# In a compact metric space, every sequence has a convergent subsequence

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.

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

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?
• 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? Sep 4 '14 at 10:15
• @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. Sep 4 '14 at 15:26