TITLE: Every first countable space is a moscow space.
QUESTION [5 upvotes]: First countable space $X$ is an example of moscow spaces.
Let $U$ is an open subset of $X$ and $x\in \overline{U}$. If $\overline{U}$ is open or even a nbhood of $x$ this proposition is immediately proved by definition of first countability of space $X$.
But now how can I prove this?
Thanks.

REPLY [5 votes]: In a first countable space each point is a $G_\delta$, so every set is a union of $G_\delta$ sets. In particular, then, $\operatorname{cl}U$ is a union of $G_\delta$ sets for any $U\subseteq X$. (Note that this argument really needs only that $X$ have countable pseudocharacter.)