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 $A=\bigcap _{n\in \mathbb N} U_n$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – bof
    Nov 9, 2018 at 5:42

3 Answers 3


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.

  • $\begingroup$ What about $\beta\Bbb N$? $\endgroup$
    – Asaf Karagila
    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 ($T_{3\frac{1}{2}}$)

  • $\begingroup$ 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). $\endgroup$ Nov 10, 2018 at 7:00
  • $\begingroup$ here's a link (33.4): web.math.ku.dk/~moller/e02/3gt/opg/S33.pdf $\endgroup$
    – Mariah
    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".


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