# Topological spaces where every closed set is a countable intersection of open sets?

Metric spaces have the following nice property: Every closed set is a countable intersection of open sets.

What other spaces have this property? Are there some nice known sufficient or necessary conditions on a topological space, for this property?

(e.g. do locally compact Hausdorff spaces satisfy it? )

The proof for metric spaces is very easy:

Let $$A\subseteq X$$ be closed. For all $$n\in \mathbb N$$ define $$U_n=\bigcup _{a\in A} B(a,\frac{1}{n}).$$ Each $$U_n$$ is open, and $$A=\bigcap _{n\in \mathbb N} U_n$$.

• A topological space in which every closed set is a $G_\delta$-set (a countable intersection of open sets) is called a perfect space (not to be confused with a perfect set, a closed set with no isolated points); a normal space with this property is called a perfectly normal space.
– bof
Nov 9, 2018 at 5:42

Such a space is called a $$G_\delta$$ space. A $$G_\delta$$ normal space is called "perfectly normal" and perfect normality is equivalent to normality plus every closed set being the vanishing set of some real-valued continuous function. Beyond metric spaces, another notable class of examples is that all CW-complexes are perfectly normal.

Not all compact Hausdorff spaces are $$G_\delta$$ spaces. For instance, the ordinal $$\omega_1+1$$ is compact Hausdorff with the order topology but the singleton $$\{\omega_1\}$$ is not a $$G_\delta$$ set.

• What about $\beta\Bbb N$? Nov 9, 2018 at 8:05

As said above, these are $$G_{\delta}$$ sets.

For a normal space $$X$$ An equivalent description of these:

$$A \subset X$$ is a $$G_{\delta}$$ set iff $$\exists f:[0,1] \to X$$, continuous, s.t $$f(a) = 0$$ on $$A$$, and else $$f(x) > 0$$.

do locally compact Hausdorff spaces satisfy it?

No, as Eric Wofsey showed above, but it is true that every locally compact Hausdorff space is a Tychonoff space ($$T_{3\frac{1}{2}}$$)

• Thanks; How do you prove that a $G_{\delta}$ set in a normal space is the zero level set of a non-negative continuous function? (I guess this is related to Urysohn's lemma somehow). Nov 10, 2018 at 7:00
• here's a link (33.4): web.math.ku.dk/~moller/e02/3gt/opg/S33.pdf Nov 11, 2018 at 20:38

A notion that is a stronger version of this is called "stratifiable": a space $$X$$ is called stratifiable, when for every closed subset $$F$$ of $$X$$ we have a (WLOG decreasing) sequence of open subsets $$U(F,n), n \in \mathbb{N}$$ such that for all $$F,G$$ closed, if $$F \subseteq G$$ then for all $$n$$, $$U(F,n) \subseteq U(G,n)$$ and for all closed $$F$$ we have $$F = \bigcap_{n \in \mathbb{N}} U(F,n)$$. So every closed set is a $$G_\delta$$ in a monotonous way.

Metric spaces are a prime example of such spaces, as we can take $$U(F,n) = \{x: d(x,F) < \frac{1}{n}\}$$ in that case.

"Just" having every closed set a $$G_\delta$$ is called "perfect" and if the space is also $$T_4$$ it's then called $$T_6$$ or "perfectly normal". It implies that $$X$$ is hereditarily normal (every subspace is normal). Stratifiable also implies a strong form of normality, namely "monotonically normal".