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solutionshw8-554-2010

solutionshw8-554-2010 - Solutions homework 8 Page 47...

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Solutions homework 8. Page 47 Problem 1. : “ ” If | a n k | ≥ k , then is clear that there can’t be an M such that | a n | ≤ M for all n (take k > M to get a contradiction). ” Assume ( a n ) is unbounded. Then for all M there exists n such that | a n | > M . Let M = 1. Then there exists n 1 such that | a n 1 | > 1. Assume we have constructed a n k such that | a n k | > k for k = 1 , · · · , m . Then the sequence ( a n ) n>n m is also unbounded (otherwise the whole sequence would be bounded). Therefore we can find n m +1 > n m such that | a n m +1 | > m + 1. By induction we get the desired subsequence. Page 47 Problem 2. (a) (b): If ( a n ) divergent then for all a it is not true that for all > 0 we have | a n - a | < ultimately. Hence for all a there exists > 0 such that | a n - a | ≥ frequently. By Theorem 3.5.6 this implies that for all a there exists > 0 and a subsequence ( a n k ) such that | a n k - a | ≥ . (b) (a): Assume ( a n ) converges to some a R . Then also every subsequence ( a n k ) converges to a , which contradicts (b). Page 47 Problem 4. True, since | x n | ≤ 1 for all n the Bolzano-Weierstrass theorem implies that there is a convergent subsequence.
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