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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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, (...
View Full Document

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.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online