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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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    @Mathematics: I think you should edit your question, because it is otherwise misleading in my opinion.2012-11-04

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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 $x$.

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    I think it should be a metric space2012-11-04
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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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    No. You really do need to use the argument that littleO gave and I repeated in an earlier comment.2012-11-05