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How might we rigorously argue that if we have a sequence $\{x_n\}\subset X$ such that every subsequence of it has a convergence subsequence that tends to $a$ and $X$ is a compact set then $\{x_n\}$ converges to $a$?

In my mind, I am thinking that if otherwise then then we can pick a subsequence with such that all terms lie at least a finite distance away from $a$ then there will be no subsequence that converges to $a$. Is this a valid argument? How might I make it "rigorous"? Thank you.

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It will work, to make it rigorous just use $\varepsilon$. You will see that in fact we don't need compactness. –  Davide Giraudo May 19 '12 at 8:30
    
@DavideGiraudo: Thank you! –  Terry May 19 '12 at 8:37

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Your idea is spot on:

Negation of definition of a limit: If $x_n$ does not converge to $a$ then there exists an $\epsilon>0$ such that $ |x_n-a| \geq \epsilon$ for arbitrarily large $n.$ So we can form a subsequence $x_{n_k}$ where this inequality holds. Any subsequence of this new sequence must certainly satisfy $| x_{n_{k_l}} - a| \geq \epsilon $ as $x_{n_k}$ satisfies it. So by definition of limit, $x_{n_{k_l}}$ can not tend to $a.$

As Davide pointed out as well, compactness is irrelevant here.

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Thank you, Ragib! –  Terry May 19 '12 at 8:37
    
@Ragib I have a proof to a similar question: math.stackexchange.com/questions/505401/…. –  user87274 Sep 26 '13 at 20:25

Your idea is fine.

From your mention of distance, I assume that you’re working in a metric space. If $\langle x_n:n\in\Bbb N\rangle$ does not converge to $a$, there is an $\epsilon>0$ such that for every $n\in\Bbb N$, there is an $m>n$ such that $d(x_m,a)\ge\epsilon$. If you wish to construct your ‘bad’ subsequence rigorously, you can argue as follows.

Choose $n_0$ so that $d(x_{n_0},a)>\epsilon$. Let $n_1$ be the smallest natural number such that (1) $n_1>n_0$ and (2) $d(x_{n_1},a)>\epsilon$; the hypothesis that $\langle x_n:n\in\Bbb N\rangle$ does not converge to $a$ ensures the existence of $n_1$. In general, given $n_k$, let $n_{k+1}$ by the smallest natural number such that (1) $n_{k+1}>n_k$ and (2) $d(x_{n_{k+1}},a)>\epsilon$; once again, the hypothesis that $\langle x_n:n\in\Bbb N\rangle$ does not converge to $a$ ensures the existence of $n_{k+1}$. Clearly the sequence $\langle x_{n_k}:k\in\Bbb N\rangle$ has no subsequence converging to $a$; this contradiction shows that $\langle x_n:n\in\Bbb N\rangle$ must actually converge to $a$.

Notice that nowhere did I use compactness: it’s not necessary. The result also does not require a metric: it’s true in arbitrary topological spaces, and the proof is essentially the same.

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Thank you, Brian! –  Terry May 19 '12 at 8:39

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