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 $X$.
Best Answer
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.