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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).

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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)

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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)

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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).

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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

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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

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1 (b) Exercises 4.5 (pages 223-225)

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2 (b)

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2 (d)

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(e)

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2 (f)

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3 (a)

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3 (b)

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(c)

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3 (d)

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(e)

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3 (f)

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4 (a)

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4 (d)

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(e)

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4 (f)

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