0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.