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Thread: 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 $\displaystyle <x_n>$ is a sequence in X such that it has no convergent subsequence. Claim: $\displaystyle A=\{x_1,x_2,x_3,...\}$ is closed in X.

    By assumption, for every sequence $\displaystyle <p_n>$ in A, $\displaystyle <p_n>$ does not converge. Otherwise, $\displaystyle <x_n>$ 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 $\displaystyle <p_n>$ in A if $\displaystyle <p_n>$ converges then $\displaystyle lim p_n \in A$.

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

    Second direction: I'll show X-A is open. Let $\displaystyle <p_n>$ be a sequence in X converging to $\displaystyle p \in X-A$. It suffices to show that there exists N such that if n>N, then $\displaystyle p_n \in X-A$. Assume $\displaystyle p_n \in X-A$ for only finitely many times, then $\displaystyle <p_n>$ is a sequence in A. By hypothesis, $\displaystyle lim p_n \in A$. But $\displaystyle lim p_n=p \in X-A$. 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 $\displaystyle <p_n>$ be the sequence $\displaystyle <x_2,x_1,x_4,x_3,...>$. Then $\displaystyle <p_n>$ is a sequence in $\displaystyle A$, but it is not a subsequence of $\displaystyle <x_n>$.

    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 $\displaystyle A$ is a subset of a metric space $\displaystyle (M,d)$ and $\displaystyle p$ is a limit point of $\displaystyle A$ then there is sequence of distinct points from $\displaystyle A$ that converges to $\displaystyle p$.
    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 $\displaystyle <p_n>$ be the sequence $\displaystyle <x_2,x_1,x_4,x_3,...>$. Then $\displaystyle <p_n>$ is a sequence in $\displaystyle A$, but it is not a subsequence of $\displaystyle <x_n>$.

    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 $\displaystyle p$ is a limit point of $\displaystyle A$, every neighborhood of $\displaystyle p$ has nonempty intersection with $\displaystyle A-\{p\}$. So, there exists $\displaystyle p_n \in B(p,1/n) \bigcap A-\{p\}$. The sequence $\displaystyle <p_n>$ 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 $\displaystyle n$ make sure that $\displaystyle \delta_n=\min\left\{\frac{1}{n},d(p,x_{n-1}\right)\}$.
    Pick $\displaystyle p_n\in\mathcal{B}(p;\delta_{n})\setminus\{p\}$
    Last edited by Plato; Mar 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 $\displaystyle <x_n>$ converges.
    So we have a contradiction.
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