# Rudin's theorem on compact sets and limit points

I have trouble in understanding the proof of a theorem on compact sets and limit points in Rudin's book.

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 thus $$K$$. This contradicts the compactness of $$K$$.

My question is: How do we know the collection $$\{V_q\}$$ is an open cover of $$K$$, given $$V_q\cap E=\emptyset$$ or $$q\in E$$? I mean, the radius of each open neighborhood around $$q$$ is fixed, because each punctured neighborhood has to satisfy certain conditions $$V_q^* \cap E=\emptyset$$, where $$*$$ denotes "punctured." In this case, how can we ensure the collection of such neighborhoods still covers $$K$$?

• There is one $V_q$ for each $q$ in $K$. Therefore, $\{V_q\}$ covers $K$. Moreover, each $V_q$ contains at most one element of $E$. Since $E$ is infinite, we cannot remove all but a finite number of the $V_q$s. Therefore, we conclude that our assumption that $K$ contains no limit points of $E$ is false. – John Douma Jul 12 '18 at 20:57

Since $V_q$ is a neighborhood of $q$, $q\in V_q$. And so$$K=\bigcup_{q\in K}\{q\}\subset\bigcup_{q\in K}V_q.$$

Correct me if wrong.

A bit of context:

Let $X$ be a metric space.

$E \subset K \subset X$, $K$ compact, $E$ infinite.

Definition:

$p \in K$ is a limit point of $E$ if every neighbourhood of $p$ contains a point $q \not = p$ where $q \in E$.

Negation: $p$ is not a limit point of $E$:

There is a neighbourhood of $p$ that does not contain a $q \not = p$, $q \in E$.

Either:

1) $p \not \in E$ , then there is a $V_p$ such that

$V_p \cap E =\emptyset.$

Or

2) $p \in E$, then $V_p \cap E = {p}$.

Recall $E \subset K =\bigcup_{p} V_p$, $p \in K$.

Since $E$ is infinite, no finite subcollection can cover $E$, contradiction to $K$ compact.

P.S. Short proof of the above statement.

Since $K$ compact there is a finite subcollection

$E \subset K \subset \bigcup_{i} V_{p_i}$ , $i=1,2,...n.$

Recall; For every $i=1,2,...,n$, $V_{p_i}$ has at most $1$ element of $E$.

Hence a contradiction to $E$ is infinite.