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

Question

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 $X$.

Answer 1

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.

Answer 2

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.

Answer 3

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.)