general-topology
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.
Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
No. Let $Y=\{0\}\cup\{1/n: n\in\Bbb N\}$ be a sequence convergent to zero, $X=Y\times\Bbb N$, $Z=\{0\}\times\Bbb N$, and $f:Z\to Y$, $(0,n)\mapsto 1/n$. It is easy to see that the space $X \sqcup_f Y$ is homeomorphic to the following Franklin’s space decribed in the following example from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989).
Moreover, let $q: X \sqcup Y\to X \sqcup_f Y$ be the quotient map. It is easy to see that any neighborhood of $q(0)$ is $X \sqcup_f Y$ contains a sequence of a form $\{q(n, m_n):n\ge N\}$ for some $N$, which has no limit points, so $X \sqcup_f Y$ is not locally compact.
Best Answer
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.