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Let $X$ be a topological space.

  1. We say it is normal if for any two disjoint closed sets $A$ and $B$, we can find open sets $U$ and $V$ such that $A \subset U$, $B \subset V$, and $U \cap V = \varnothing$.

  2. We say $X$ is regular $G_\delta$, if for any closed set $A$ in $X$, we can find a countable collection of open sets $\{U_n: n \in \mathbb N\}$ such that $$A = \bigcap_{n\in \mathbb N} \overline U_n,\quad \mathrm{and}\quad \forall n \in \mathbb N: A \subset U_n.$$

I'm trying to prove that if $X$ is regular $G_\delta$, then $X$ is normal. Here is my attempted proof outline (which doesn't work!)

Let $A$ and $B$ be disjoint closed subsets of $X$. Then $A$ and $B$ are regular $G_\delta$ sets, so let $\{U_n:n\in \mathbb N\}$ be a collection of open sets each containing $A$, the intersection of whose closures is $A$, and $\{V_m:m\in \mathbb N\}$ a collection of open sets each containing $B$, the intersection of whose closures is $B$. Without loss of generality we assume each $U_n$ is disjoint from $B$, and each $V_n$ is disjoint from $A$. (We can do this by taking intersections with $X\setminus B$ and $X\setminus A$ respectively.) Since $A$ and $B$ are disjoint, there must exist $n \in \mathbb N$ such that $\overline U_n \cap B = \varnothing$. Similarly, pick $m\in\mathbb N$ such that $\overline V_m \cap A = \varnothing$. Then $U_n\setminus \overline{V}_m$ is an open set containing $A$, and $V_m \setminus \overline U_n$ is an open set containing $B$, whose intersection is empty. Thus $X$ is normal.

The reason it doesn't work is because the existence of $n$ such that $\overline U_n \cap B = \varnothing$ is not true - If I try to prove by contradiction that such an $n$ exists, I will need to prove that a countable intersection of nested non-empty closed sets is non-empty, but this is not true in general.

I've also attempted to disprove this. The Moore plane is an example of a non-normal topological space such that every closed set is $G_\delta$. However, I was unable to show that every closed set is regular $G_\delta$.

Could someone provide some guidance? Thank you

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For future reference - if anybody has the same question: I posted the question to MathOverflow and got an answer https://mathoverflow.net/questions/309139/does-regular-g-delta-imply-normal

The crux is: use the characterisation

$X$ is normal iff for each closed set $F$ 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 $n$, $\overline{W_n} \subseteq O$.

Then one can answer my question in the affirmative: Regular $G_\delta$ implies normal.

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  • $\begingroup$ And thus perfectly normal and hence hereditarily normal (or completely normal, as it's also called). $\endgroup$ – Henno Brandsma Aug 26 '18 at 21:41

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