Show that a locally compact subspace of a locally compact Hausdorff space is open in its closure

Question

Let $Y$ be a locally compact Hausdorff space and let $X$ be a locally compact subspace of $Y$. Show that $X$ is open in $\overline{X}$.

It seems I need to show that there exists an open set $U$ of $Y$ such that $X=U\cap\overline{X}$. Alternatively, since $Y$ is Hausdorff, I could show that $\overline{X}\setminus{X}$ is compact or closed. However, I can't think of a way to do either of these things.

By definition of locally compact, given $x\in Y$, there exists an open set $U$ and a compact subspace $C$ such that $x\in U\subset C$. Since $Y$ is Hausdorff, this definition is equivalent to saying given $x\in Y$ and a neighborhood $U$ around $x$, there exists a neighborhood $V$ around $x$ such that $\overline{V}\subset U$ and $\overline{V}$ is compact. I believe this equivalent definition could be useful, but I am not sure how it will come into play.

Answer 1

We prove the following more general result:

Each locally compact subspace $X$ of a Hausdorff space $Y$ is open in $\overline{X}$.

It suffices to consider the case $\overline{X} = Y$ and to show that $X$ is open in $Y$.

Each $x \in X$ has an open neighborhood $U$ in $X$ such that the set $\overline{U}^X = \overline{U} \cap X$ is compact and in particular closed in $Y$. But $U \subset \overline{U} \cap X$, hence $\overline{U} \subset \overline{U} \cap X \subset X$. Let $W$ be open in $Y$ such that $W \cap X = U$. Then

$$x \in W \subset \overline{W} = \overline{W \cap X} = \overline{U} \subset X .$$

Here we used a well-known fact: If $\overline{X} = Y$ and $W$ is open in $Y$, then $\overline{W} = \overline{W \cap X}$. This is true because if we have an open $V \subset Y$ such that $V' = V \cap W \ne \emptyset$, then $V \cap (W \cap X) = V' \cap X \ne \emptyset$.

Answer 2

Let $x\in X$, since $Y$ is locally compact and $X$ is a locally compact subset of $Y$. There exists an open neighborhood $V$ of $x$ whose adherence in $X$ is compact. We can write $V=U\cap X$ where $U$ is an open subset of $Y$. Let $y\in \bar X\cap U$. Since $Y$ is Hausdorff, there exists a sequence $x_n\in X$ such that $lim_nx_n=y$, since $x_n\in V$, we deduce that $y=lim_nx_n\in X$, since $V$ is a compact subset of $X$. It results that $\bar X\cap U=X\cap U$.

Answer 3

Hint: The following are equivalent for a subset $A$ of a topological space $X$:

• $A$ is open in $\overline A$,
• there is an open $U\subseteq X$ such that $A$ is closed in $U$,
• there is an open $U\subseteq X$ and a closed $F\subseteq X$ such that $F\cap U=A$.

(Such sets are called locally closed.)