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Math Help - Closed set

  1. #1
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    Closed set

    I am working on this problem, and I am not 100% sure with my argument. Could someone help me verify it?

    Let (X,d) be a metric space. Suppose that <x_n> is a sequence in X such that it has no convergent subsequence. Claim: A=\{x_1,x_2,x_3,...\} is closed in X.

    By assumption, for every sequence <p_n> in A, <p_n> does not converge. Otherwise, <x_n> has a convergent subsequence. This implies that A is closed since I'll show that a set A is closed in (X,d) iff for every sequence <p_n> in A if <p_n> converges then lim p_n \in A.

    First direction: Suppose A is closed. Let <p_n> be a convergent sequence in A. Assume lim p_n=p \notin A, then lim p_n \in X-A. But X-A is open and lim p_n \in X-A implies that there exists N such that if n>N, then p_n \in X-A. Contradicts to our assumption that <p_n> is a sequence in A.

    Second direction: I'll show X-A is open. Let <p_n> be a sequence in X converging to p \in X-A. It suffices to show that there exists N such that if n>N, then p_n \in X-A. Assume p_n \in X-A for only finitely many times, then <p_n> is a sequence in A. By hypothesis, lim p_n \in A. But  lim p_n=p \in X-A. Contradiction. So, X-A is open, ie A is closed.
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  2. #2
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    I think there is a technicality that you need to be careful of here.

    Let <p_n> be the sequence <x_2,x_1,x_4,x_3,...>. Then <p_n> is a sequence in A, but it is not a subsequence of <x_n>.

    Thus your statement following "by assumption" is not in fact by assumption, but must be proved.
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  3. #3
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    THEOREM If A is a subset of a metric space (M,d) and p is a limit point of A then there is sequence of distinct points from A that converges to p.
    If you can prove that theorem the you have done this question.
    i.e. The set has no limits points.
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  4. #4
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    Quote Originally Posted by DrSteve View Post
    I think there is a technicality that you need to be careful of here.

    Let <p_n> be the sequence <x_2,x_1,x_4,x_3,...>. Then <p_n> is a sequence in A, but it is not a subsequence of <x_n>.

    Thus your statement following "by assumption" is not in fact by assumption, but must be proved.
    Thanks a lot for your correction, DrSteve. I should have seen that.
    Thanks Plato. I think I can prove that statement. Could you verify my argument for me? Since p is a limit point of A, every neighborhood of p has nonempty intersection with A-\{p\}. So, there exists p_n \in B(p,1/n) \bigcap A-\{p\}. The sequence <p_n> converges to p. But how can I be sure that this is a sequence of distinct points. Also, I don't see how proving this means I'm done with the original question. Could you elaborate on that please?
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  5. #5
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    Quote Originally Posted by jackie View Post
    how can I be sure that this is a sequence of distinct points.
    For each n make sure that \delta_n=\min\left\{\frac{1}{n},d(p,x_{n-1}\right)\}.
    Pick p_n\in\mathcal{B}(p;\delta_{n})\setminus\{p\}
    Last edited by Plato; March 16th 2011 at 05:06 PM.
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  6. #6
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    From a PM.
    You know that no subsequence of <x_n> converges.
    So we have a contradiction.
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