If X is local compact, then it holds: A is closed $\iff$ $A\cap K$ is compact for all compact K

Question

Prove: Show that for every local compact space X holds the following:

A $\subseteq$ X is closed $\iff$ $A \cap K$ is compact, for all compact sets K.

I use the following definition of local compactness:

Definition: A hausdorff space is local compact, if every $x \in X$ has a neighborhood $U\subseteq X$ which lies in a compact set.

Answer 1

One implication is clear. For the other note that for $x ∈ \overline{A} \setminus A$ and $N$ neighborhood of $x$ we have $x ∈ \overline{(A ∩ N)} \setminus (A ∩ N)$ in $N$.

Of course we use the facts that closed subspace of a compact space is compact and that compact subspace is closed in a Hausdorff space.

Answer 2

In a Hausdorff space, every compact set is closed (How to prove that a compact set in a Hausdorff topological space is closed?).

This establishes the direction "$\Rightarrow$", because $A\cap K$ is then closed as an intersection of closed sets.

In the other direction, let $x \in \overline{A}$ be arbitrary. Then there is a neighbourhood $U \subset X$ of $x$ such that $\overline{U}$ is compact (because $U \subset K$ for some compact (hence closed) set $K$, so that $\overline{U} \subset K$ is compact as a closed subset of a compact set (Closed subsets of compact sets are compact).

But then $A \cap \overline{U}$ is closed and $x \in \overline{A \cap \overline{U}}$, because for each neighbourhood $V$ of $x$, we have

$$ A \cap \overline{U} \cap V \supset A \cap (U \cap V) \neq \emptyset, $$

because $U\cap V$ is a neighbourhood of $x$ and $x \in \overline{A}$.

Thus, $x \in \overline{A \cap \overline{U}} = A \cap \overline{U} \subset A$, i.e. $\overline{A} \subset A$, so that $A$ is closed.