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Math Help - Subsequences and Subsequential Limits

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    Subsequences and Subsequential Limits

    One of my book problems is:
    Let sn be a bounded sequence and let S denote the set of subsequential limits of sn. Prove that S is closed.

    Here's my attempt:

    Let S' be the set of accumulation points (limit points) of S.
    If S' = \varnothing, then S' \subseteq S, so S is closed. Otherwise,
    Given any x \in S', where N*(x, \epsilon) represents the deleted neighborhood of x of length \epsilon:
    For any \epsilon > 0, N*(x, \epsilon/2) \cap S \neq \varnothing, so there is some y in the preceding intersection.
    y \in S, so for some subsequence tn of sn, tn converges to y. Then for some M, n > M \rightarrow |tn - y| < \epsilon/2.
    We know that 0 < |y - x| < \epsilon/2, so 0 < |y - x| + |tn - y| < \epsilon.
    Then, by triangle inequality, |tn - x| < \epsilon

    What I don't understand here is that this seems to show that tn converges to x which can't be true because tn converges to y by definition, and x \neq y.

    My goal here was to show that within any \epsilon x, there's some other subsequential limit y within \epsilon/2 of x since x is an accumulation point (limit point), and within \epsilon/2 of y, there has to be some element of some subsequence of sn. Thus, within any \epsilon of x, we can always find some element of sn that is \epsilon away from x, so x must be a subsequential limit and therefore in S.

    Any help in finding my error is greatly appreciated!
    Last edited by Lord Voldemort; April 17th 2013 at 10:55 PM.
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    Re: Subsequences and Subsequential Limits

    Quote Originally Posted by Lord Voldemort View Post
    One of my book problems is:
    Let sn be a bounded sequence and let S denote the set of subsequential limits of sn. Prove that S is closed.

    Here's my attempt: Let S' be the set of accumulation points (limit points) of S.
    If S' = \varnothing, then S' \subseteq S, so S is closed. Otherwise,
    Given any x \in S', where N*(x, \epsilon) represents the deleted neighborhood of x of length \epsilon:
    For any \epsilon > 0, N*(x, \epsilon/2) \cap S \neq \varnothing, so there is some y in the preceding intersection.
    y \in S, so for some subsequence tn of sn, tn converges to y. Then for some M, n > M \rightarrow |tn - y| < \epsilon/2.
    We know that 0 < |y - x| < \epsilon/2, so 0 < |y - x| + |tn - y| < \epsilon.
    Then, by triangle inequality, |tn - x| < \epsilon

    What I don't understand here is that this seems to show that tn converges to x which can't be true because tn converges to y by definition, and x \neq y.
    I don't think that you have an error in that work.
    It well known that is a Hausdorff space that the set of accumulation points of a set is closed.
    The only possible difficulty that I see is showing that if t is an accumulation point of S then t is subsequential limit of s_n.
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