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.