1. 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).

2. Proof of (a) Let b  f(C  D). Then b = f(a) for some a  C  D. a  C /\ a  D b  f(C)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)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)

3. 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)

4. 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  DC  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).

5. 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

6. 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

7. 1  (b) Exercises 4.5 (pages 223-225)

8. 2  (b)

9. 2  (d)

10.  (e)

11. 2  (f)

12. 3  (a)

13. 3  (b)

14.  (c)

15. 3  (d)

16.  (e)

17. 3  (f)

18. 4  (a)

19. 4  (d)

20.  (e)

21. 4  (f)

22. 6 6 