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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3Compactness seems irrelevant here. Try the contrapositive: Assume that ${x_n}$ does not converge to $x$, and see where that leads. – Harald Hanche-Olsen Nov 04 '12 at 09:51
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1definitely relevant. Consider what happens if S is not compact – Mathematics Nov 04 '12 at 09:53
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Compactness gives that each subsequence has a convergent subsequence. Now proceed as @Harald suggested. – martini Nov 04 '12 at 09:55
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1@HaraldHanche-Olsen Try $x_n=n$ in $S=\mathbb R$. There is no convergent subsequence hence the hypothesis holds. – Did Nov 04 '12 at 10:04
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What does "every convergent subsequence of $x_{k(n)}$" mean? What is $k(n)$? – wj32 Nov 04 '12 at 10:06
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$k(n): \mathbb{N}\rightarrow \mathbb{N}$ – Mathematics Nov 04 '12 at 10:13
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@Mathematics: No, did you mean "every convergent subsequence $x_{k(n)}$" or is $k(n)$ fixed? – wj32 Nov 04 '12 at 10:20
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Ooops, I had missed the word “convergent” in the problem statement. – Harald Hanche-Olsen Nov 04 '12 at 10:22
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the former one, $k(n)$ is not fixed – Mathematics Nov 04 '12 at 10:22
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1@Mathematics: I think you should edit your question, because it is otherwise misleading in my opinion. – wj32 Nov 04 '12 at 10:29
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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 $x$.
littleO
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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?
Mathematics
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1I’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 04 '12 at 16:48
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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 05 '12 at 06:52
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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 05 '12 at 07:03
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What about if i use the theorem that any compact subset is complete, is it work in this way? – Mathematics Nov 05 '12 at 07:25
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No. You really do need to use the argument that littleO gave and I repeated in an earlier comment. – Brian M. Scott Nov 05 '12 at 07:35