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Unformatted text preview: Math 3110 Homework 12 Solutions Exercise 13.1.1 (a) At least one of the intervals must contain infinitely many terms of the sequence, for otherwise, the sequence would only have finitely many terms in each of finitely many intervals, and hence only finitely many terms in total. Hence, there must be a subsequence consisting of points in a single compact interval. By Theorem 13.1, this interval is sequentially compact, so the subsequence consisting of terms of the original sequence in this interval has a convergent subsequence. This is a convergent subsequence of the original sequence, so S is sequentially combat. (b) Let a n = n for all natural numbers n and let S consist of all intervals of the form [ n . 1 , n + . 1] for all natural numbers n . Then a n is in the nth interval, but no two terms of the sequence are at a distance less than 1 from each other, so there can be no convergent subsequence....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Math

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