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

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 $K$.

• Without loss of generality, each $V_q$ is open. $(V_q)_{q\in K}$ covers $K$. – Daniel Fischer Jul 22 '13 at 20:46
• What is your definition of "compact"? – GEdgar Jul 22 '13 at 20:48
• If you find one open covering of $K$ that doesn't admit a finite subcovering, then $K$ is not compact. – egreg Jul 22 '13 at 23:35

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 $E$.
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.