DEFINITION Let f : A B and let X A and Y B. The image (set) of X is

f(X) = {y B : y = f(x) for some x X}

and the inverse image of Y is

f –1(Y) = {x A : f(x) Y}.

Look at the examples and comments on pages 220, 221, and 222.

Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E

and F be subsets of B. Then

(a) f(C D) f(C) f(D) ,

(b) f(C D) = f(C) f(D) ,

(c) f –1(E F) = f –1(E) f –1(F) ,

(d) f –1(E F) = f –1(E) f –1(F) .

Proof of (a)

Let b f(C D). Then b = f(a) for some a C D.

a C /\ a D

b f(C)

Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E

and F be subsets of B. Then

(a) f(C D) f(C) f(D) ,

(b) f(C D) = f(C) f(D) ,

(c) f –1(E F) = f –1(E) f –1(F) ,

(d) f –1(E F) = f –1(E) f –1(F) .

___________________________

___________________________

b f(D) ___________________________

definition of C D

a C /\ b = f(a)

a D /\ b = f(a)

b f(C) f(D) ___________________________definition of f(C) f(D)

f(C D) f(C) f(D) b f(C D) b f(C) f(D)

Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E

and F be subsets of B. Then

(a) f(C D) f(C) f(D) ,

(b) f(C D) = f(C) f(D) ,

(c) f –1(E F) = f –1(E) f –1(F) ,

(d) f –1(E F) = f –1(E) f –1(F) .

Proof of (b)

Let b f(C D). Then b = f(a) for some a C D.

[a C /\ b = f(a)] \/ [a D /\ b = f(a)]

[b f(C)] \/ [b f(D)]

___________________________

___________________________

b f(C) f(D) ___________________________

definition of C D

definition of image

definition of f(C) f(D)

f(C D) f(C) f(D) b f(C D) b f(C) f(D)

Proof of (b)

Let b f(C D). Then b = f(a) for some a C D.

[a C /\ b = f(a)] \/ [a D /\ b = f(a)]

[b f(C)] \/ [b f(D)]

___________________________

___________________________

b f(C) f(D) ___________________________

definition of C D

definition of image

definition of f(C) f(D)

f(C D) f(C) f(D) b f(C D) b f(C) f(D)

Let b f(C) f(D). Then b f(C) or b f(D).

f(C) f(D) f(C D), since b f(C) f(D) b f(C D)

If b f(C), then b = f(a) for some a C C D.

If b f(D), then b = f(a) for some a D C D.

In either case, we can say b = f(a) for some a _________ .C D

Since f(C D) f(C) f(D) and f(C) f(D) f(C D), we have f(C D) = f(C) f(D).

Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E

and F be subsets of B. Then

(a) f(C D) f(C) f(D) ,

(b) f(C D) = f(C) f(D) ,

(c) f –1(E F) = f –1(E) f –1(F) ,

(d) f –1(E F) = f –1(E) f –1(F) .Proof of (c)a f –1(E F)f(a) E F

___________________________definition of inverse image

f(a) E /\ f(a) F

___________________________definition of intersection

a f –1(E) /\ a f –1(F)

___________________________definition of inverse image

a f –1(E) f –1(F)

___________________________definition of intersection

Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E

and F be subsets of B. Then

(a) f(C D) f(C) f(D) ,

(b) f(C D) = f(C) f(D) ,

(c) f –1(E F) = f –1(E) f –1(F) ,

(d) f –1(E F) = f –1(E) f –1(F) .Proof of (d)a f –1(E F)f(a) E F

___________________________definition of inverse image

f(a) E \/ f(a) F

___________________________definition of union

a f –1(E) \/ a f –1(F)

___________________________definition of inverse image

a f –1(E) f –1(F)

___________________________definition of union

1 (b)

Exercises 4.5 (pages 223-225)

2 (b)

2 (d)

(e)

2 (f)

3 (a)

3 (b)

(c)

3 (d)

(e)

3 (f)

4 (a)

4 (d)

(e)

4 (f)

6