I have been trying to prove the following theorem

If $\{p_n\}$ is a sequence in a compact metric space X, then sub-sequence of $\{p_n\}$ converges to a point in $X$

My version of the prof is as follows: As $X$ is compact every open cover of $X$ has a finite sub-covering. So if $\{V_{\alpha}\}_{\alpha \in I}\supset X$ $\implies \exists\alpha_1,\alpha_2...\alpha_n $ s.t $\bigcup_{\alpha \in\{1,2,..n\}} V_\alpha \supset X$

Now since the finite open cover $V_\alpha$ covers X which contains a sequence which by definition has infinite elements, then there must be an open set for some $\alpha$ s.t it contains infinitely many terms of the sequence $\{p_n\}$

We take this open say say $O$ and the closure $\overline{O}$ is closed and therefore $\overline{O}\cap X$ is compact as $X$ is compact and $\overline{O}$ is closed

Now we can apply the same argument as above and obtain another open set say $O_1$ such that it contains infinitely many elements of the the sequence $\{p_n\}$

We can continue this process over and over again and clearly we can choose a sub-sequence $\{p_{n_k}\}$ such that the terms of this sequence get arbitrarily closed to a point in X.

Now the problem is I do not understand this clearly Can you help me figure out the gap in the reasoning? Or is my reasoning flawed? Edit: Ofcourse if the range of $\{p_n\}$ is finite then the problem is trivial.I want to understand the proof of the case when the range is infinite.

  • $\begingroup$ I feel you are complicating matters. I suggest you to use the Heine-Borel condition for compactness, ie closed and boundedness of $X.$ $\endgroup$
    – creative
    Sep 4 '14 at 8:05

Yes, steps marked "clearly" are often the hardest. The problem is that you start with a simgle open cover, and you may not get a really new cover in the subsequent steps. The key is to start right away with a suitable open cover and then use compactness.

What does it mean that a subsequence converges to $p\in X$? It means that each open neighbourhood contains infinitely many $p_n$. So if no subsequence converges to $p$, there exists an open neighbourhood $V_p$ of $p$ such that $p_n\in V_p$ holds only for finitely many $n$. Clearly (this time justified), $p\in V_p$ and hence $X\subseteq \bigcup_{p\in X}$. Now what contradiction do you get from a finite subcover?

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    $\begingroup$ So if I understand correctly , if we choose the finite sub-cover which contains finitely many elements of the sub-sequence from the over cover formed by $\{V_p\}_{p\in X}$, it will cover X(as X is compact). But clearly a finite cover ,with each open set having finitely many $p_n$ cannot contain a sequence(which by definition is infinite. Hence the contracdiction which completed the proof. Right? $\endgroup$ Sep 4 '14 at 10:15
  • $\begingroup$ @user3503589 Yes, you are right. Any finite subcover of this open cover cannot contain all of $X$ because it will only contain finitely many elements of the sequence. $\endgroup$
    – layman
    Sep 4 '14 at 15:26

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