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I have to prove the fact that, given a metric space $(X,d)$ and a subset $K$ of $X$ compact, taking a closed subset $C$ of $K$, this $C$ is compact too.

I have used the characterization of closed sets with sequences, stating that there exist a generic sequence $Xn$ in $C$ that converges to an $x$ belonging to the same set $C$.

Then, using the fact that, since a sequence converges, every subsequence converges to the same limit, I stated that, under the Bolzano-Weierstrass property, set $C$ is compact since a generic sequence of its has a convergent subsequence to a some point of the set.

Now my question is: is that reasoning correct? Since in the solution of the question, the professor uses the fact that $K$ is compact to state the existence of a subsequence that converges.

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    $\begingroup$ What are you trying to prove? That every sequence in C has a convergent subsequence? (Note: the proof that C is also compact may be easier using the open cover equivalence ;) ) $\endgroup$ – Guillermo Mosse Oct 8 '15 at 14:09
  • $\begingroup$ I found a result that states that every convergent sequence in a set has convergent subsequences and then I tried to apply the Bolzano-Weierstrass property to check for compactness. $\endgroup$ – PhDing Oct 8 '15 at 14:40
  • $\begingroup$ I'll just note that this has nothing to do with metric spaces: a closed subset of a compact space is compact. $\endgroup$ – Justin Young Oct 8 '15 at 15:22
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Your argument is not correct. Every sequence in $C$ has a convergent subsequence because $K$ is compact. However, the subsequential limit is in $K$. You need to show that every subsequential limit is in $C$.

Let $\{x_n\}$ be a sequence in $C$. Because, $K$ is compact, $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ which converges to a point in $K$. Let $x=\lim_{k\to\infty}x_{n_k}$. So, $x\in\overline C$. But $C$ is closed, therefore $x\in C$.

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  • $\begingroup$ Ok that's clear. But does this mean that it is not true that every convergent sequence has convergent subsequences? $\endgroup$ – PhDing Oct 8 '15 at 14:41
  • $\begingroup$ Yes, that is true, but to show that $C$ is compact, you need to show that every sequence in $C$, convergent or not, has a subsequence which converges in $C$. $\endgroup$ – Tim Raczkowski Oct 8 '15 at 14:45
  • $\begingroup$ Ok, so just keeping the convergent ones, I am excluding the non convergent; it is not like taking a generic one, is it too specific? $\endgroup$ – PhDing Oct 8 '15 at 14:54
  • $\begingroup$ Yes. That's the idea. $\endgroup$ – Tim Raczkowski Oct 8 '15 at 14:55

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