# Does a nonempty open subset of a regular space contain the closure of some nonempty open set?

I'm including the additional requirement that single point sets be closed for a space to be regular.

I know this holds for normal spaces, and locally compact Hausdorff spaces, but what about regular spaces?

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As a minor remark, Brian's proof below doesn't use that points are closed in the topology of X, showing this is not really necessary. – Salvo Tringali Dec 7 '11 at 22:41

Yes. Let $X$ be a $T_3$-space, and let $V\,$ be a non-empty open set in $X$. Pick $x\in V$. Then $X\setminus V\,$ is a closed set not containing $x$, so there are disjoint open sets $U$ and $W$ such that $x\in U$ and $X\setminus V\subseteq W$. Clearly $\operatorname{cl}U\subseteq X\setminus W\subseteq X\setminus(X\setminus V)=V$, i.e., $V$ contains the closure of the non-empty open set $U$.
Indeed, regularity of a space $X$ is often defined by the condition that for any non-empty open set $V\,$ in $X$ and any point $x\in V\,$ there is an open set $U$ such that $x\in U\subseteq\operatorname{cl}U\subseteq V$.