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In Engelking's General topology, in the exercises section, there is Ju. M. Smirnov's characterization of normal spaces:

A $T_1$ space is normal iff the following properties hold (both):

  1. Every closed $G_\delta$ set is zero-set;
  2. for every $F$ closed set and $G$ open set, such that $F$ is in $G$, there exist $M$ closed $G_\delta$ set, such that $F$ is in $M$ and $M$ is in $G$.

This equivalency is not hard to prove. Then there is written that, neither of the properties by itself imply the normality of $X$. For 2, maybe for example one may use the co-finite topology on a countable set, because any closed set is also $G_\delta$ (Also, for that example one may use Niemetzki plane by the same argument).

But I can't find an example of $T_1$ space, which has the property 1 and is not normal. Thanks for any help.

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    $\begingroup$ Maybe I'm missing something, but what about the cofinite topology on an uncountable set $X$? The only closed $G_\delta$ sets are $\emptyset$ and $X$, which are zero sets. $\endgroup$ Jun 11, 2021 at 20:54
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    $\begingroup$ @NateEldredge That is a nice simple counterexample. Zero-sets are almost always talked about not in the context of all topological spaces but instead in the context of completely regular $T_{0}$ spaces (or in completely regular frames), so I decided to give a completely regular counterexample. $\endgroup$ Jun 12, 2021 at 13:55

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Observe that every function $f:\omega_{1}\rightarrow\mathbb{R}$ is eventually constant and $\omega_{1}$ is normal. Observe also that if $A\subseteq\omega_{1}$ is a closed $G_{\delta}$ set, then the characteristic function $\chi_{A}$ of $A$ is eventually constant.

Let $X=((\omega_{1}+1)\times(\omega+1))\setminus\{(\omega_{1},\omega)\}$. $X$ is known as the Tychonoff plank. $X$ is a typical example of a completely regular space that is not normal since the closed sets $\omega_{1}\times\{\omega\}$ and $\{\omega_{1}\}\times\omega$ cannot be separated by an open set.

Now, if $A$ is a closed $G_{\delta}$ subset of $X$, then $X\cap(\omega_{1}+1)\times\{n\}$ is eventually constant for each $n$. Therefore, there is a subset $D\subseteq\omega+1$, and an ordinal $\alpha<\omega_{1}$ and a closed subset $C\subseteq \omega+1$ such that $A=D\cup(\alpha,\omega+1]\times C\setminus\{(\omega_{1},\omega)\}$. One can show that $A$ is a zero set of some continuous function $f:X\rightarrow[0,1]$. Thus, every closed $G_{\delta}$ set in $X$ is a zero set.

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    $\begingroup$ Recall that a P-space is a regular space where every $G_{\delta}$-set is open (every regular P-space is automatically zero-dimensional and hence completely regular). In a P-space, every closed $G_{\delta}$-set will be clopen and hence a zero set. There are certainly examples of P-spaces that are not normal (you can generalize the Tychonoff plank example to get a P-space). $\endgroup$ Jun 12, 2021 at 14:07

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