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The below is the proof for the theorem that if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

The only part I can't accept in the proof is the last sentence. So, it's shown that no finite subcollection of $\{V_q\}$ can cover $K$. However, how does this contradict the compactness of $K$?

Thank you!

$\mathbf{2.37}\,\,$ *Theorem*$\,\,\,\,$*If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.*

*Proof*$\quad$If no point of $K$ where a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$ (namely, $q$, if $q\in E$). 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 $K$.

InfimumMaximum
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2 Answers2

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Each $q \in K$ has a neighbourhood $V_q$ that contains only finitely many points of $E$. The family $\left(\overset{\circ}{V}_q\right)_{q\in K}$ is an open cover of $K$, hence has a finite subcover, since $K$ is compact.

That contradicts the fact that $E$ is infinite, since any finite subfamily of the $V_q$ can only cover finitely many points of $E$.

Daniel Fischer
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Definition: A set $K$ is compact if whenever $\mathscr{U}$ is a family of open sets that covers $K$, then some finite subfamily of $\mathscr{U}$ covers $K$.

Here $\mathscr{V}=\{V_q:q\in K\}$ is a family of open sets covering $K$, and $K$ is compact, so by the definition of compactness some finite subfamily of $\mathscr{V}$ must cover $K$. But the argument shows that this is not the case, so we have a contradiction.

It’s possible that you’ve been given some other definition of compactness rather than the standard one. If so, however, you should already have seen a proof that it implies the standard definition that I gave above.

Brian M. Scott
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