Math 117

Math 117 - Math 117, F 09, HW#2 (sample solutions) P.55,...

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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....
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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.

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Math 117 - Math 117, F 09, HW#2 (sample solutions) P.55,...

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