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?
 A: 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.


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*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.


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*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 $\mathscr{V}$.

