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If I have a compact set $A$ and a closed subset $\Sigma \subset A$ which only contains isolated points (that is, none of them is a limit point). Does the compactness of $A$ then force $\Sigma$ to have finite cardinality ?

Here is my attempt at a proof that the above question can be answered in the positive:

Suppose for contradiction that $\Sigma$ contains infinitely many distinct points.

EDIT : Then we can construct a sequence of points in $\Sigma$ which consists of distinct points.

By compactness of A, this sequence must have a convergent subsequence, and by the fact that $\Sigma$ is closed, this limit lies in $\Sigma$. But then it cannot be a limit point, because all points in $\Sigma$ are isolated. So the subsequence must eventually constant and equal to the limit, contrary to the construction of the sequence.

Is the reasoning above correct ? If no, what did go wrong ?

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3 Answers 3

up vote 5 down vote accepted

What you did seems correct if you can show that you can indeed extract a converging subsequence. In fact, you don't need to do it by contradiction: for each $x\in \Sigma$, pick $U_x$ an open neighborhood which doesn't meet any point of $\Sigma$ except $x$. Then $\{U_x\}_{x\in\Sigma}$ is an open cover of $\Sigma$, which is compact as a closed set of a compact, so you can extract a finite sub-cover.

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Can I ask a question? Is it false that Given a topological space X, A is the set of all isolated points, then A is closed? I have not thought of any counter example. A should be infinite, so the union is not closed. –  user1412 May 10 '12 at 20:15
    
@Long Mai: No, it's false in general. What is always true is that $A$ is open. –  Brian M. Scott May 10 '12 at 20:20
    
If we just have X compact, A contains all isolated points, and A is not closed, can we have a counter example that A is not finite? –  user1412 May 10 '12 at 20:27
    
@Long Mai: Sure: let $X=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$, with the topology inherited from $\Bbb R$. This space is compact, since every open set containing $0$ contains all but finitely many of the other points, but the set of isolated points is the infinite set $\{2^{-n}:n\in\Bbb N\}$. –  Brian M. Scott May 10 '12 at 20:29
    
@BrianM.Scott: Thank you very much. –  user1412 May 10 '12 at 20:34

What you did is not quite correct even assuming that you're working in a first countable space, so that compactness is equivalent to sequential compactness. Assuming that $\Sigma$ is infinite to get a contradiction is fine (if unnecessary), but that doesn't make $\Sigma$ a sequence: it has no ordering, and it might even be uncountable. However, if $\Sigma$ is infinite there must be a sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points of $\Sigma$. Then this sequence must have a convergent subsequence, and your argument goes through from there. It's a small point, but it's a good idea to get into the habit of precision in order to avoid confusing yourself when things get more complicated.

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Thanks very much for pointing this out! I edited the post accordingly, hope this is now more accurate ! –  harlekin May 10 '12 at 21:06
    
@harlekin: Yes, that's much better. –  Brian M. Scott May 10 '12 at 21:10

Compactness does not mean that sequences have convergent subsequences in a general topological space. See sequential compactness at Wikipedia.

However, your statement is still true. Just stick more closely to the definition of compactness.

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thanks for your comment ! The case that led me to think about my question above takes place in Euclidean space, you're absolutely right for the general situation and I am glad you reminded me of that! This also shows that Davide's argument is the one I should adopt going forward –  harlekin May 10 '12 at 20:17

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