# Can the closure of a set be written as the intersection of open neighborhoods in a non-metrizable space?

Let $$\Omega$$ be a topological space. If $$A\subseteq\Omega$$, then $$\overline A=\bigcap_{\substack{B\supseteq A\\B\text{ is closed}}}B\tag1.$$ If the topology on $$\Omega$$ is generated by a metric $$d$$, then $$B_\varepsilon:=\left\{x\in\Omega:d(x,A)<\varepsilon\right\}$$ is open for all $$\varepsilon>0$$ and $$\overline A=\bigcap_{\varepsilon>0}B_\varepsilon=\bigcap_{n\in\mathbb N}B_{\frac1n}\tag2.$$

By $$(2)$$, the closure of an arbitrary subset of a metric space can be written as the (countable) intersection of open neighborhoods. Does the same hold in an arbitrary topological space?

• What condition $(2)$ states is that $x\in\overline{A}$ if and only if every open neighborhood of $x$ intersects $A$. The same thing happens in every topological space. (however in general open neighborhoods are more complicated than in metric spaces, so you can't write it in such a nice way like with metric spaces)
– Mark
Commented Mar 8, 2021 at 10:36

No, the property that every closed set is a countable intersection of open sets (which is what your property comes down to) is called being perfectly normal (usually in combination with being $$T_4$$ as well). I've also seen it called a $$G_\delta$$ space. All metric spaces have this property, but not all spaces do:

The Michael line, and the lexicographically ordered square are some examples of non-metrisable but nice (herditarily normal etc) spaces that do not satisfy this. And $$[0,1]^I$$ with uncountable $$I$$ is compact Hausdorff where even singleton sets are not $$G_\delta$$'s.

No. Consider a set $$X$$ with $$p \in X$$ and $$\vert X \vert > 1$$ with the excluded point topology w.r.t. $$p$$. Then $$\{p\}$$ is closed since its complement is open, but the intersection of all of its neighbourhood is $$X$$, since it is the only set containing $$p$$ and being open i.e. the closure is strictly contained in the intersection of open neighbourhoods.

It can fail in the other direction as well, as can be shown with the particular point topology w.r.t. $$p$$. i.e. the intersection of open neighbourhoods can be strictly contained in the closure.