Are closed subsets of a locally compact space, locally compact?

Question

I know that there are open subsets of locally compact topological spaces that are not locally compact ($\mathbb{Q}$ in the Alexandroff's compactification).

I wonder if any closed subset of a locally compact space is always locally compact.

Definition. $(X,\tau)$ is locally compact if for each $x\in X$, there is a neighborhood of $x, A_x\in \tau:closure(A_x)$ is compact.

Answer 1

(Credit to my prof, not myself) Let $Y$ closed subspace of locally compact space $X$, and let $y\in Y$. Since $X$ is locally compact, there exists compact subset $K \subseteq X$ and open $U\subseteq X$ such that $y \in U \subseteq K$. Now let $$ V= Y \cap U \subseteq Y $$ $$ and $$ $$ K_0 = Y \cap K \subseteq Y$$ so that $y \in V \subseteq K_0$. Notice $V$ is open by definition of the subspace topology, so we are left to show $K_0$ is compact.

Note that since $Y$ is closed in $X$, $K_0$ is a closed subset of $K$ when equipped with the subspace topology. Hence $K_0$ is compact, as a closed subset of the compact space $K$. Since $y$ was arbitrary, $Y$ is locally compact

Answer 2

Take $F $ to be a closed subset of a locally compact space $X $. Then, for every $U $, open subset of $F$, there exists $V $, open subset of $X $ such that $U $ is the intersection of $V $ and $F $ by definition of the induced topology. Let $x \in F $ and $V $ be a compact neighborhood of $x $ in $X $. $V \cap F $ is compact and is a neighborhood of $x $ in $F $ because $V$ contains an open subset of $X $ $U $ containing $x $, so $U \cap F $ is open in $F$. Thus $V \cap F $ fits. So it is true.