# Closed subspaces of a locally compact space are locally compact

I need to show that any closed subspace of a locally compact space is locally compact.

My definition of a locally compact space $$S$$ is that for each point in $$S$$ there exists a compact neighborhood $$U$$ in $$S$$. Now let $$K$$ be a closed subspace in $$X$$, then for each $$x \in K$$, there exists a compact neighbourhood $$U_x \subset X$$. Then $$x \in U \cap K$$ is a neighbourhood of $$x$$ in $$K$$, but how do I show that it is compact as well?

• Can you show a closed subset of a compact space is compact? Jan 29, 2013 at 4:48
• Klara, have you tried taking any open cover of $U_x\cap K$ and union $U_x-K$? Jan 29, 2013 at 4:48
• @Clayton No I have not, but i guess the union of U_x with intK would be one right? Jan 29, 2013 at 4:57
• @Klara: You are close, I think. The goal is that the open cover of $U_x\cap K$ unioned with $U_x-K$ would be an open cover of $U_x$, which is compact, hence...what can we say about $U_x\cap K$? Jan 29, 2013 at 4:59
• @ Jacob, I can take that for granted since it is proved in the text, hard to follow but I can use your statement for free..I just do not see where? Jan 29, 2013 at 5:11

You want to show that if $K$ is a closed subset of $X$, $x\in K$, and $U$ is a compact nbhd of $x$ in $X$, then $U\cap K$ is a compact nbhd of $x$ in $K$. You’ve done everything except show that $U\cap K$ is compact.
To show this, let $\mathscr{V}$ be an open cover of $U\cap K$; you want to find a finite subcover. $K$ is closed in $X$, so $X\setminus K$ is open; let $\mathscr{W}=\mathscr{V}\cup\{X\setminus K\}$. $\mathscr{W}$ is a collection of open sets.
• Show that $\mathscr{W}$ covers $U$.
Then you’ll know that some finite $\mathscr{W}_0\subseteq\mathscr{W}$ covers $U$, since $U$ is compact. Clearly $\mathscr{W}_0$ covers $U\cap K$. If $X\setminus K\notin\mathscr{W}_0$, then $\mathscr{W}_0\subseteq\mathscr{V}$, and you have the subcover that you wanted.
• Show that even if $X\setminus K\in\mathscr{W}_0$, you can throw it away, and what’s left will still cover $U\cap K$. That is, the family $\mathscr{W}_0\setminus\{X\setminus K\}$ still covers $U\cap K$ and is the desired subcover of $\mathscr{V}$.