tx|| : ||x||=1 } < ||T||.

Is this possible?

Let ||T|| = inf{ m:-|R : /\x:-X ||Tx|| <= m*||x|| }.

Suppose that sup{ ||Tx|| : ||x||=1 } < ||T||.

Is this possible?

Yes.

Let X = Y = |R. Let T(x) = x*x.

Then sup{ ||Tx|| : ||x||=1 } = 1 < oo = ||T||.

If T is linear, then it cannot happen.

page 151 in OLDTIMER

Let X = Y = |R. Let T(x) = x*x.

Then sup{ ||Tx|| : ||x||=1 } = 1 < oo = ||T||.

If T is linear, then it cannot happen.

page 151 in OLDTIMER

Let X,Y be normed vector spaces.

Let T:X->Y be linear and non-constant.

{ ||T(x)|| : x:-X } = ???

Let T:X->Y be linear and non-constant.

{ ||T(x)|| : x:-X } = ???

{ ||T(x)|| : x:-X } = [0,oo)

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F.

(7) 1(A[n])->0 in measure.

What can we conclude?

Let A[n] be a sequence of sets in F.

(7) 1(A[n])->0 in measure.

What can we conclude?

lim(n->oo) y(A[n]) = 0

page 85 in MSW measure

page 85 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F.

(1) lim(n->oo) y(A[n]) = 0;

(2) 1(A[n])->0 in measure.

Is it true that (1)=>(2) ?

Let A[n] be a sequence of sets in F.

(1) lim(n->oo) y(A[n]) = 0;

(2) 1(A[n])->0 in measure.

Is it true that (1)=>(2) ?

Yes.

page 85 in MSW measure

page 85 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F.

(1) 1(A[n])->0 in measure;

(2) lim(n->oo) y(A[n]) = 0.

Is it true that (1)=>(2) ?

Let A[n] be a sequence of sets in F.

(1) 1(A[n])->0 in measure;

(2) lim(n->oo) y(A[n]) = 0.

Is it true that (1)=>(2) ?

Yes.

page 85 in MSW measure

page 85 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F. Let A:-F.

(1) lim(n->oo) y(A[n]+A) = 0;

(2) 1(A[n])->1(A) in measure.

Is it true that (1)=>(2) ?

Let A[n] be a sequence of sets in F. Let A:-F.

(1) lim(n->oo) y(A[n]+A) = 0;

(2) 1(A[n])->1(A) in measure.

Is it true that (1)=>(2) ?

Yes.

page 86 in MSW measure

page 86 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F. Let A:-F.

(1) 1(A[n])->1(A) in measure;

(2) lim(n->oo) y(A[n]+A) = 0;

Is it true that (1)=>(2) ?

Let A[n] be a sequence of sets in F. Let A:-F.

(1) 1(A[n])->1(A) in measure;

(2) lim(n->oo) y(A[n]+A) = 0;

Is it true that (1)=>(2) ?

Yes.

page 86 in MSW measure

page 86 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence of sets in F. Let A:-F.

(1) 1(A[n])->1(A) in measure.

What can we conclude?

Let A[n] be a sequence of sets in F. Let A:-F.

(1) 1(A[n])->1(A) in measure.

What can we conclude?

(2) lim(n->oo) y(A[n]+A) = 0

page 86 in MSW measure

page 86 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure;

(2) f[n]->f pointwise almost everywhere;

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f in measure;

(2) f[n]->f pointwise almost everywhere;

Is it true that (1)=>(2) ?

No.

Consider this sequence of sets { A[n] } :

[0,1], [0,1/2], [1/2,1], [0,1/3], [1/3,2/3], [2/3,1], [0,1/4], [1/4,2/4], [2/4,3/4], [3/4,1], ...

Put f[n](x) = 1(A[n])(x).

f[n]->0 in measure because lim(n->oo) y(A[n])=0.

And { f[n](x) } diverges for every x in [0,1].

Consider this sequence of sets { A[n] } :

[0,1], [0,1/2], [1/2,1], [0,1/3], [1/3,2/3], [2/3,1], [0,1/4], [1/4,2/4], [2/4,3/4], [3/4,1], ...

Put f[n](x) = 1(A[n])(x).

f[n]->0 in measure because lim(n->oo) y(A[n])=0.

And { f[n](x) } diverges for every x in [0,1].

Let (X,F,y) be a measure space.

Let f[n]:X->|R be measurable for all n:-|N. Let f,g : X->|R.

(1) /\n:-|N /\x:-X |f[n](x)| <= g(x)

(2) g is integrable

(3) /\x:-X lim(n->oo) f[n](x) = f(x)

Decide if we can conclude that:

(4) lim(n->oo) Leb*(X,|f[n]-f|,dy) = 0.

Let f[n]:X->|R be measurable for all n:-|N. Let f,g : X->|R.

(1) /\n:-|N /\x:-X |f[n](x)| <= g(x)

(2) g is integrable

(3) /\x:-X lim(n->oo) f[n](x) = f(x)

Decide if we can conclude that:

(4) lim(n->oo) Leb*(X,|f[n]-f|,dy) = 0.

Yes.

|f[n]|<=g, |f|<=g, hence |f[n]-f|<=2g

Apply Lebesgue Dominated Convergence Theorem to { |f[n]-f| }.

page 13 in 2nd measure

|f[n]|<=g, |f|<=g, hence |f[n]-f|<=2g

Apply Lebesgue Dominated Convergence Theorem to { |f[n]-f| }.

page 13 in 2nd measure

Let f:|R->|R be unbounded.

Can it be uniformly approached by simple functions?

Can it be uniformly approached by simple functions?

No.

The limit of a uniformly convergent sequence of bounded functions is bounded.

The limit of a uniformly convergent sequence of bounded functions is bounded.

Consider the Lebesgue measure on [0,1].

Let f[n](x)=x^n for x:-[0,1], and n:-|N.

Does this sequence converge nearly uniformly?

Let f[n](x)=x^n for x:-[0,1], and n:-|N.

Does this sequence converge nearly uniformly?

Yes.

page 82 in MSW measure

page 82 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

(1) sequence f[n] is fundamental in measure;

Prove that there exist k:|N->|N strictly increasing and f:X->|R

such that f[k(n)] converges nearly uniformly to f as n->oo.

Let f[n]:X->|R be a sequence of measurable functions.

(1) sequence f[n] is fundamental in measure;

Prove that there exist k:|N->|N strictly increasing and f:X->|R

such that f[k(n)] converges nearly uniformly to f as n->oo.

page 87 in MSW measure

Suppose that a net of real numbers converges to a real number.

Does the net have to be bounded?

Does the net have to be bounded?

No.

hint: T = (1,2,3,4,5,6,7,...,1,2,3,4,5,6,7,...)

hint: T = (1,2,3,4,5,6,7,...,1,2,3,4,5,6,7,...)

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of functions. Let f,g:X->|R be functions.

(1) f[n]->f nearly uniformly

(2) g~f

Decide if we can conclude that

Let f[n]:X->|R be a sequence of functions. Let f,g:X->|R be functions.

(1) f[n]->f nearly uniformly

(2) g~f

Decide if we can conclude that

Yes.

page 89 in MSW measure

page 89 in MSW measure

Let a[n] be a sequence of real numbers other than zero.

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Suppose that A!=0 and B!=0.

Can we conclude that A=B ?

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Suppose that A!=0 and B!=0.

Can we conclude that A=B ?

No.

1, -1, 1, -1, 1, -1, ...

A = 1

B = -1

1, -1, 1, -1, 1, -1, ...

A = 1

B = -1

Let a[n] be a sequence of positive numbers.

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Can we conclude that A=B ?

Let A = 1 / lim_sup a[n].

Let B = lim_inf 1/a[n].

Can we conclude that A=B ?

Yes.

page 184 in OLDTIMER

page 184 in OLDTIMER

Let a[n] be a sequence of positive numbers.

Let A = lim_inf (a[n])^(1/n).

Let B = lim_inf a[n+1]/a[n].

What can we conclude?

Let A = lim_inf (a[n])^(1/n).

Let B = lim_inf a[n+1]/a[n].

What can we conclude?

B <= A

page 185 in OLDTIMER

page 185 in OLDTIMER

lim(n->oo) ( (n!)^(1/n) )/n = ???

lim(n->oo) ( (n!)^(1/n) )/n = 1/e

page 186 in OLDTIMER

page 186 in OLDTIMER

Let a[n] be a sequence of positive numbers.

Let A = lim_inf a[n+1]/a[n].

Let B = lim_inf (a[n])^(1/n).

Let C = lim_sup (a[n])^(1/n).

Let D = lim_sup a[n+1]/a[n].

Is it possible that 0 < A < B=C < D < oo ?

Let A = lim_inf a[n+1]/a[n].

Let B = lim_inf (a[n])^(1/n).

Let C = lim_sup (a[n])^(1/n).

Let D = lim_sup a[n+1]/a[n].

Is it possible that 0 < A < B=C < D < oo ?

Yes.

1,2,1,2,1,2,1,2,1,2,...

page 271 in OLDTIMER

1,2,1,2,1,2,1,2,1,2,...

page 271 in OLDTIMER

Let a[n] be a sequence of positive numbers.

Let A = lim_inf a[n+1]/a[n].

Let B = lim_inf (a[n])^(1/n).

Let C = lim_sup (a[n])^(1/n).

Let D = lim_sup a[n+1]/a[n].

Is it possible that A < B < C < D ?

Let A = lim_inf a[n+1]/a[n].

Let B = lim_inf (a[n])^(1/n).

Let C = lim_sup (a[n])^(1/n).

Let D = lim_sup a[n+1]/a[n].

Is it possible that A < B < C < D ?

Yes.

2, 3, 2^2, 3^2, 2^3, 3^3, 2^4, 3^4, 2^5, 3^5, ..., 2^n, 3^n, ...

0 < 2 < 3 < oo

page 271 in OLDTIMER

2, 3, 2^2, 3^2, 2^3, 3^3, 2^4, 3^4, 2^5, 3^5, ..., 2^n, 3^n, ...

0 < 2 < 3 < oo

page 271 in OLDTIMER

Find a diffable function g:|R->|R such that g'(x) = x*exp(x).

g(x) = x*exp(x) - exp(x)

Find a diffable function g:|R->|R such that g'(x) = x*x*exp(x).

g(x) = x*x*exp(x) - 2x*exp(x) + 2exp(x)

hint: [x*exp(x) - exp(x)]' = x*exp(x)

hint: [x*exp(x) - exp(x)]' = x*exp(x)

lim(n->oo) ( (n! * n!)^(1/n) )/n = ???

lim(n->oo) ( (n! * n!)^(1/n) )/n = oo

A similar thing can be found on page 186 in OLDTIMER.

A similar thing can be found on page 186 in OLDTIMER.

Let a[n] be a sequence of positive numbers.

Suppose that lim(n->oo) a[n+1]/a[n] exists (real or +oo).

This implies the existence of which limit?

Suppose that lim(n->oo) a[n+1]/a[n] exists (real or +oo).

This implies the existence of which limit?

lim(n->oo) (a[n])^(1/n) = lim(n->oo) a[n+1]/a[n]

page 186 in OLDTIMER

page 186 in OLDTIMER

(1) /\(n:-|N) W[n] is a ring of sets in X

(2) /\(n:-|N) W[n] c W[n+1].

Can we conclude that U(n:-|N)_W[n] is also a ring in X?

(2) /\(n:-|N) W[n] c W[n+1].

Can we conclude that U(n:-|N)_W[n] is also a ring in X?

Yes.

U(n:-|N)_W[n] is a ring in X

U(n:-|N)_W[n] is a ring in X

Let W[n] be a ring of sets in X for all n:-|N.

Can we conclude that U(n:-|N)_W[n] is also a ring in X?

Can we conclude that U(n:-|N)_W[n] is also a ring in X?

No.

W[n] = { O, {n} }

W = U(n:-|N)_W[n] = { O, {1}, {2}, {3}, ..., {n}, ... }.

W is not a ring, {1} u {2} #- W.

W[n] = { O, {n} }

W = U(n:-|N)_W[n] = { O, {1}, {2}, {3}, ..., {n}, ... }.

W is not a ring, {1} u {2} #- W.

Let W[n] be a s-ring of sets in X for all n:-|N.

Can we conclude that U(n:-|N)_W[n] is also a s-ring in X?

Can we conclude that U(n:-|N)_W[n] is also a s-ring in X?

No.

W[n] = { O, {n} }

W = U(n:-|N)_W[n] = { O, {1}, {2}, {3}, ..., {n}, ... }.

W is not a s-ring, {1} u {2} #- W.

W[n] = { O, {n} }

W = U(n:-|N)_W[n] = { O, {1}, {2}, {3}, ..., {n}, ... }.

W is not a s-ring, {1} u {2} #- W.

Find a diffable function g:(0,oo)->|R such that g'(x) =
log(x)/x/x.

g(x) = - (1+log(x)) / x

hint1: [log(x)/x]'

hint2: log(x)/x/x = log(1/x)*(1/x)'

==============================

MRW 2001.05.04 : I don't understand these hints.

I gave a correct answer to this item and then I pondered

hint1: [log(x)/x]'

hint2: log(x)/x/x = log(1/x)*(1/x)'

==============================

MRW 2001.05.04 : I don't understand these hints.

I gave a correct answer to this item and then I pondered

(1) J c (0,oo), J connected, 1:-J

(2) y:J->|R diffable

(3) /\t:-J t*y'(t) + y(t) = y(t)*y(t)*log(t)

(4) y(1)=1

Find all such functions.

(2) y:J->|R diffable

(3) /\t:-J t*y'(t) + y(t) = y(t)*y(t)*log(t)

(4) y(1)=1

Find all such functions.

J c (1/e ; oo)

y(t) = 1 / (1+log(t))

hint: u=1/y

hint: [ (1+log(x))/x ]' = - log(x)/(x^2)

y(t) = 1 / (1+log(t))

hint: u=1/y

hint: [ (1+log(x))/x ]' = - log(x)/(x^2)

Can an infinitely countable subset of |R be compact?

Yes.

Take any convergent sequence and add its limit to the set.

For example, {0} u {1/n : n:-|N}.

Take any convergent sequence and add its limit to the set.

For example, {0} u {1/n : n:-|N}.

Let X be a topological space. Let AcX.

What can we say about the set Int(Clo(A)\A) ?

What can we say about the set Int(Clo(A)\A) ?

Int(Clo(A)\A) is always empty

page 84 in gen top

page 84 in gen top

Let X be a topological space. Let A,BcX, x:-X.

(1) x:-Clo(A)

(2) x:-B

(3) B is open

What can we conclude?

(1) x:-Clo(A)

(2) x:-B

(3) B is open

What can we conclude?

BnA is not empty

page 23 in gen top

page 23 in gen top

Let X be a topological space. Let A,BcX, x:-X.

(1) x:-Clo(A)

(2) x:-B

(3) BnA = O

What can we conclude?

(1) x:-Clo(A)

(2) x:-B

(3) BnA = O

What can we conclude?

B is not open

page 23 in gen top

page 23 in gen top

Let X be a topological space. Let A,BcX, x:-X.

(1) x:-Clo(A)

(2) BnA = 0

(3) B is open

What can we conclude?

(1) x:-Clo(A)

(2) BnA = 0

(3) B is open

What can we conclude?

x #- B (x does not belong to B)

page 23 in gen top

page 23 in gen top

Let X be a topological space. Let A,BcX, x:-X.

(1) B is open

(2) x:-B

(3) BnA = O

What can we conclude?

(1) B is open

(2) x:-B

(3) BnA = O

What can we conclude?

x #- Clo(A) (x does not belong to the closure of A)

x :- Int(X\A)

page 23 in gen top

x :- Int(X\A)

page 23 in gen top

Let X be a topological space. Let AcX, x:-X.

(1) /\BcX [ (B is open and x:-B) => BnA != O ]

What can we conclude?

(1) /\BcX [ (B is open and x:-B) => BnA != O ]

What can we conclude?

x:-Clo(A)

page 23 in gen top

page 23 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Int( U(t:-T)_A[t] ) c U(t:-T)_Int(A[t])

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Int( U(t:-T)_A[t] ) c U(t:-T)_Int(A[t])

No.

T={1,2}

A[1] = |Q

A[2] = |R \ |Q

page 13 in gen top

T={1,2}

A[1] = |Q

A[2] = |R \ |Q

page 13 in gen top

Let X be a topological space. Let AcX.

(1) Int(Clo(A)) c A

Is this generally true?

(1) Int(Clo(A)) c A

Is this generally true?

No.

Int(Clo(|Q)) = |R

Int(Clo(|Q)) = |R

Let Ac|R be connected.

Let f:A->|R be monotonic.

Let a:-|R.

What can we conclude about this set {x:-A : f(x)>a} ?

Let f:A->|R be monotonic.

Let a:-|R.

What can we conclude about this set {x:-A : f(x)>a} ?

It is connected.

Let X be a topological space. Let A,B c X.

Suppose that Int(A) c Int(B) and Int(A) != Int(B).

Can we conclude that A c B ?

Suppose that Int(A) c Int(B) and Int(A) != Int(B).

Can we conclude that A c B ?

No.

A = {0} u [1,2]

B = [1,3]

A = {0} u [1,2]

B = [1,3]

Let W1,W2 be topologies on X.

(1) W1-convergence implies W2-convergence

What can we conclude?

(1) W1-convergence implies W2-convergence

What can we conclude?

(2) W2 c W1

page 83 in gen top

page 83 in gen top

Let W1,W2 be topologies on X.

(1) W1 c W2

What can we conclude?

(1) W1 c W2

What can we conclude?

(2) W2-convergence implies W1-convergence

page 83 in gen top

page 83 in gen top

Let W1,W2 be topologies on X.

(1) W1-convergence implies W2-convergence

(2) ???

Give an equivalent condition.

Do not prove the equivalence in this item.

(1) W1-convergence implies W2-convergence

(2) ???

Give an equivalent condition.

Do not prove the equivalence in this item.

2) W2 c W1

page 83 in gen top

page 83 in gen top

Let X be a normed vector space.

Let BcX satisfy:

(1) B is countable

(2) lin(B) is dense in X

Can we conclude that X contains a countable dense subset?

Let BcX satisfy:

(1) B is countable

(2) lin(B) is dense in X

Can we conclude that X contains a countable dense subset?

Yes.

page 187 in OLDTIMER

page 187 in OLDTIMER

Let X be a normed vector space.

Let BcX satisfy:

(1) B is countable

(2) lin(B) is dense in X

Can we conclude that lin(B) is a countable dense subset of X?

Let BcX satisfy:

(1) B is countable

(2) lin(B) is dense in X

Can we conclude that lin(B) is a countable dense subset of X?

No.

X = |R

B = |Q

lin(B) = |R

X = |R

B = |Q

lin(B) = |R

Let X be a normed vector space.

Suppose that X contains a countable dense subset.

Decide if we can conclude that X contains a subset B satisfying:

(1) B is countable;

(2) lin(B) is dense in X.

Suppose that X contains a countable dense subset.

Decide if we can conclude that X contains a subset B satisfying:

(1) B is countable;

(2) lin(B) is dense in X.

Yes.

The countable dense subset which exists satisfies these conditions.

page 187 in OLDTIMER

The countable dense subset which exists satisfies these conditions.

page 187 in OLDTIMER

Let X,Y be normed vector spaces.

(1) X is complete

(2) /\(n:-|N) T[n]:X->Y linear and continuous

(3) /\x:-X \/m:-|N /\n:-|N ||T[n](x)|| <= m

What can we conclude?

(1) X is complete

(2) /\(n:-|N) T[n]:X->Y linear and continuous

(3) /\x:-X \/m:-|N /\n:-|N ||T[n](x)|| <= m

What can we conclude?

(4) \/m:-|N /\x:-X /\n:-|N ||T[n](x)|| <= m*||x||

page 189 in OLDTIMER

page 189 in OLDTIMER

Consider a vector space over |R or over |C.

Let x[1],x[2], ... ,x[n] be linearly dependent vectors.

Prove that there exist scalars a[1],a[2], ... ,a[n] such that:

(1) a[1]*x[1] + ... + a[n]*x[n] = 0

(2) |a[1]| + |a[2]| + ... + |a[n]| = 1

Let x[1],x[2], ... ,x[n] be linearly dependent vectors.

Prove that there exist scalars a[1],a[2], ... ,a[n] such that:

(1) a[1]*x[1] + ... + a[n]*x[n] = 0

(2) |a[1]| + |a[2]| + ... + |a[n]| = 1

page 190 in OLDTIMER

Let X be a normed vector space over |R or over |C.

Let x[1],x[2], ... ,x[n] be n vectors.

Let y[k]:|N->X, for k=1,2, ... ,n.

(1) /\m:-|N y[1](m), y[2](m), ... ,y[n](m) are linearly dependent

(2) lim(m->oo) y[k](m) = x[k], for k=1,2, ... ,n.

What can we conclude?

Let x[1],x[2], ... ,x[n] be n vectors.

Let y[k]:|N->X, for k=1,2, ... ,n.

(1) /\m:-|N y[1](m), y[2](m), ... ,y[n](m) are linearly dependent

(2) lim(m->oo) y[k](m) = x[k], for k=1,2, ... ,n.

What can we conclude?

x[1],x[2], ... ,x[n] are linearly dependent

page 191 in OLDTIMER

page 191 in OLDTIMER

Let (X,d) be a metric space containing a countable dense subset.

Let AcX satisfy:

(1) /\x:-A \/e>0 /\y:-A [ d(x,y)<e => x=y ].

Can we conclude that A is countable?

Let AcX satisfy:

(1) /\x:-A \/e>0 /\y:-A [ d(x,y)<e => x=y ].

Can we conclude that A is countable?

Yes.

page 193 in OLDTIMER

page 193 in OLDTIMER

Let (X,d) be a metric space containing a countable dense subset.

Let AcX satisfy:

(1) \/e>0 /\x,y:-A [ d(x,y)<e => x=y ].

What can we conclude about set A?

Let AcX satisfy:

(1) \/e>0 /\x,y:-A [ d(x,y)<e => x=y ].

What can we conclude about set A?

A is countable

page 193 in OLDTIMER

page 193 in OLDTIMER

Let (X,d) be a metric space.

Let AcX satisfy:

(1) /\x:-A \/e>0 /\y:-A [ d(x,y)<e => x=y ]

(2) A is uncountable

What can we conclude?

Let AcX satisfy:

(1) /\x:-A \/e>0 /\y:-A [ d(x,y)<e => x=y ]

(2) A is uncountable

What can we conclude?

X does not contain a countable dense subset.

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

Let (X,d) be a metric space.

(1) /\AcX [ /\x:-A \/e>0 /\y:-A d(x,y)<e => x=y ] => A is countable

(2) /\AcX [ \/e>0 /\x:-A /\y:-A d(x,y)<e => x=y ] => A is countable

Are these conditions equivalent?

(1) /\AcX [ /\x:-A \/e>0 /\y:-A d(x,y)<e => x=y ] => A is countable

(2) /\AcX [ \/e>0 /\x:-A /\y:-A d(x,y)<e => x=y ] => A is countable

Are these conditions equivalent?

Yes.

(1)=>(2) is easy.

(2)=>(1) is convoluted.

(2) implies that X contains a countable dense subset and that in turn implies (1)

page 193 in OLDTIMER

(1)=>(2) is easy.

(2)=>(1) is convoluted.

(2) implies that X contains a countable dense subset and that in turn implies (1)

page 193 in OLDTIMER

Let X be the set of all bounded sequences of real numbers.

Let's equip X with the sup norm.

Does X contain a countable dense subset?

Let's equip X with the sup norm.

Does X contain a countable dense subset?

No.

Let A be the subset of X containing all sequences whose only terms are 0 or 1.

This set is uncountable, equinumerous with |R.

And /\a,b:-A [ a!=b => ||a-b|| = 1 ].

Hence X cannot contain a countable dense subset.

page 193 in OLDTIMER

Let A be the subset of X containing all sequences whose only terms are 0 or 1.

This set is uncountable, equinumerous with |R.

And /\a,b:-A [ a!=b => ||a-b|| = 1 ].

Hence X cannot contain a countable dense subset.

page 193 in OLDTIMER

Consider an uncountable subset of |R.

Does it have to have a point of accumulation (= a cluster point = a limit point) ?

Does it have to have a point of accumulation (= a cluster point = a limit point) ?

Yes.

Two proofs:

(1) page 193 in OLDTIMER (|R contains a countable dense subset)

===

(2) /\k:-|N An[-k,k] is finite, hence U(k:-|N)_An[-k,k] = A is countable

Two proofs:

(1) page 193 in OLDTIMER (|R contains a countable dense subset)

===

(2) /\k:-|N An[-k,k] is finite, hence U(k:-|N)_An[-k,k] = A is countable

If A,B are sets, let A<B mean that AcB and A!=B. (A is properly
contained in B.)

Suppose B<E and not(A<E).

What can we conclude?

Suppose B<E and not(A<E).

What can we conclude?

A is not contained in B

Let X be a topological space. Let A,B c X.

Suppose that Clo(B) < Clo(A), (properly contained).

Is it possible that A c B?

Suppose that Clo(B) < Clo(A), (properly contained).

Is it possible that A c B?

No.

Suppose that A c B. Then Clo(A) c Clo(B).

And since Clo(B) < Clo(A), we have Clo(A)=Clo(B).

Contradiction.

Suppose that A c B. Then Clo(A) c Clo(B).

And since Clo(B) < Clo(A), we have Clo(A)=Clo(B).

Contradiction.

Let X be a metric space.

Let x[n] be a convergent sequence in X.

Let B = \\//(k=1 to oo) //\\(n=k to oo) B(x[n]; 1/n).

Is it possible that B is empty?

Let x[n] be a convergent sequence in X.

Let B = \\//(k=1 to oo) //\\(n=k to oo) B(x[n]; 1/n).

Is it possible that B is empty?

Yes.

Let X = [0,7] with the Euclidean metric.

Let x[n]=1/n.

Let X = [0,7] with the Euclidean metric.

Let x[n]=1/n.

Let X be a metric space. Let AcX. Let x:-X.

Let e[n] be a sequence of positive numbers, converging to zero.

Let x[n] be a sequence in X, converging to x.

(1) /\n:-|N B(x[n],e[n]) n A != O

What can we conclude?

Let e[n] be a sequence of positive numbers, converging to zero.

Let x[n] be a sequence in X, converging to x.

(1) /\n:-|N B(x[n],e[n]) n A != O

What can we conclude?

x:-Clo(A)

Let "A<B" mean that set A is properly contained in set B.

Let X be a non-empty set. Let WcP(X). Let AcX.

Can we conclude that U{E:-W : E<A} < A ?

Let X be a non-empty set. Let WcP(X). Let AcX.

Can we conclude that U{E:-W : E<A} < A ?

No.

1) U{ |N\{n} : n:-|N } = |N

2) U{ {1,2,...,n} : n:-|N } = |N

3) {1} u {2} = {1,2}

1) U{ |N\{n} : n:-|N } = |N

2) U{ {1,2,...,n} : n:-|N } = |N

3) {1} u {2} = {1,2}

Let X be a topological space.

(1) /\(t:-T) A[t] c X

U(t:-T)_Int(A[t]) c ???

(1) /\(t:-T) A[t] c X

U(t:-T)_Int(A[t]) c ???

U(t:-T)_Int(A[t]) c Int( U(t:-T)_A[t] )

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

U(t:-T)_Int(A[t]) c Int( U(t:-T)_A[t] )

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

U(t:-T)_Int(A[t]) c Int( U(t:-T)_A[t] )

Yes.

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

??? c Int( U(t:-T)_A[t] )

(1) /\(t:-T) A[t] c X

??? c Int( U(t:-T)_A[t] )

U(t:-T)_Int(A[t]) c Int( U(t:-T)_A[t] )

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

??? c //\\(t:-T) Clo(A[t])

(1) /\(t:-T) A[t] c X

??? c //\\(t:-T) Clo(A[t])

Clo( //\\(t:-T) A[t] ) c //\\(t:-T) Clo(A[t])

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Clo( //\\(t:-T) A[t] ) c //\\(t:-T) Clo(A[t])

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Clo( //\\(t:-T) A[t] ) c //\\(t:-T) Clo(A[t])

Yes.

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

//\\(t:-T) Clo(A[t]) c Clo( //\\(t:-T) A[t] )

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

//\\(t:-T) Clo(A[t]) c Clo( //\\(t:-T) A[t] )

No.

T={1,2}

A[1] = |Q

A[2] = |R \ |Q

page 13 in gen top

T={1,2}

A[1] = |Q

A[2] = |R \ |Q

page 13 in gen top

Let X be a topological space. Let A,B c X.

Int(A) u Int(B) c ???

Int(A) u Int(B) c ???

Int(A) u Int(B) c Int(A u B)

page 13 in gen top

page 13 in gen top

Let X be a topological space. Let A,B c X.

Decide if it is generally true that:

Int(A u B) c Int(A) u Int(B)

Decide if it is generally true that:

Int(A u B) c Int(A) u Int(B)

No.

A = |Q

B = |R \ |Q

page 13 in gen top

A = |Q

B = |R \ |Q

page 13 in gen top

Let X be a topological space. Let A,B c X.

??? c Int(A u B)

??? c Int(A u B)

Int(A) u Int(B) c Int(A u B)

page 13 in gen top

page 13 in gen top

Let X be a topological space. Let A,B c X.

Clo(A n B) c ???

Clo(A n B) c ???

Clo(A n B) c Clo(A) n Clo(B)

page 13 in gen top

page 13 in gen top

Let X be a topological space. Let A,B c X.

??? c Clo(A) n Clo(B)

??? c Clo(A) n Clo(B)

Clo(A n B) c Clo(A) n Clo(B)

page 13 in gen top

page 13 in gen top

Let X be a topological space. Let A,B c X.

Decide if it is generally true that:

Clo(A) n Clo(B) c Clo(A n B)

Decide if it is generally true that:

Clo(A) n Clo(B) c Clo(A n B)

No.

1) A = |Q; B = |R \ |Q

2) A = [0,1); B = (1,2]

page 13 in gen top

1) A = |Q; B = |R \ |Q

2) A = [0,1); B = (1,2]

page 13 in gen top

Let X be a topological space. Let A,B c X.

Decide if it is generally true that:

Clo(A n B) c Clo(A) n Clo(B)

Decide if it is generally true that:

Clo(A n B) c Clo(A) n Clo(B)

Yes.

page 13 in gen top

page 13 in gen top

Let X be a topological space. Let A,B c X.

Decide if it is generally true that:

Int(A) u Int(B) c Int(A u B)

Decide if it is generally true that:

Int(A) u Int(B) c Int(A u B)

Yes.

page 13 in gen top

page 13 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Int( //\\(t:-T) A[t] ) c ???

(1) /\(t:-T) A[t] c X

Int( //\\(t:-T) A[t] ) c ???

Int( //\\(t:-T) A[t] ) c //\\(t:-T) Int( A[t] )

page 14 in gen top

page 14 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

??? c //\\(t:-T) Int( A[t] )

(1) /\(t:-T) A[t] c X

??? c //\\(t:-T) Int( A[t] )

Int( //\\(t:-T) A[t] ) c //\\(t:-T) Int( A[t] )

page 14 in gen top

page 14 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Int( //\\(t:-T) A[t] ) c //\\(t:-T) Int( A[t] )

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

Int( //\\(t:-T) A[t] ) c //\\(t:-T) Int( A[t] )

Yes.

page 14 in gen top

page 14 in gen top

Let X be a topological space.

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

//\\(t:-T) Int( A[t] ) c Int( //\\(t:-T) A[t] )

(1) /\(t:-T) A[t] c X

Decide if it is generally true that:

//\\(t:-T) Int( A[t] ) c Int( //\\(t:-T) A[t] )

No.

Consider |R with the Euclidean topology. Let A[n] = [-1/n, 1/n).

Then the left-hand side is {0}, and the right-hand side is empty.

page 14 in gen top

Consider |R with the Euclidean topology. Let A[n] = [-1/n, 1/n).

Then the left-hand side is {0}, and the right-hand side is empty.

page 14 in gen top

Let X be a topological space. Let A[t] c X for all t:-T.

U(t:-T) Clo(A[t]) c ???

U(t:-T) Clo(A[t]) c ???

U(t:-T) Clo(A[t]) c Clo( U(t:-T) A[t] )

page 14 in gen top

page 14 in gen top

Let X be a topological space. Let A[t] c X for all t:-T.

??? c Clo( U(t:-T) A[t] )

??? c Clo( U(t:-T) A[t] )

U(t:-T) Clo(A[t]) c Clo( U(t:-T) A[t] )

page 14 in gen top

page 14 in gen top

Let X be a topological space. Let A[t] c X for all t:-T.

Decide if it is generally true that:

U(t:-T) Clo(A[t]) c Clo( U(t:-T) A[t] )

Decide if it is generally true that:

U(t:-T) Clo(A[t]) c Clo( U(t:-T) A[t] )

Yes.

page 14 in gen top

page 14 in gen top

Let X be a topological space. Let A[t] c X for all t:-T.

Decide if it is generally true that:

Clo( U(t:-T) A[t] ) c U(t:-T) Clo(A[t])

Decide if it is generally true that:

Clo( U(t:-T) A[t] ) c U(t:-T) Clo(A[t])

No.

Consider X=[0,1] with the Euclidean topology. Let A[n] = [1/n , 1].

The left-hand side is [0,1]. The right-hand side is (0,1].

page 14 in gen top

Consider X=[0,1] with the Euclidean topology. Let A[n] = [1/n , 1].

The left-hand side is [0,1]. The right-hand side is (0,1].

page 14 in gen top

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of finite functions. Let f:X->|R* be a function.

Suppose that f[n] converges to f nearly uniformly.

Notice that f may assume infinite values.

Does there exist a finite function g:X->|R\{-oo,oo}

Let f[n]:X->|R be a sequence of finite functions. Let f:X->|R* be a function.

Suppose that f[n] converges to f nearly uniformly.

Notice that f may assume infinite values.

Does there exist a finite function g:X->|R\{-oo,oo}

Yes.

page 90 in MSW measure

page 90 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of finite functions. Let f:X->|R* be a function.

(1) f[n]->f nearly uniformly;

(2) f is finite almost everywhere

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of finite functions. Let f:X->|R* be a function.

(1) f[n]->f nearly uniformly;

(2) f is finite almost everywhere

Is it true that (1)=>(2) ?

Yes.

page 83 in MSW measure

page 83 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be a sequence of measurable functions.

(1) f[n] is fundamental in measure

(2) \/(f:X->|R) f[n]->f in measure

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of measurable functions.

(1) f[n] is fundamental in measure

(2) \/(f:X->|R) f[n]->f in measure

Is it true that (1)=>(2) ?

Yes.

page 91 in MSW measure

page 91 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n],g[n]:X->|R be two sequences of functions. Let f:X->|R.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->f nearly uniformly

What can we conclude?

Let f[n],g[n]:X->|R be two sequences of functions. Let f:X->|R.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->f nearly uniformly

What can we conclude?

(3) g[n]->f nearly uniformly

page 92 in MSW measure

page 92 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n],g[n]:X->|R be two sequences of functions. Let f:X->|R.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->f in measure

What can we conclude?

Let f[n],g[n]:X->|R be two sequences of functions. Let f:X->|R.

(1) /\n:-|N f[n]~g[n]

(2) f[n]->f in measure

What can we conclude?

(3) g[n]->f in measure

page 92 in MSW measure

page 92 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise everywhere

(2) y(X) < oo

Decide if we can conclude that

(3) f[n]->f nearly uniformly

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise everywhere

(2) y(X) < oo

Decide if we can conclude that

(3) f[n]->f nearly uniformly

Yes.

page 93 in MSW measure

page 93 in MSW measure

(1) /\n:-|N 0 <= a[n] <= b[n]

(2) \/k:-|N a[k] < b[k]

(3) +(n=1 to oo) b[n] < oo

Decide if we can conclude that:

(4) +(n=1 to oo) a[n] < +(n=1 to oo) b[n]

(2) \/k:-|N a[k] < b[k]

(3) +(n=1 to oo) b[n] < oo

Decide if we can conclude that:

(4) +(n=1 to oo) a[n] < +(n=1 to oo) b[n]

Yes.

(1) /\n:-|N 0 <= a[n] <= b[n]

(2) \/k:-|N a[k] < b[k]

Decide if we can conclude that:

(3) +(n=1 to oo) a[n] < +(n=1 to oo) b[n]

(2) \/k:-|N a[k] < b[k]

Decide if we can conclude that:

(3) +(n=1 to oo) a[n] < +(n=1 to oo) b[n]

No.

a[n] = 1/(n+1)

b[n] = 1/sqrt(n)

a[n] = 1/(n+1)

b[n] = 1/sqrt(n)

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let A[n] be a sequence in F such that +(n=1 to oo) y(A[n]) < oo.

Can we conclude that 1(A[n])->0 pointwise almost everywhere?

Let A[n] be a sequence in F such that +(n=1 to oo) y(A[n]) < oo.

Can we conclude that 1(A[n])->0 pointwise almost everywhere?

Yes.

page 132 in 2nd measure

page 132 in 2nd measure

(1) f:|R->|R

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is nowhere diffable

Is this possible?

(2) h:|R->|R is diffable

(3) g:|R->|R, g(x)=h(f(x))

(4) g is diffable

(5) f is nowhere diffable

Is this possible?

Yes.

If x is rational, let f(x) = -x.

If x is irrational, let f(x) = x.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is nowhere diffable.

If x is rational, let f(x) = -x.

If x is irrational, let f(x) = x.

Let h(x)=x*x.

Then g(x) = x*x.

Now, h,g are diffable, but f is nowhere diffable.

Let X,Y be normed vector spaces.

(1) X is complete

(2) /\a:-A T[a]:X->Y linear and continuous

(3) /\(x:-X) \/(m:-|N) /\(a:-A) ||T[a](x)|| <= m

Prove that:

(4) /\e>0 \/d>0 /\x:-X [ ||x||<d => /\(a:-A) ||T[a](x)||<e ]

(1) X is complete

(2) /\a:-A T[a]:X->Y linear and continuous

(3) /\(x:-X) \/(m:-|N) /\(a:-A) ||T[a](x)|| <= m

Prove that:

(4) /\e>0 \/d>0 /\x:-X [ ||x||<d => /\(a:-A) ||T[a](x)||<e ]

page 7 in redcrest

Let X,Y be normed vector spaces.

Let T[n]:X->Y be linear for all n:-|N.

Let T:X->Y satisfy:

(1) /\(x:-X) T(x) = lim(n->oo) T[n](x).

What can we conclude about T?

Let T[n]:X->Y be linear for all n:-|N.

Let T:X->Y satisfy:

(1) /\(x:-X) T(x) = lim(n->oo) T[n](x).

What can we conclude about T?

T is linear

Let X,Y be normed vector spaces.

(1) X is complete

(2) /\(n:-|N) T[n]:X->Y is linear and continuous

(3) T:X->Y; /\(x:-X) T(x) = lim(n->oo) T[n](x)

What can we conclude about T?

(1) X is complete

(2) /\(n:-|N) T[n]:X->Y is linear and continuous

(3) T:X->Y; /\(x:-X) T(x) = lim(n->oo) T[n](x)

What can we conclude about T?

T is linear and continuous

page 8 in redcrest

page 8 in redcrest

Let X,Y be normed vector spaces.

(1) AcX

(2) /\n:-|N T[n]:X->Y linear and continuous

(3) /\(x:-A) { T[n](x) } converges in Y

What can we conclude?

(1) AcX

(2) /\n:-|N T[n]:X->Y linear and continuous

(3) /\(x:-A) { T[n](x) } converges in Y

What can we conclude?

(4) /\x:-lin(A) { T[n](x) } converges in Y

page 9 in redcrest

page 9 in redcrest

(1) /\(n:-|N) A[n] c X

(2) A[1] is non-empty

(3) //\\(n:-|N)_A[n] is empty

What can we conclude?

(2) A[1] is non-empty

(3) //\\(n:-|N)_A[n] is empty

What can we conclude?

(4) U(n:-|N) A[n]\A[n+1] is non-empty

page 135 in 2nd measure

page 135 in 2nd measure

(1) /\(n:-|N) A[n] c X

(2) A[1] is non-empty

(3) //\\(n:-|N)_A[n] is empty

Decide if we can conclude that

(4) U(n:-|N) A[n]\A[n+1] is non-empty

(2) A[1] is non-empty

(3) //\\(n:-|N)_A[n] is empty

Decide if we can conclude that

(4) U(n:-|N) A[n]\A[n+1] is non-empty

Yes.

page 135 in 2nd measure

page 135 in 2nd measure

Let X,Y be normed vector spaces.

(1) AcX, A is dense in X

(2) /\(n:-|N) T[n]:X->Y linear and continuous

(3) /\(x:-A) { T[n](x) } converges in Y

(4) \/M>0 /\n:-|N /\x:-X ||T[n](x)|| <= M*||x||

What can we conclude?

(1) AcX, A is dense in X

(2) /\(n:-|N) T[n]:X->Y linear and continuous

(3) /\(x:-A) { T[n](x) } converges in Y

(4) \/M>0 /\n:-|N /\x:-X ||T[n](x)|| <= M*||x||

What can we conclude?

(5) /\x:-X { T[n](x) } is Cauchy in Y

page 9 in redcrest

page 9 in redcrest

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise everywhere

(2) f[n]->f nearly uniformly

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise everywhere

(2) f[n]->f nearly uniformly

Is it true that (1)=>(2) ?

No.

Consider [0,oo) with the Lebesgue measure. Let f[n](x) = x/n.

But if we additionally suppose that y(X)<oo, then (2) follows.

page 93 in MSW measure

Consider [0,oo) with the Lebesgue measure. Let f[n](x) = x/n.

But if we additionally suppose that y(X)<oo, then (2) follows.

page 93 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise almost everywhere

(2) y(X) < oo

Decide if we can conclude that

(3) f[n]->f nearly uniformly

Let f[n]:X->|R be a sequence of finite measurable functions. Let f:X->|R.

(1) f[n]->f pointwise almost everywhere

(2) y(X) < oo

Decide if we can conclude that

(3) f[n]->f nearly uniformly

Yes.

page 94 in MSW measure

page 94 in MSW measure

Let (X,F,y) be a measure space. Let y(X) < oo.

Let f[n]:X->|R be a sequence of measurable functions. Let f:X- >|R be measurable.

Let S = {a:|N->|N : a is strictly increasing}.

(1) f[n]->f in measure

(2) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

Let f[n]:X->|R be a sequence of measurable functions. Let f:X- >|R be measurable.

Let S = {a:|N->|N : a is strictly increasing}.

(1) f[n]->f in measure

(2) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

(1)<=>(2)

page 100 in MSW measure

page 100 in MSW measure

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of functions. Let f:X->|R.

Let S = {a:|N->|N : a is strictly increasing}.

(1) f[n]->f pointwise almost everywhere

(2) /\(a:-S) f[a(n)]->f pointwise almost everywhere as n->oo

Is it true that (1)=>(2) ?

Let f[n]:X->|R be a sequence of functions. Let f:X->|R.

Let S = {a:|N->|N : a is strictly increasing}.

(1) f[n]->f pointwise almost everywhere

(2) /\(a:-S) f[a(n)]->f pointwise almost everywhere as n->oo

Is it true that (1)=>(2) ?

Yes.

page 101 in MSW measure

page 101 in MSW measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Prove that /\(f,g:-S) d(f,g)=0 <=> ???

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Prove that /\(f,g:-S) d(f,g)=0 <=> ???

/\(f,g:-S) d(f,g)=0 <=> f~g

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

(1) /\(f,g,h:-S) d(f,g) <= d(f,h) + d(h,g)

Is (1) true?

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

(1) /\(f,g,h:-S) d(f,g) <= d(f,h) + d(h,g)

Is (1) true?

Yes.

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Prove that /\(f,g:-S) /\(a:-|R,|a|>0) d(f,g) = ??? .

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Prove that /\(f,g:-S) /\(a:-|R,|a|>0) d(f,g) = ??? .

/\(f,g:-S) /\(a:-|R,|a|>0) d(f,g) = d(af,ag)

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

page 129 in 2nd measure

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

(1) lim(n->oo) d(f[n],f) = 0

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

(1) lim(n->oo) d(f[n],f) = 0

Yes.

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

NOTICE THAT:

d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r} = y{x: f(x)!=g(x)}

Let (X,F,y) be a measure space. Let y(X)<oo.

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {b:|N->|N : b is strictly increasing}.

(1) not ( f[n]->f pointwise almost everywhere )

(2) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {b:|N->|N : b is strictly increasing}.

(1) not ( f[n]->f pointwise almost everywhere )

(2) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

Yes.

page 133 in 2nd measure

page 133 in 2nd measure

Let (X,F,y) be a measure space. Let y(X)<oo.

Let S = {f:X->|R : f is measurable}.

Does there exist a topology on S such that convergence in this topology

is equivalent to pointwise-almost-everywhere convergence?

Let S = {f:X->|R : f is measurable}.

Does there exist a topology on S such that convergence in this topology

is equivalent to pointwise-almost-everywhere convergence?

No.

(1) page 133 in 2nd measure

(2) page 85 in gen top

The conclusion flows from (1) and (2) put together.

(1) page 133 in 2nd measure

(2) page 85 in gen top

The conclusion flows from (1) and (2) put together.

Let (X,F,y) be a measure space. Let y(X)<oo.

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {b:|N->|N : b is strictly increasing}.

(1) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

(2) f[n]->f pointwise almost everywhere

Let f[n]:X->|R be measurable for all n:-|N. Let f:X->|R be measurable.

Let S = {b:|N->|N : b is strictly increasing}.

(1) /\(a:-S) \/(b:-S) f[a(b(n))]->f pointwise almost everywhere as n->oo

(2) f[n]->f pointwise almost everywhere

No.

page 133 in 2nd measure

page 133 in 2nd measure

Let X be a topological space. Let (T,>) be a directed set.

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) x does not converge to p

(2) /\(a:-S) \/(b:-S) x[a(b(t))] -> p (t:-T)

Is this possible?

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) x does not converge to p

(2) /\(a:-S) \/(b:-S) x[a(b(t))] -> p (t:-T)

Is this possible?

No.

page 85 in gen top

page 85 in gen top

Let X be a topological space. Let (T,>) be a directed set.

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) /\(a:-S) \/(b:-S) x[a(b(t))] -> p (t:-T)

What can we conclude?

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) /\(a:-S) \/(b:-S) x[a(b(t))] -> p (t:-T)

What can we conclude?

x[t] -> p (t:-T) (net x converges to p)

page 85 in gen top

page 85 in gen top

Let X be a topological space. Let (T,>) be a directed set.

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) \/(a:-S) /\(b:-S) not ( x[a(b(t))] -> p (t:-T) )

What can we conclude?

Let S = {b:T->T : /\(t:-T) b(t)>t}.

Let x:T->X be a net in X, let p:-X.

(1) \/(a:-S) /\(b:-S) not ( x[a(b(t))] -> p (t:-T) )

What can we conclude?

net x does not converge to p

page 85 in gen top

page 85 in gen top

Let (X,F,y) be a measure space.

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise everywhere

Can we conclude that f[n]->f in measure ?

Let f[n]:X->|R be a sequence of measurable functions.

Let f:X->|R be a measurable function.

(1) f[n]->f pointwise everywhere

Can we conclude that f[n]->f in measure ?

No.

Let X = |R. Let f[n](x) = x/n.

Notice y(X)=oo.

If y(X)<oo, then we could conclude that f[n]->f in measure.

see page 77 in MSW measure

Let X = |R. Let f[n](x) = x/n.

Notice y(X)=oo.

If y(X)<oo, then we could conclude that f[n]->f in measure.

see page 77 in MSW measure

f:X->Y

(1) /\(AcX) f-1(f(A)) c A

What can we conclude about function f ?

(1) /\(AcX) f-1(f(A)) c A

What can we conclude about function f ?

it is injective (1-1)

(2) /\(x,u:-X) f(x)=f(u) => x=u

page 86 in gen top

(2) /\(x,u:-X) f(x)=f(u) => x=u

page 86 in gen top

Let A,B be different subsets of |N. Let m:-|N.

(1) min(A+B) > m

(2) A n [1,m] = B n [1,m]

Prove that (1)=>(2).

(1) min(A+B) > m

(2) A n [1,m] = B n [1,m]

Prove that (1)=>(2).

page 151 in palace

Let A,B be different subsets of |N. Let m:-|N.

(1) min(A+B) > m

(2) A n [1,m] = B n [1,m]

Prove that (2)=>(1).

(1) min(A+B) > m

(2) A n [1,m] = B n [1,m]

Prove that (2)=>(1).

page 151 in palace

f:(0,oo)->|R

f(x) = -log(x)/(x^2)

Find F'=f.

f(x) = -log(x)/(x^2)

Find F'=f.

F(x) = (1+log(x))/x

hint: -log(x) = 1 - ( 1+log(x) )

hint: -log(x) = 1 - ( 1+log(x) )

Consider a uniformly convergent sequence of bounded functions.

What can we conclude about the limit of this sequence?

What can we conclude about the limit of this sequence?

it is bounded

page 11 in redcrest

page 11 in redcrest

Let X be a normed vector space.

Let A be a linear subspace in X.

Suppose that A contains a ball.

What can we conclude?

Let A be a linear subspace in X.

Suppose that A contains a ball.

What can we conclude?

A = X

page 12 in redcrest

page 12 in redcrest

Let X be a normed vector space.

Let A be a linear subspace in X.

Suppose that A != X.

What can we conclude?

Let A be a linear subspace in X.

Suppose that A != X.

What can we conclude?

no ball is contained in A

page 12 in redcrest

page 12 in redcrest

Let X be a normed vector space.

Let E be a proper subset of X.

Suppose that E contains a ball.

What can we conclude?

Let E be a proper subset of X.

Suppose that E contains a ball.

What can we conclude?

E is not a linear subspace in X

page 12 in redcrest

page 12 in redcrest

What is a convex set?

Let (X,K) be a vector space. Let A c X.

A is convex <=(def)=> /\(x,y:-A) /\(a,b>0, a+b=1) ax+by :- A

page 12 in redcrest

A is convex <=(def)=> /\(x,y:-A) /\(a,b>0, a+b=1) ax+by :- A

page 12 in redcrest

Let X be a normed vector space.

Let E = {x:-X : ||x|| <= 1}.

Does E have to be convex?

Let E = {x:-X : ||x|| <= 1}.

Does E have to be convex?

Yes.

page 13 in redcrest

page 13 in redcrest

Let X be a normed vector space.

Let E = {x:-X : ||x|| < 1}.

Does E have to be convex?

Let E = {x:-X : ||x|| < 1}.

Does E have to be convex?

yes

page 13 in redcrest

page 13 in redcrest

Let X be a vector space. Let A be a convex subset of X.

Let p:-X. Let E = { p+a: a:-A }. What can we conclude?

Let p:-X. Let E = { p+a: a:-A }. What can we conclude?

E is convex

page 13 in redcrest

page 13 in redcrest

Let X be a vector space. Let A be a convex subset of X.

Let g be a scalar. Let E = { g*a: a:-A }. What can we conclude?

Let g be a scalar. Let E = { g*a: a:-A }. What can we conclude?

E is convex

page 14 in redcrest

page 14 in redcrest

k:-|N ; 0<a<1

+(n=k to oo) a^n = ???

+(n=k to oo) a^n = ???

+(n=k to oo) a^n = a^k / (1-a)

k:-|N

+(n=k to oo) 1/(2^n) = ???

+(n=k to oo) 1/(2^n) = ???

+(n=k to oo) 1/(2^n) = 1 / 2^(k-1)

Let X be a normed vector space.

Let a:-X, p>0.

Let A = {x:-X : ||x-a|| <= p}.

Let B = {x:-X : ||x-a|| < p}.

Is it generally true that A c Clo(B) ?

Let a:-X, p>0.

Let A = {x:-X : ||x-a|| <= p}.

Let B = {x:-X : ||x-a|| < p}.

Is it generally true that A c Clo(B) ?

yes

page 42 in redcrest

page 42 in redcrest

(1) A c |R connected

(2) /\(n:-|N) g[n]:A->|R diffable

(3) /\(x:-A) { g[n](x) } is a Cauchy sequence

(4) { g[n]' } converges uniformly on A

Decide if we can conclude that

(5) { g[n] } converges uniformly on A

(2) /\(n:-|N) g[n]:A->|R diffable

(3) /\(x:-A) { g[n](x) } is a Cauchy sequence

(4) { g[n]' } converges uniformly on A

Decide if we can conclude that

(5) { g[n] } converges uniformly on A

NO

A=[0,oo); g[n](x) = x/n

A=[0,oo); g[n](x) = x/n

Let A be a connected subset of |R.

Let W be the set of all sequences { g[n] } such that

(1) /\(n:-|N) g[n]:A->|R diffable

(2) { g[n]' } converges uniformly on A

(3) \/(x:-A) { g[n](x) } is Cauchy.

Suppose that every sequence in W converges uniformly on A.

What can we conclude?

Let W be the set of all sequences { g[n] } such that

(1) /\(n:-|N) g[n]:A->|R diffable

(2) { g[n]' } converges uniformly on A

(3) \/(x:-A) { g[n](x) } is Cauchy.

Suppose that every sequence in W converges uniformly on A.

What can we conclude?

A is bounded

If A is not bounded, then put g[n](x)=x/n, for x:-A, n:-N.

If A is not bounded, then put g[n](x)=x/n, for x:-A, n:-N.

(1) h:[0,oo)->|R diffable

(2) h(0)=0

(3) /\(x>0) h(x)*h'(x) = 1 + h(x)*h(x)

Find all such functions.

(2) h(0)=0

(3) /\(x>0) h(x)*h'(x) = 1 + h(x)*h(x)

Find all such functions.

There are no such functions.

The only candidates would be h(x)=(+/-)sqrt(exp(2x)-1),

but they are not diffable at 0.

The only candidates would be h(x)=(+/-)sqrt(exp(2x)-1),

but they are not diffable at 0.

Let a[n] be a decreasing sequence of positive numbers.

(1) the series a[n] converges

(2) the series 2^k * a[2^k] converges

Prove that (1) => (2).

(1) the series a[n] converges

(2) the series 2^k * a[2^k] converges

Prove that (1) => (2).

page 58 in the first analysis notebook

Let 0 <= p.

Consider the series 1 / n*(log(n))^p.

Does it converge or diverge?

How does the answer depend on p ?

Consider the series 1 / n*(log(n))^p.

Does it converge or diverge?

How does the answer depend on p ?

converges for p > 1

diverges for 0 <= p <= 1

page 43 in redcrest

diverges for 0 <= p <= 1

page 43 in redcrest

Consider the series 1 / n*log(log(n)).

Does it converge or diverge?

Does it converge or diverge?

diverges

hint: 2^k*a[2^k]

hint: 2^k*a[2^k]

f:(0;1)u(1;oo)->|R, f(x)=1/(x*log(x))

Find a function F:(0;1)u(1;oo)->|R such that F'=f.

Find a function F:(0;1)u(1;oo)->|R such that F'=f.

F(x)=log( |log(x)| )

Let J be a connected subset of |R.

(1) g:J->|R continuous

(2) y:J->|R diffable

(3) /\(t:-J) y'(t) + a(t)*y(t) = 0

(4) \/(t:-J) y(t) != 0

(5) E = {t:-J : y(t)=0}

What can we conclude?

(1) g:J->|R continuous

(2) y:J->|R diffable

(3) /\(t:-J) y'(t) + a(t)*y(t) = 0

(4) \/(t:-J) y(t) != 0

(5) E = {t:-J : y(t)=0}

What can we conclude?

E is empty

(1) A,B c |R

(2) f:AxB->|R

(3) \/(L>0) /\(t:-A) /\(x,y:-B) |f(t,x) - f(t,y)| <= L * |x-y|

Can we conclude that f is continuous?

(2) f:AxB->|R

(3) \/(L>0) /\(t:-A) /\(x,y:-B) |f(t,x) - f(t,y)| <= L * |x-y|

Can we conclude that f is continuous?

No. In fact, f can even be discontinuous everywhere.

f:|Rx|R->|R

f(t,x)=0 if t is rational,

f(t,x)=1 if t is irrational.

f:|Rx|R->|R

f(t,x)=0 if t is rational,

f(t,x)=1 if t is irrational.

A = { (-oo;a) : a>0 }

A c P(|R)

Does |Q belong to the s-algebra generated by A ?

A c P(|R)

Does |Q belong to the s-algebra generated by A ?

No.

Let M = (-7 ; -1).

Let W = {Ec|R : McE or MnE=O}.

W is a s-algebra in |R (page 142 in 2nd measure)

A c W

s(A) c s(W) = W

|Q does not belong to W

Let M = (-7 ; -1).

Let W = {Ec|R : McE or MnE=O}.

W is a s-algebra in |R (page 142 in 2nd measure)

A c W

s(A) c s(W) = W

|Q does not belong to W

Let X be a set. Let W be a topology on X other than P(X).

Is it possible that W is uncountable and closed under taking arbitrary intersections?

Is it possible that W is uncountable and closed under taking arbitrary intersections?

Yes.

Let W = {O} u { Ec|R : [0,1] c E }.

Let W = {O} u { Ec|R : [0,1] c E }.

Let X be a set and let M be a subset of X.

Let W = { EcX : McE or MnE=O }.

Prove that W is closed under taking arbitrary unions and arbitrary intersections.

Let W = { EcX : McE or MnE=O }.

Prove that W is closed under taking arbitrary unions and arbitrary intersections.

page 142 in 2nd measure

Decide if there exist a subset of the Euclidean plane having
the following properties:

(1) it is connected

(2) it is dense

(3) it has empty interior

(1) it is connected

(2) it is dense

(3) it has empty interior

Yes.

{ (x,y) : x:-|Q and y=7 }

{ (x,y) : x:-|Q and y=7 }

f:X->Y

(1) /\(AcX) f-1(f(A)) c A

(2) /\(x,u:-X) f(x)=f(u) => x=u

Prove or disprove: (1) => (2)

(1) /\(AcX) f-1(f(A)) c A

(2) /\(x,u:-X) f(x)=f(u) => x=u

Prove or disprove: (1) => (2)

(1)=>(2) OK

page 86 in gen top

page 86 in gen top

f:X->Y

(1) /\(AcX) f-1(f(A)) c A

(2) /\(x,u:-X) f(x)=f(u) => x=u

Prove or disprove: (2) => (1)

(1) /\(AcX) f-1(f(A)) c A

(2) /\(x,u:-X) f(x)=f(u) => x=u

Prove or disprove: (2) => (1)

(2)=>(1) OK

page 86 in gen top

page 86 in gen top

f:X->Y

(1) /\(AcX) f-1(f(A)) c A

Can we conclude that Y=f(X) ?

(1) /\(AcX) f-1(f(A)) c A

Can we conclude that Y=f(X) ?

NO

f:{1}->{2,3}

f:{1}->{2,3}

Suppose that f:X->Y is injective (1-1).

Which of the two can we conclude:

(1) /\(AcX) f-1(f(A)) c A

(2) /\(BcY) B c f(f-1(B))

Which of the two can we conclude:

(1) /\(AcX) f-1(f(A)) c A

(2) /\(BcY) B c f(f-1(B))

We can conclude (1).

We cannot conclude (2).

The counter-example is f:{1}->{2,3}, f(1)=2, B = {2,3}.

We cannot conclude (2).

The counter-example is f:{1}->{2,3}, f(1)=2, B = {2,3}.

f:X->Y

(1) f(X)=Y

(2) /\(BcY) B c f(f-1(B))

Prove or disprove: (1) => (2)

(1) f(X)=Y

(2) /\(BcY) B c f(f-1(B))

Prove or disprove: (1) => (2)

(1)=>(2) OK

page 87 in gen top

page 87 in gen top

f:X->Y

(1) f(X)=Y

(2) /\(BcY) B c f(f-1(B))

Prove or disprove: (2) => (1)

(1) f(X)=Y

(2) /\(BcY) B c f(f-1(B))

Prove or disprove: (2) => (1)

(2)=>(1) OK

page 87 in gen top

page 87 in gen top

f:X->Y

(1) /\(BcY) B c f(f-1(B))

Can we conclude that function f is surjective, Y=f(X) ?

(1) /\(BcY) B c f(f-1(B))

Can we conclude that function f is surjective, Y=f(X) ?

it is surjective, Y=f(X)

page 87 in gen top

page 87 in gen top

f:X->Y

(1) /\(BcY) B c f(f-1(B))

Can we conclude that function f is injective (1-1) ?

(1) /\(BcY) B c f(f-1(B))

Can we conclude that function f is injective (1-1) ?

NO

f:{1,2}->{3}

f:{1,2}->{3}

Suppose that f:X->Y is surjective, Y=f(X).

Which of the two can we conclude:

(1) /\(AcX) f-1(f(A)) c A

(2) /\(BcY) B c f(f-1(B))

Which of the two can we conclude:

(1) /\(AcX) f-1(f(A)) c A

(2) /\(BcY) B c f(f-1(B))

We can conclude (2).

We cannot conclude (1).

The counter-example is: f:{1,2}->{3}, A={1}.

We cannot conclude (1).

The counter-example is: f:{1,2}->{3}, A={1}.

We are interested in whether a given function is bijective (1-1
and onto).

What are the two characteristics of bijective functions involving images and inverse images? (Give a detailed answer without proof.)

What are the two characteristics of bijective functions involving images and inverse images? (Give a detailed answer without proof.)

f:X->Y

(1) f is bijective

(2) /\(AcX) f-1(f(A)) c A

(3) /\(BcY) B c f(f-1(B))

thesis: (1) <=> (2)and(3)

-----------------------------

(2) <=> f is injective

(1) f is bijective

(2) /\(AcX) f-1(f(A)) c A

(3) /\(BcY) B c f(f-1(B))

thesis: (1) <=> (2)and(3)

-----------------------------

(2) <=> f is injective

Let X,Y be two topological spaces. Let f:X->Y be bijective (1-1
and onto).

(1) f and f-1 are both continuous

(2) /\(AcX) A is open in X <=> f(A) is open in Y

Prove or disprove: (1) => (2)

(1) f and f-1 are both continuous

(2) /\(AcX) A is open in X <=> f(A) is open in Y

Prove or disprove: (1) => (2)

(1) => (2) OK

page 88 in gen top

page 88 in gen top

Let X,Y be two topological spaces. Let f:X->Y be bijective (1-1
and onto).

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Prove or disprove: (1) => (2)

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Prove or disprove: (1) => (2)

(1) => (2) OK

page 88 in gen top

page 88 in gen top

Let X,Y be two topological spaces. Let f:X->Y be bijective (1-1
and onto).

(1) /\(BcY) B is open in Y <=> f-1(B) is open in X

(2) f and f-1 are both continuous

Prove or disprove: (1) => (2)

(1) /\(BcY) B is open in Y <=> f-1(B) is open in X

(2) f and f-1 are both continuous

Prove or disprove: (1) => (2)

(1)=>(2) OK

page 88 in gen top

page 88 in gen top

Let X,Y be two topological spaces. Let f:X->Y be continuous.

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is bijective, f-1 exists ?

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is bijective, f-1 exists ?

NO

We can have that all sets are open.

Then conditions (1) and (2) have no power over f.

------

Let f be a constant function from X into Y,

both spaces having such topologies that every set is open.

We can have that all sets are open.

Then conditions (1) and (2) have no power over f.

------

Let f be a constant function from X into Y,

both spaces having such topologies that every set is open.

Let X,Y be two connected subsets of |R.

Consider X,Y as topological spaces with the Euclidean topology.

f:X->Y

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is bijective, f-1 exists ?

Consider X,Y as topological spaces with the Euclidean topology.

f:X->Y

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is bijective, f-1 exists ?

NO

X=|R, Y=(0,1]

f(x)=1/(1+x^2)

X=|R, Y=(0,1]

f(x)=1/(1+x^2)

Let f be a continuous function from X onto Y. ( Y=f(X) )

Decide if it is generally true that:

(1) /\(AcX) A is open in X => f(A) is open in Y.

Decide if it is generally true that:

(1) /\(AcX) A is open in X => f(A) is open in Y.

NO

Let X = [0,5], Y = [7,10].

Consider these intervals as topological spaces with the Euclidean topology.

Let f:X->Y be defined as follows.

If 0 <= x <= 3, let f(x) = 10-x.

If 3 <= x <= 5, let f(x) = x+4.

Let X = [0,5], Y = [7,10].

Consider these intervals as topological spaces with the Euclidean topology.

Let f:X->Y be defined as follows.

If 0 <= x <= 3, let f(x) = 10-x.

If 3 <= x <= 5, let f(x) = x+4.

Let f be a continuous function from X onto Y. ( Y=f(X) )

Decide if it is generally true that:

(1) /\(AcX) A is closed in X => f(A) is closed in Y.

Decide if it is generally true that:

(1) /\(AcX) A is closed in X => f(A) is closed in Y.

NO

Let X = (1,5], Y = [7,9].

Consider these intervals as topological spaces with the Euclidean topology.

Let f:X->Y be defined as follows.

If 1 < x <= 3, let f(x) = x+6.

If 3 <= x <= 5, let f(x) = 12-x.

Let X = (1,5], Y = [7,9].

Consider these intervals as topological spaces with the Euclidean topology.

Let f:X->Y be defined as follows.

If 1 < x <= 3, let f(x) = x+6.

If 3 <= x <= 5, let f(x) = 12-x.

Let X,Y be two topological spaces. Let f:X->Y.

What does it mean that f is a homeomorphism?

What does it mean that f is a homeomorphism?

(1) f is bijective (1-1 and onto)

(2) f is continuous

(3) f-1 is continuous

========================================

(1) f is bijective (1-1 and onto)

(2) /\(AcX) A is open in X <=> f(A) is open in Y

======================================== page 89 in gen top

(2) f is continuous

(3) f-1 is continuous

========================================

(1) f is bijective (1-1 and onto)

(2) /\(AcX) A is open in X <=> f(A) is open in Y

======================================== page 89 in gen top

Let X,Y be two topological spaces. Let f:X->Y be bijective (1-1
and onto).

We have (1) /\(AcX) A is open in X <=> f(A) is open in Y.

Is this enough to conclude that f is a homeomorphism?

We have (1) /\(AcX) A is open in X <=> f(A) is open in Y.

Is this enough to conclude that f is a homeomorphism?

Yes.

page 89 in gen top

page 89 in gen top

Let X,Y be two topological spaces. Let f:X->Y be bijective (1-1
and onto).

We have (2) /\(BcY) B is open in Y <=> f-1(B) is open in X.

Is this enough to conclude that f is a homeomorphism?

We have (2) /\(BcY) B is open in Y <=> f-1(B) is open in X.

Is this enough to conclude that f is a homeomorphism?

Yes.

page 89 in gen top

homeomorphism (hoh mee uh mor'fiz uhm)

= a mathematical function between two topological spaces that is continuous, one-to-one, and onto, and the inverse of which is continuous

page 89 in gen top

homeomorphism (hoh mee uh mor'fiz uhm)

= a mathematical function between two topological spaces that is continuous, one-to-one, and onto, and the inverse of which is continuous

Let X,Y be two topological spaces. Let f:X->Y be continuous.

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is a homeomorphism?

(1) /\(AcX) A is open in X <=> f(A) is open in Y

(2) /\(BcY) B is open in Y <=> f-1(B) is open in X

Can we conclude that f is a homeomorphism?

NO

We can have that all sets are open.

Then conditions (1) and (2) have no power over f.

------

Let f be a constant function from X into Y,

both spaces having such topologies that every set is open.

We can have that all sets are open.

Then conditions (1) and (2) have no power over f.

------

Let f be a constant function from X into Y,

both spaces having such topologies that every set is open.

Let X,Y be two topological spaces.

What does it mean that they are homeomorphic?

What does it mean that they are homeomorphic?

There exists f:X->Y such that it is a homeomorphism.

Or, there exists g:Y->X such that it is a homeomorphism.

homeomorphism (hoh mee uh mor'fiz uhm)

= a mathematical function between two topological spaces that is continuous, one-to-one, and onto, and the inverse of which is continuous

Or, there exists g:Y->X such that it is a homeomorphism.

homeomorphism (hoh mee uh mor'fiz uhm)

= a mathematical function between two topological spaces that is continuous, one-to-one, and onto, and the inverse of which is continuous

Consider the sets |R and |Q.

Is it possible to equip them with such topologies that they become homeomorphic?

Is it possible to equip them with such topologies that they become homeomorphic?

NO

A homeomorphism between these two sets would have to be a bijective function.

Since they are not equinumerous, there cannot be a bijective function between them.

Hence, they cannot be homeomorphic.

A homeomorphism between these two sets would have to be a bijective function.

Since they are not equinumerous, there cannot be a bijective function between them.

Hence, they cannot be homeomorphic.

Let (X,G) be a topological space.

Consider a proper subset of X, say A, with the induced topology, call it G(A).

Is it possible that (A,G(A)) is homeomorphic with (X,G) ?

Consider a proper subset of X, say A, with the induced topology, call it G(A).

Is it possible that (A,G(A)) is homeomorphic with (X,G) ?

Yes.

The open interval (-1;1) is homeomorphic with (-7;7).

The open interval (-1;1) is homeomorphic with (-7;7).

(X,F,y) is a measure space

f:X->[0,oo] is measurable

y(X)<oo; 0 < M < oo

Leb*(X,f,dy) < M*y(X)

Can we conclude that y{x:f(x)>M}=0 ?

f:X->[0,oo] is measurable

y(X)<oo; 0 < M < oo

Leb*(X,f,dy) < M*y(X)

Can we conclude that y{x:f(x)>M}=0 ?

NO

f:[0,1]->[0,oo)

f(x)=3M/2, for 0<=x<=1/2

f(x)=0, for 1/2<x<=1

Leb([0,1],f,dy)=3M/4 < M

f:[0,1]->[0,oo)

f(x)=3M/2, for 0<=x<=1/2

f(x)=0, for 1/2<x<=1

Leb([0,1],f,dy)=3M/4 < M

0 < M, f:X->[M,oo)

Decide if we can conclude that

\/(a>0) /\(x:-X) a < f(x)/(1+f(x))

Decide if we can conclude that

\/(a>0) /\(x:-X) a < f(x)/(1+f(x))

hint1: a=M/(M+1)

hint2: argue by contradiction

hint2: argue by contradiction

0<M, f:X->[M,oo)

Prove that

/\(x:-X) M/(M+1) <= f(x)/(1+f(x))

Prove that

/\(x:-X) M/(M+1) <= f(x)/(1+f(x))

hint: g(x)=x/(1+x) is increasing

A > 0

Does there exist an open and dense subset of |R having Lebesgue measure less than A?

Does there exist an open and dense subset of |R having Lebesgue measure less than A?

yes

page 143 in 2nd measure`

page 143 in 2nd measure`

0 < A < 1

Does there exist a closed subset of [0,1] having empty interior

and having Lebesgue measure greater than A?

Does there exist a closed subset of [0,1] having empty interior

and having Lebesgue measure greater than A?

yes

page 143 in 2nd measure

page 143 in 2nd measure

(X,F,y) measure space

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (1)=>(2).

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (1)=>(2).

(1)=>(2) OK

page 144 in 2nd measure

page 144 in 2nd measure

(X,F,y) measure space

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (2)=>(1).

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (2)=>(1).

Consider |R with the Lebesgue measure.

f[n]=1/n

f[n]->0 in measure

but the integrals are always equal to infinity

-----

we need to suppose y(X)<oo in order that (2)=>(1)

page 145 in 2nd measure

f[n]=1/n

f[n]->0 in measure

but the integrals are always equal to infinity

-----

we need to suppose y(X)<oo in order that (2)=>(1)

page 145 in 2nd measure

(X,F,y) measure space, y(X) < oo

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (2)=>(1).

f[n]:X->[0,oo) sequence of measurable functions

(1) lim(n->oo) Leb( X, f[n] / (1+f[n]) , dy ) = 0

(2) f[n]->0 in measure

Prove or disprove (2)=>(1).

(2)=>(1) OK

page 145 in 2nd measure

page 145 in 2nd measure

f,g:[a,b]->|R Riemann integrable

g is nonnegative

m = inf f([a,b]), M = sup f([a,b])

Prove that there exists a number K, m <= K <= M, such that

Integral(a,b,f(x)g(x),dx) = K * Integral(a,b,g(x),dx)

g is nonnegative

m = inf f([a,b]), M = sup f([a,b])

Prove that there exists a number K, m <= K <= M, such that

Integral(a,b,f(x)g(x),dx) = K * Integral(a,b,g(x),dx)

page 36 in 2nd measure

If |A,|B are collections of sets, let |A & |B := s{AxB : A:-|A
, B:-|B}.

(the s-algebra generated by rectangles)

Let X be a countable set. Let Y be an arbitrary set. Both sets non-empty.

Prove that

(the s-algebra generated by rectangles)

Let X be a countable set. Let Y be an arbitrary set. Both sets non-empty.

Prove that

page 138 in 2nd measure

Let F be an algebra of sets in X.

Let y be a finite nonnegative additive function on F.

Let A,B belong to F and y(A)=y(B)=y(X)<oo.

What can we conclude about the value of y(AnB) ?

Let y be a finite nonnegative additive function on F.

Let A,B belong to F and y(A)=y(B)=y(X)<oo.

What can we conclude about the value of y(AnB) ?

y(AnB) = y(X)

page 139 in 2nd measure

page 139 in 2nd measure

W c P(X)

(1) A,B:-W => A\B:-W

(2) A,B:-W and AnB=O => AuB:-W

Can we conclude that A,B:-W => AuB:-W ?

(1) A,B:-W => A\B:-W

(2) A,B:-W and AnB=O => AuB:-W

Can we conclude that A,B:-W => AuB:-W ?

yes

page 140 in 2nd measure

page 140 in 2nd measure

Let S be a ring of sets in X. Let T be a ring of sets in Y.

R = { \\*//(k=1 to n) (A[k] x B[k]) : n:-|N, A[k]:-S, B[b]:-T }.

(finite disjoint unions of "SxT" rectangles)

Can we conclude that R is a ring?

R = { \\*//(k=1 to n) (A[k] x B[k]) : n:-|N, A[k]:-S, B[b]:-T }.

(finite disjoint unions of "SxT" rectangles)

Can we conclude that R is a ring?

page 140 in 2nd measure

Let F be a s-algebra in X. Let y:F->[0,oo] be monotone,
countably subadditive, y(O)=0.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Contemplate this definition.

Let S = {f:X->|R : f is measurable}.

Define d:SxS->[0,oo], by d(f,g) = sup(r>0) y{x:|f(x)-g(x)|>r}.

Contemplate this definition.

d(f,g) = y{ x: f(x)!=g(x) }

page 129 in 2nd measure

page 129 in 2nd measure

Prove that

lim(n->oo) n^(1/n) = 1

lim(n->oo) n^(1/n) = 1

page 85 in 1st analysis

Let X be an arbitrary non-empty set.

Let f:X->P(X) be an arbitrary function.

What can we conclude about f ?

Let f:X->P(X) be an arbitrary function.

What can we conclude about f ?

f is not onto = f is not surjective

f(X) != P(X)

page 172 in 2nd analysis

f(X) != P(X)

page 172 in 2nd analysis

Let X be a topological space. Let A,B c X.

Int(A\B) c ???

Int(A\B) c ???

Int(A\B) c Int(A) \ Int(B)

Let X be a topological space. Let A,B c X.

Decide if it is generally true that

Int(A) \ Int(B) c Int(A\B)

Decide if it is generally true that

Int(A) \ Int(B) c Int(A\B)

no

A=|R

B=|Q

A=|R

B=|Q

Let X be a topological space. Let A,B c X.

Decide if it is generally true that

Clo(A) \ Clo(B) c Clo(A\B)

Decide if it is generally true that

Clo(A) \ Clo(B) c Clo(A\B)

yes

page 68 in topologia (the old one, brickwall cover)

page 68 in topologia (the old one, brickwall cover)

Let X be a topological space. Let A,B c X.

Decide if it is generally true that

Clo(A\B) c Clo(A) \ Clo(B)

Decide if it is generally true that

Clo(A\B) c Clo(A) \ Clo(B)

NO

A=|R

B=|Q

A=|R

B=|Q

Prove that any two completions of one metric space are isometric.

page 31 in OLDTIMER

D c |C open

h:D->|R diffable

a,b:-D, [a,b] c D, a!=b

Prove that there exists p:-]a,b[ such that h(b)-h(a) = h'(p)*(b- a).

h:D->|R diffable

a,b:-D, [a,b] c D, a!=b

Prove that there exists p:-]a,b[ such that h(b)-h(a) = h'(p)*(b- a).

page 32 in OLDTIMER

Let S be an infinitely countable set.

Let F(S) be the set of all real-valued functions on S.

Does there exist a metric on F(S) which determines pointwise convergence?

Let F(S) be the set of all real-valued functions on S.

Does there exist a metric on F(S) which determines pointwise convergence?

YES

page 34 in OLDTIMER

page 34 in OLDTIMER

Let S be an uncountable set.

Let F(S) be the set of all real-valued functions on S.

Does there exist a metric on F(S) which determines pointwise convergence?

Let F(S) be the set of all real-valued functions on S.

Does there exist a metric on F(S) which determines pointwise convergence?

NO

page 56 in redcrest

page 35 in OLDTIMER

page 56 in redcrest

page 35 in OLDTIMER

Let (X,d) be a metric space.

Let x[n] be a sequence contained in this space.

Let S be the set of all subsequential limits of this sequence.

S = { x:-X : /\e>0 /\k \/(n>k) d(x[n],x) < e }

What can we say about set S ?

Let x[n] be a sequence contained in this space.

Let S be the set of all subsequential limits of this sequence.

S = { x:-X : /\e>0 /\k \/(n>k) d(x[n],x) < e }

What can we say about set S ?

S is closed

page 55 in redcrest

page 66 in golden gate

page 55 in redcrest

page 66 in golden gate