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hw5_solutions

hw5_solutions - Problem Set 5 Solutions Problem 1(The given...

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Problem Set 5 Solutions Problem 1: (The given statement is not true, so we will state and prove its negation.) 1. Statement: There are sets A, B, X , with X A , and a function f : A B , such that f 1 ( f ( X )) is not a subset of X . 2. Proof: Let A = { a 1 , a 2 } , X = { a 1 } , B = { b } . Then X A . Let f : A B be defined by f ( a 1 ) = b and f ( a 2 ) = b . Since a 1 X and f ( a 1 ) = b , b f ( X ). And since b f ( X ) and f ( a 2 ) = b , a 2 f 1 ( f ( X )). Since a 2 f 1 ( f ( X )) but a 2 X , f 1 ( f ( X )) X . Problem 2 Lemma 1: f ( f 1 ( Y )) Y . Proof of Lemma 1: Let b Y . Since f is onto, a A such that f ( a ) = b . Since f ( a ) = b and b Y , a f 1 ( Y ). This implies that f ( a ) f ( f 1 ( Y )). But since b = f ( a ), this means that b f ( f 1 ( Y )). Lemma 2: f ( f 1 ( Y )) Y . Proof of Lemma 2: Let b f ( f 1 ( Y )). Then a f 1 ( Y ) such that f ( a ) = b . Since a f 1 ( Y ), f ( a ) Y . Since f ( a ) = b , this means that b Y . Claim: f ( f 1 ( Y )) = Y . Proof of Claim: Since f ( f 1 ( Y )) Y (by Lemma 1) and f ( f 1 ( Y )) Y (by Lemma 2), f ( f 1 ( Y )) = Y .
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