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.

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    $\begingroup$ In general $\overline X\backslash X$ may not be compact. For example if $Y= \Bbb R$ and $X=\Bbb R \backslash \Bbb Z$. $\endgroup$ – DanielWainfleet Aug 7 '18 at 19:33
  • $\begingroup$ You’re absolutely right. I realized that a little after seeing the solution. $\endgroup$ – Anonymous Aug 8 '18 at 16:22

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 $V \cap (W \cap X) = V' \cap X \ne \emptyset$.

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    $\begingroup$ +1... Once we know that $\overline U^X$ is closed in Y, the rest is equivalent to the not-well-known fact that in any space $Y$, with $\overline X=Y,$ if $U$ is open in $X$ and $\overline U^X$ is closed in $Y$ then $U$ is open in $Y$ and $\overline U\subset X.$ This has other uses. E.g. if $id:X\to cX$ is a compactification of a Tychonoff space $X$, and $V$ is a nbhd in $cX$ of some $p\in cX\backslash X$ then $Cl_X(X\cap V)$ is not compact. $\endgroup$ – DanielWainfleet Aug 7 '18 at 19:51
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    $\begingroup$ @DanielWainfleet That is interesting, in fact I was not aware of it. $\endgroup$ – Paul Frost Aug 7 '18 at 22:05
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    $\begingroup$ I was looking at a problem about compactifications and wanted $Cl_X(X\cap V) $to be non-compact and found it. Specifically if $X=\Bbb R$ and if $p$ is an isolated point of the space $cX\backslash X$ then there is an open $V$ of the space $cX$ such that $V=\{p\}\cup (V\cap X)$ and $Cl_{cX}V=\{p\}\cup Cl_X(V\cap X).$... I wanted to show that $U=V\cap X$ is an unbounded open subset of $X=\Bbb R$ whose boundary in $\Bbb R$ is compact, implying that $U$ (and hence $V$) covers a half-line of $\Bbb R.$ $\endgroup$ – DanielWainfleet Aug 8 '18 at 2:02

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.)


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 $\bar X\cap U=X\cap U$.

  • $\begingroup$ There are numerous parts of your proof which appear to make unjustified jumps. $\endgroup$ – Anonymous Aug 7 '18 at 3:02

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