In a Locally Compact Hausdorff and second countable space every compact subset is a $G_\delta$ set

Question

Let $X$ be a second countable LCH (Locally Compact Hausdorff) then every compact subset of $X$ is a $G_{\delta}$ set.

A set $E$ is $G_{\delta}$ if is a countable intersection of open sets.

I have two questions,

How to prove this result,
I am also wondering if that would be true for the closed sets in this space, or what would be necessary to be true for closed sets.

Edit:

I posted a resolution, which I found very interesting, of question 1.

Follow these: Locally compact Hausdorff space is regular $\longrightarrow$ A topological space is separable and metrizable if and only if it is regular, Hausdorff and second-countable $\longrightarrow$ A closed set in a metric space is $G_δ$ — Sumanta, Jan 14 '21 at 03:13

score 2 · Answer 1 · answered Jan 14 '21 at 07:08

If $X$ is LCH and C-II, as indicated, $K:=\alpha (X)$, the Aleksandrov (one-point) compactification is compact and metrisable. In a metric space, all closed sets are $G_\delta$ and a compact subset of $X$ is still closed in $K$. Then relativize to $X$ again.

score 1 · Accepted Answer · answered Jan 21 '21 at 23:17

Let $\mathcal{B}$ be a countable base for the topology of $X$, and let $K\subseteq X$ be a compact. We have that $X$ is Hausdorff, so for each $x\in K^c$ and $y\in K$, there is an open $U_{xy}$ such that $y\in U_{xy}$ but $x\notin U_{xy}$. Moreover, there exists $V_xy \in \mathcal{B}$ such that $y\in V_{xy}\subseteq U_{xy}$

Since $K$ is compact, for each $x\in K^c$ there is a finite set $Y_x \subseteq K$ such that,

$$K\subseteq \bigcup_{y\in Y_x} V_{xy} \qquad (*)$$

We have,

$$K=\bigcap_{x\in K^c}\left(\,\bigcup_{y\in Y_x}V_{xy} \right)$$ since $x\notin V_{xy}$ for all $x\in K^c$ and $y\in Y_x$.

Since the set, $$\left\{ \bigcup_{U\in \mathcal{A}} U : \mathcal{A}\subseteq \mathcal{B} \text{ is finite}\right\}$$

is a countable collection of open sets, to demonstrate this, just prove that the set of all finite subsets of $\mathbb{N}$ is countable.

Therefore the intersection in (*) is countable, because, although the intersection is indexed by a possible uncontable quantity, there is only an countable number of distinct sets, by the previous result.

This shows that $K$ is $G_\delta$.

score 0 · Answer 3 · answered Jan 14 '21 at 04:04

This is not an answer but recommendation for you to see the following theorem Theorem In a locally compact Hausdorff space, the intersection of an open set with a closed set is locally compact. Conversely, a locally compact subset of a Hausdorff space is the intersection of an open set and a closed set.

You can find the proof Willard's book page 130.

In a Locally Compact Hausdorff and second countable space every compact subset is a $G_\delta$ set

3 Answers3