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posted by  lizardking on 8/2/2009 10:06:23 PM  |  status: Closed  |  Earned Karma: 802

analysis

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Let Xn be a bounded sequence of integers.  Show that Xn has a subsequence that is eventually constant. 
So, i know that by the Bolzano-Weierstrass Thm for sequences that since Xn is bounded that it must  have a convergent subsequnce, but i dont really know what to do from there...any help would be greatly appreciated
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posted by lawrence_tutor on 8/3/2009 8:29:50 AM  |  status: Live
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If L is the limit of the subsequence, then L must be an integer.  If it weren't, then for E < (distance from L to the nearest integer), there would be no elements of the whole sequence within a distance E of L.  Thus the subsequence could not converge to L.

If you let E = 1/2, and let L be the limit of the subsequence, then for some N, you have | x_n - L | < 1/2 for every n < N, such that x_n is in the subsequence.  But the x_n are all integers, and the only integer within a distance of 1/2 from L is L itself.  So all these x_n are equal to L.  So, the subsequence is eventually constant.
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