# Is every nowhere dense closed set intersection of closures of non-intersecting open sets?

This is a double-cross-post: I first asked a question on cs.se, then posted a particularly mathematical part of it to mathoverflow; after about month and a half I realised that I don't know answer to a simpler question which looks like an exercise from a general topology textbook. So I decided to ask it here.

Given two non-intersecting open sets, $$U\cap V=\varnothing$$, it is easy to show that the intersection $$F=\bar U\cap\bar V$$ of their closures is nowhere dense (i. e. has empty interior, that is, does not contain any open set).

Is the converse true? That is, is any nowhere dense closed set $$F$$ intersection of closures of two non-intersecting open sets?

• As far as general topology goes, there is a reasonable class of topological spaces (called irreducible) where every two non-empty open sets have non-empty intersection, so this cannot be the case in general. This is the case for, say, $\Bbb C^n$ endowed with Zariski topology. However, these spaces aren't usually metrizable (because they aren't Hausdorff as soon as there are at least two points).
– user239203
Commented Dec 19, 2020 at 19:01
• What if we add $T_2$ to the problem statement? Commented Dec 19, 2020 at 19:03

Let $$X=\Bbb N$$, and let $$\tau=\{\varnothing\}\cup\{U\subseteq\Bbb N:0\in U\}$$; $$\tau$$ is a topology on $$X$$. $$\{1\}$$ is closed and nowhere dense but it cannot be the intersection of the closures of two disjoint, non-empty open sets, because the space doesn’t have two disjoint, non-empty open sets.