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I'm trying to prove that if every closed set in a topological space is regular $G_\delta$, then the space is normal. By regular $G_\delta$, I mean for any closed set $A$,

  1. there exists a countable collection $\{U_n:n\in\mathbb N\}$ such that for each $n$, $A \subset U_n$.
  2. $A = \bigcap_{n\in\mathbb N}\overline{U_n}$.

When I try to prove this, I encounter "a countable intersection of nested nonempty closed sets" and I require this to be nonempty for my proof to work. Of course, this is not true in general.

I've also looked for counter examples - The Moore plane is almost a counter example, since it is not normal but every closed set is $G_\delta$. However, I haven't been able to show that every closed set is regular $G_\delta$.

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A well-known characterisation of normality is useful here (proposition 1.5.15 in Engelking's General Topology), I wrote its proof here:

$X$ is normal iff for each closed set $A$ of $X$ and each open set $O$ with $F \subseteq O$, there are open sets $W_n$, $n \in \mathbb{N}$ of $X$ such that $F \subseteq \bigcup_n W_n$ and for all $n$, $\overline{W_n} \subseteq O$.

The condition of regular $G_\delta$ dualised to open sets says that every open set of $X$ is the countable union of closed sets whose interiors also cover $O$. Using these interiors as the $W_n$ for an open $O$, I think shows that regular-$G_\delta$ spaces are normal.

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  • $\begingroup$ The link is not working now. (I am not sure whether it has moved, stopped working, or it is temporarily down.) Still, here is a link to the article in the Wayback Machine. $\endgroup$
    – Martin Sleziak
    Commented May 15, 2020 at 20:55
  • $\begingroup$ @MartinSleziak I noticed. I’ve asked for the links to be redirected. Thx for finding the internet archive version. $\endgroup$
    – Henno Brandsma
    Commented May 15, 2020 at 20:57
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    $\begingroup$ @MartinSleziak Most stuff from bbqa is back up again. $\endgroup$
    – Henno Brandsma
    Commented Mar 3, 2021 at 10:02
  • $\begingroup$ Is $A$ the same set as $F$? $\endgroup$
    – Arvid Samuelsson
    Commented Jun 20, 2022 at 16:04
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Edit: I made a mistake in disagreeing with Henno, who after all is right. But perhaps someone can learn from my mistake.

It is worth noting that a space in which every closed set is "regular $G_\delta$" is not the same as a "regular $G_\delta$ space." (That is, a regular space that is also $G_\delta$.)

Indeed if the question were asked about the latter type of space, the answer would be negative:

Dan Ma has shown that the Sorgenfrey plane (which is well known to be regular but not normal) is $G_\delta$ (which he calls "perfect").

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  • $\begingroup$ I haven't read through the proofs on the page you link to yet, but it doesn't look to me like Dan Ma is claiming the Sorgenfrey plane has the property described in the question. The claim in Dan Ma's blog is that the Sorgenfrey plane is "perfect", and he defines this to mean that every closed set is $G_\delta$. But the property in the question is a little stronger. $\endgroup$
    – Will Brian
    Commented Sep 3 at 20:15
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    $\begingroup$ For a specific example, let $B$ be a Bernstein subset of the reals, and let $A = \{ (x,-x) :\, x \in B\}$. This is a closed subset of the Sorgenfrey plane. I can see how to write it as a $G_\delta$ set (so it's not a counterexample to what Dan Ma's blog claims about the Sorgenfrey plane being perfect). But I cannot see how to show that $A$ is a regular $G_\delta$ set in the sense of the question. $\endgroup$
    – Will Brian
    Commented Sep 3 at 20:18
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    $\begingroup$ Oh you are right, I thought he meant a space that is both regular and $G_\delta.$ I should not have been so hasty. After reading the notion of "regular $G_\delta$" I actually agree with Henno. I will update my response, perhaps my response should be taken down, but perhaps it should be left up to dispell confusion over the issue of "regular $G_\delta$"$\neq$"regular"+"$G_\delta$" $\endgroup$
    – Nick Mendler
    Commented Sep 3 at 20:51

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