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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