Let $X$ be a locally compact Hausdorff space, and $A$ a closed subspace. Show that $A$ is a locally compact Hausdorff space.
Here is what I have for a proof. Will I need to clarify anything else?
This previous theorem will be related:
Closed subspace of a compact topological space is compact
Proof:
First we note that $X$ is locally compact if for all $x \in X$, there exists a compact set $C$ that contains an open neighborhood of $x$.
Suppose $a \in A$, and $C$ is a compact subset of $X$ that contains an open neighborhood of $a$ in $X$.
We would like to show that $C \cap A$ is a compact subset of $A$, and $U \cap A$ is open in $A$, for $U$ being an open cover of $A$.
To show that $C \cap A$ is compact, we need to show that it's closed, since it's a Hausdorff space.
We know that $C$ is closed in $X$, so $X-C$ is open, which implies that $(X-C) \cap A$ is open in $A$, therefore $C \cap A$ is closed in $C$, therefore $C \cap A$ is compact.
To show that $U \cap A$ is open in $A$: Since $U$ is open, by the definition of a subspace topology, $U \cap A$ is open in $A$.
Therefore $A$ is a locally compact Hausdorff space.
Q.E.D.
Thanks in advance!
Best Answer
Looks fine, for the most part, though I have no idea what you could mean by $U\cap A$ if $U$ is an open cover of $A$. I suspect that you instead mean that $C$ is a compact subset of $X$ containing an open neighborhood $U$ of $x.$ Also, closed subsets of Hausdorff spaces need not be compact (consider $\Bbb R$ as a subset of itself, for example), though compact subsets of Hausdorff spaces will be closed.
Alternately, you can use your previous result to proceed even more directly in showing that $C\cap A$ is compact. Since $A$ is closed, then $C\cap (X\setminus A)$ is open in $C,$ so $C\cap A=C\setminus \bigl(C\cap(X\setminus A)\bigr)$ is closed in $C,$ so is compact by your previous result since $C$ is compact.