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I'm reading principles of mathematical analysis and have a question about a theorem 2.37.

Theorem 2.37

If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

The proof is

If no point of $K$ were a limit point of $E$, each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$. It is clear that no finite subcollection of $\{V_q\}$ can cover $E$. The same is true of $K$, since $E \subset K$. This contradicts the compactness of $K$.

I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.

Thanks in advance.

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If we would have finite cover of $K$ it immediately will give us finite cover of $E$. – Norbert Aug 12 '12 at 22:29
Is it because $E$ is closed? At first, I thought the author is saying that any subset of a compact set is compact, and I was confused. – Tengu Aug 13 '12 at 1:23
@Tengu $E$ is closed if no point of $K$ were a limit point of $E$, for $\overline E=E\cup E^d$ and here $E^d=\emptyset$. Therefore, $E$ must be compact because any closed subset of a compact space is compact. – Paul Aug 13 '12 at 2:09
@Paul I got it! Thanks! – Tengu Aug 13 '12 at 2:15
@WacDonald's: your comment is misinformed and may confuse others (myself included). In fact, there is no assumption that $K$ lives in a metric space. Yes, that means this proof shows that compact implies limit point compact. I am not sure why you believe this statement to be false. If you don't believe this proof, it also follows from the implications compact implies countably compact implies limit point compact. Check out the first two properties listed on – pre-kidney Aug 28 '13 at 17:44
up vote 3 down vote accepted

A set with no limit points is necessarily a closed set. Being a closed subset of a compact space, it is compact. On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.

Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$. That set is open, since as noted above, $E$ is closed. Now you've got an open cover of $K$. It must therefore have a finite subcover. But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction.

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I wasn't really caring about the closedness of $E$. If $E$ has a limit point, then it should be in $K$ since $K$ is closed. Even if $E$ doesn't have limit points, $E$ is closed and thus $E^c$ is open. You solved my confusion thanks! – Tengu Aug 13 '12 at 1:21

Since for all $q$, $V_q \cap E$ has at most one element, for any finite subset $\{q_1,\ldots,q_n\}$,

$\bigcup_{i=1}^n V_{q_i} \cap E$ has at most $n$ elements, i.e., is finite. But $E$ is assumed to be infinite, so we cannot have $\bigcup_{i=1}^n V_{q_i} \supset E$. Since $K \supset E$, therefore we cannot have $\bigcup_{i=1}^n V_{q_i} \supset K$, contradicting the compactness of $K$.

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Here is my comment, it maybe a little far from the topic, however, I hope it will be helpful for you.

The compactness condition is too strong, we only need the countable extent condition in the Theorem.

Countable extent= The cardinality of any closed discrete subspace must be countable.

So, in fact, we have

If $E$ is an infinite subset of a space $K$ which is countable extent, then $E$ has a limit point in $K$.

There are many topological space which is countable extent but not compact. For example, countably compact space, lindelof space etc. It is far from countable extent to compactness.

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@tomasz: The uncountable discrete space in the compactification is not closed. I remind you to notice the definition of countable extent. – Paul Aug 13 '12 at 2:43
Mea culpa. I confused extent and spread. – tomasz Aug 13 '12 at 2:45

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