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Consider a sequence $\{x_n\}$ in $S$. Where $S$ is a metric subspace. Given that every convergent subsequence of $\{x_{k(n)}\}$ converges to the same point say $x$. Prove that if S is compact, show that $\{x_n\}$ converges to $x$. Is my answer correct?

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Compactness seems irrelevant here. Try the contrapositive: Assume that $\{x_n\}$ does not converge to $x$, and see where that leads. –  Harald Hanche-Olsen Nov 4 '12 at 9:51
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definitely relevant. Consider what happens if S is not compact –  Mathematics Nov 4 '12 at 9:53
    
Compactness gives that each subsequence has a convergent subsequence. Now proceed as @Harald suggested. –  martini Nov 4 '12 at 9:55
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@HaraldHanche-Olsen Try $x_n=n$ in $S=\mathbb R$. There is no convergent subsequence hence the hypothesis holds. –  Did Nov 4 '12 at 10:04
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@Mathematics: I think you should edit your question, because it is otherwise misleading in my opinion. –  wj32 Nov 4 '12 at 10:29
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2 Answers 2

up vote 4 down vote accepted

I'll assume we're in a metric space with metric $d$.

Suppose $\{x_n\}$ does not converge to $x$. Then $\exists \, \epsilon > 0$ and a sequence of positive integers $m_1,m_2,\ldots$ such that $d(x, x_{m_i})\geq \epsilon$ for all positive integers $i$. The sequence $\{ x_{m_i}\}$ has a convergent subsequence because $S$ is compact. This subsequence converges to $x$, but that's a contradiction because no term in this subsequence is within $\epsilon$ of $x$.

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I think it should be a metric space –  Mathematics Nov 4 '12 at 10:36
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Every subsequent converges to the same point if S is compact and hence every subsequence are converges and converge to the same points and hence the sequence also convergent to x. Is the proof valid?

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I’m afraid not; I’m not sure that it even makes sense. Compactness of $S$ is needed to ensure that $\langle x_k:k\in\Bbb N\rangle$ has any convergent subsequences at all. The hypothesis then says that all of these convergent subsequences have the same limit, $x$. Finally, one argues as @littleO did: if the sequence doesn’t converge to $x$, it has a subsequence bounded away from $x$, and compactness of $S$ implies that this subsequence has a convergent limit. This limit must be different from $x$, and we get a contradiction. –  Brian M. Scott Nov 4 '12 at 16:48
    
Compactness ensures that the subsequence of $\{x_n\}$ is convergent and from the question mentioned, every convergent subsequence converge to the same point. That's why every convergent subsequence converge to the same point and hence the sequence$\{x_n\}_n$ –  Mathematics Nov 5 '12 at 6:52
    
You’re still trying to make it simpler than it actually is. Compactness only ensures that $\langle x_n:n\in\Bbb N\rangle$ has at least one convergent subsequence. Then the hypotheses of the problem tell you that all of its convergent subsequences converge to $x$. But it still takes some work to prove that $\langle x_n:n\in\Bbb N\rangle$ itself is convergent. You really do need littleO’s argument or something equivalent to it; you can’t just declare that $\langle x_n:n\in\Bbb N\rangle$ converges. –  Brian M. Scott Nov 5 '12 at 7:03
    
What about if i use the theorem that any compact subset is complete, is it work in this way? –  Mathematics Nov 5 '12 at 7:25
    
No. You really do need to use the argument that littleO gave and I repeated in an earlier comment. –  Brian M. Scott Nov 5 '12 at 7:35
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