This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 117, F 09, HW#2 (sample solutions) P.55, #14. **Note the position of Theorem 1.4 in the development it occurs before Theorem 1.7, and in fact it is used in the proof of Theorem 1.7. So Theorem 1.4 needs a proof that does not rely on Theorem 1.7! Let { a n } n =1 be a Cauchy sequence. Then (choosing = 1), there exists a positive integer N such that  a n a m  < 1 for all n,m N . In particular, taking m = N , we get that  a n a N  < 1 for all n N . Hence, a N 1 < a n < a N + 1 for all n N . Now set L = min { a 1 ,a 2 ,...,a N 1 ,a N 1 } U = max { a 1 ,a 2 ,...,a N 1 ,a N + 1 } . We have chosen L so that L a n for n = 1 , 2 ,...,N 1, and in addition, L a N 1 < a n for all n N . Thus L a n for all n . Similarly, a n U for all n . Therefore { a n } n =1 is bounded. P.55, #18. For instance, take S = { 1 k 1  k N } . Then S has exactly two accumulation points, 1 and 1. P.55, #19. For example, we could use the set in Problem 21. A little bit simpler is the set S = { n + 1 k  n,k N } . Then the accumulation points of S are precisely the positive integers, and we know that the set of positive integers is countably infinite....
View
Full
Document
This note was uploaded on 08/27/2011 for the course MATH 117 taught by Professor Akhmedov,a during the Fall '08 term at UCSB.
 Fall '08
 Akhmedov,A
 Math

Click to edit the document details