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


1 Answer 1


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.

  • 1
    $\begingroup$ And thus perfectly normal and hence hereditarily normal (or completely normal, as it's also called). $\endgroup$ Aug 26, 2018 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.