# Every open set in a locally compact Hausdorff space contains an open set whose closure is compact and contained in the open set

I know that if X is locally compact and Hausdorff, then any non-empty open set $S$ contains a non-empty closed set. I know this to be the case because a locally compact space is a regular space, in which the claim holds.

But why does any open $S$ contain a non-empty open set whose closure is compact and contained in $S$?

Context: a version of this is claimed at the beginning of the proof of the Baire category theorem in Rudin's Functional Analysis text.

Edit: the definition of local compactness that I'm familiar with: $X$ is locally compact if for all $x \in X$ there exists an open set containing $x$ whose closure is compact in $X$.

• Any singleton is compact and closed...but I'm not sure you've stated what you want correctly. – Eric Wofsey Dec 30 '17 at 1:09
• What is your definition of locally compact? – Qiaochu Yuan Dec 30 '17 at 1:10
• If there's a compact set, it will be closed automatically, see my answer. Or I missed your point? – Ivo Terek Dec 30 '17 at 1:12
• I upvoted Eric Wofsey's comment because it completely answers the question as stated. But you probably intended for that compact closed set to have nonempty interior. Some formulations of the definition of "locally compact" make that result immediate. So @QiaochuYuan's comment becomes relevant. To get an answer more useful than "just look at the definition of local compactness", you'll need to say what definition of local compactness you're using. – Andreas Blass Dec 30 '17 at 1:21
• The definition of local compactness that I'm familiar with is: for all $x \in X$ there exists an open set containing $x$ whose closure is compact in $X$. I should also mention that local compactness does not seem to be defined in Rudin's book; he only defines local compactness for topological vector spaces. – theQman Dec 30 '17 at 2:08

It is more basic: because every compact subset of an Hausdorff space is closed. Follow your nose: if $K$ is said compact and $X$ is the space, take $p \in X \setminus K$. By Hausdorffness, for each $x \in K$ there are open sets $U_x$ and $V_x$ such that $x \in U_x$, $p \in V_x$ and $U_x \cap V_x = \varnothing$. Then $\{ U_x \}_{x \in K}$ covers $K$, and by compactness it suffices to take a finite subcover $\{U_{x_i}\}_{i \in F}$, with $F$ finite. Then $\bigcap_{i \in F} V_{x_i}$ is an open set containing $p$, and $\bigcap_{i \in F} V_{x_i}\subseteq X \setminus K$. So $X \setminus K$ is open and $K$ is closed.

To justify why $\bigcap_{i \in F}V_{x_i} \subseteq X \setminus K$, assume that $y \in K \cap V_{x_i}$ for all $i \in F$. Then $y \not\in U_{x_i}$ for any $i$, and so $y \not\in \bigcup_{i \in F}U_{x_i}$. So $y \not\in K$, contradiction.

The closure bar denotes closure in $X$.

Take $p\in S.$ Let $T$ be an open set containing $p$ such that $\overline T$ is compact. Let $U=S\cap T.$

Method (I).

(i). If $\overline U=U$ then $U$ is a non-empty open subset of $S$ whose closure is compact (because $U=\overline U$ is closed in the compact subspace $\overline T$) and $\overline U=U\subset S.$

(ii). If $\overline U\ne U$ then for each $q\in \overline U\setminus U$ let $V_q$ and $W_q$ be open and disjoint with $q\in V_q$ and $p\in W_q.$

Now $\overline U\setminus U$ is compact ( because it is closed in $\overline T$) so let $Y$ be a non-empty finite subset of $\overline U \setminus U$ such that $\cup_{q\in Y}V_q\supset \overline U\setminus U.$

Let $W=\cap_{q\in Y}W_y.$ Then $W$ is a non-empty open set and $\overline W\subset \overline U .$ But $\overline W$ is disjoint from $\cup_{q\in Y}V_q,$ so $\overline U$ is disjoint from $\overline U\setminus U$, so $\overline W\subset U\subset S.$ And $\overline W$ is a closed subset of $\overline T$ so $\overline W$ is compact.

Method (II).

$\;\overline U$ is a compact $T_2$ space so it is a $T_3$ space. So let $V, W$ be open subsets of $X$ such that $V\cap \overline U\supset \overline U\setminus U$ and $p\in W$ and $(\overline U\cap V)\cap (\overline U\cap W)=\emptyset.$

Note that $W\cap (\overline U\setminus U)=\emptyset.$ So $W\cap \overline U=W\cap U$ is open in $X$. And $\overline {W \cap U}$ is a subset of $\overline U$ that is disjoint from $\overline U \cap V.$ So $\overline { W \cap U}$ is disjoint from $\overline U\setminus U,$ so $\overline {W\cap U} \subset U.$ (...And $p\in W\cap U$ so $W\cap U$ is not empty.)

Assume x in open U. By local compactness there is an
open V with x in V subset U and compact K = $\overline V$.

Thus some open W with x in W, L = $\overline W$ subset V.
L is the compact, closed set that answers the question.