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??? 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