# sols2 - M114S, Solutions to HW #2 x2.3. We know P ( N ) = c...

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Unformatted text preview: M114S, Solutions to HW #2 x2.3. We know P ( N ) = c R from Theorem 2.26 together with Lemmas 2.24 and 2.25. Now we show that for all n 2, = c n . (Since R = c , this gives R n = c n , and then we are done by Proposition 2.2.) The bijection between and n is ( a ,a 1 ,... ) mapsto (( a ,a n ,a 2 n ,... ) , ( a 1 ,a n +1 ,a 2 n +1 ,... ) ,... , ( a n- 1 ,a 2 n- 1 ,a 3 n +1 ,... )) . That is, it unzips a single infinite sequence into n infinite sequences. It is easy to check that this really is a bijection. x2.4. The map ( f mapsto G f ) is an injection of ( A B ) into A B ; because if G f = G g , then for each x A , ( x,f ( x )) G f , and hence ( x,f ( x )) G g which implies that g ( x ) = f ( x ). Thus we have ( A B ) c P ( A B ). x2.5. On the one hand, ( N N ) c P ( N N ) = c P ( N ) by the preceding problem and Lemma 2.16; and on the other, P ( N ) = c ( N { , 1 } ) by the remarks following Exercise 2.23, and since ( N { , 1 } ) ( N N ), we have P ( N ) c ( N N ). Thus by the Schroder-Bernstein Theorem, (...
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## This homework help was uploaded on 04/16/2008 for the course MATH 114S taught by Professor Moschovakis during the Winter '08 term at UCLA.

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sols2 - M114S, Solutions to HW #2 x2.3. We know P ( N ) = c...

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