[Math] Open subspaces of locally compact Hausdorff spaces are locally compact

compactnessgeneral-topology

Let $X$ be locally compact and Hausdorff. I want to prove the following:

If $Y\subset X$ is open then $Y$ is locally compact.

I have proved that closed subsets of $X$ are locally compact, but how can one prove this?

I also want to use these two lemmas to conclude the following:

If $Y\subset X$ is locally closed then $Y$ is locally compact.

Best Answer

Let $X$ be a locally compact Hausdorff space.

To show that an open subspace $Y$ of $X$ is itself locally compact, given $x \in Y$ fix open neighbourhoods $U$ and $V$ of $x$ in $X$ such that

  • $\overline{U}$ is compact. (It is possible to do this because $X$ is locally compact.)
  • $\overline{V} \subseteq Y$. (It is possible to do this because locally compact Hausdorff spaces are regular.)

Now $U \cap V$ is an open neighbourhood of $x$ in $X$ (and also in $Y$). Furthermore

  • as $\overline{U \cap V} \subseteq \overline{V} \subseteq Y$, then $\operatorname{cl}_Y ( U \cap V ) = \overline{U \cap V} \cap Y = \overline{U \cap V}$.
  • as $\overline{U \cap V}$ is a closed subset of the compact $\overline{U}$, it is also compact.

Since you already know that closed subspaces of $X$ are locally compact, it now easily follows that locally closed subspaces of $X$ are locally compact. Recall that being locally closed in $X$ is equivalent to being of the form $U \cap F$ where $U \subseteq X$ is open and $F \subseteq X$ is closed. So $F$ is locally compact as a closed subspace of $X$, and $U \cap F$ is locally compact as an open subspace of $F$.

It may be worth mentioning that locally compact subspaces of Hausdorff spaces are locally closed. (See this question and its answer.) And so in a locally compact Hausdorff space, being a locally compact subspace and being a locally closed subspace are equivalent notions.

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