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I'm having a problem with the proof of this theorem in Rudin's Principles of Mathematical Analysis.

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

Proof: If no point of $K$ were a limit point of $E$, then 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 $E$; and the same is true of $K$, since $E\subset K$. This contradicts the compactness of $K$.

My problem is this: Doesn't what is shown only disprove the compactness of $E$ (given the assumption that no limit point exists)? For it to contradict the compactness of $K$, I thought that we would first have to prove that $\cup \{ V_q \}$ is an open cover of $K$, because then we would have found an open cover of K with no finite subcover, which clearly contradicts the compactness of $K$. I can imagine a counterexample where $\cup \{ V_q \}$ is not an open cover of $K$, so What am I missing?

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It is an open cover of $K$. The cover is the collection $\{\,V_q \mid q\in K\,\}$. –  David Mitra Apr 15 '12 at 16:00

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up vote 3 down vote accepted

This follows from the construction: You choose a neigbourhood $V_q$ for each point $q \in K$. So, since $q \in V_q$, we have $K \subseteq \bigcup_{q \in K} V_q$.

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I see that I was thinking about the proof in a completely incorrect way (only considering the $q\in E$). Thanks! –  NeuroFuzzy Apr 15 '12 at 16:11

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