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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!

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  • $\begingroup$ How exactly do you conclude that $C\cap A$ is compact (in $A$)? Not every closed subset of a Hausdorff space is compact. $\endgroup$ – Thomas Mar 3 '14 at 20:34
  • $\begingroup$ I just saw your other question/posting, which would answer my question. Without that the question would remain. From a formal point of view you should add a corrsponding remark. $\endgroup$ – Thomas Mar 3 '14 at 20:38
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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.

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