# A locally compact subset of a locally compact Hausdorff space is locally closed

Let $X$ be a locally compact and Hausdorff space. Show that if $Y \subset X$ is locally compact, then $Y$ is locally closed, in essence $Y$ is an open subset of $\overline{Y}$, where $\overline{Y}$ has the subspace topology inherited by $X$.

I really don’t have any ideas how to go about this. Any help will be most appreciated!

#### Solutions Collecting From Web of "A locally compact subset of a locally compact Hausdorff space is locally closed"

You can actually get away with $X$ being just Hausdorff.

Note that it suffices to show that every $y \in Y$ has an open neighbourhood $U$ (in $X$) such that $U \cap \overline{Y} \subseteq Y$.

By local compactness of $Y$ (and definition of the subspace topology) $y \in Y$ has an open neighbourhood $U$ in $X$ such that $\mathrm{cl}_Y ( U \cap Y ) = \overline{U} \cap Y$ is compact, and hence closed in $X$. As $U \cap Y \subseteq \overline{U} \cap Y$ it follows that $\overline{U \cap Y } \subseteq \overline{U} \cap Y \subseteq Y$ and so
$$U \cap \overline{Y} \subseteq \overline{ U \cap \overline{Y} } = \overline{ U \cap Y} \subseteq Y,$$ as required.

For the equality $\overline{ U \cap \overline{Y} } = \overline{ U \cap Y}$, note that $\supseteq$ is trivial. For the opposite inclusion given $x \in \overline{ U \cap \overline{Y} }$ and any neighbourhood $V$ of $x$ it must be that $(V \cap U ) \cap \overline{Y} = V \cap ( U \cap \overline{Y} ) \neq \emptyset$, and as $U \cap V$ is open it must be that $V \cap ( U \cap Y ) = ( U \cap V ) \cap Y \neq \emptyset$. Thus $x \in \overline{ U \cap Y }$.