hw5_solutions

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

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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 ,w ith 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 deFned 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 )suchthat 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
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This note was uploaded on 11/21/2011 for the course CSE 20 taught by Professor Foster during the Spring '08 term at UCSD.

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

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