**Problem.** Let $\mathcal F$ be a locally finite (or even discrete) family of (closed) $G\_\delta$-sets in a topological space $X$. Is the union $\cup\mathcal F$ a $G\_\delta$-set in $X$?


**Remark.** The answer to this problem is affirmative in [perfect spaces](https://arxiv.org/abs/1708.01755).
A topological space $X$ is *perfect* if each open subset of $X$ can be written as the countable union of closed sets. I suspect that in general topological spaces a counterexample should exist but I cannot find it, unfortunately.




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**Added in Edit.** This question has a counterexample "living" in a second-countable Hausdorff space, which is not regular. So, it remains to find a (completely) regular example.




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**Added in a Next Edit.** A regular counterexample was constructed by Mathieu Baillif in his answer to [this question](https://mathoverflow.net/questions/293714/is-there-a-first-countable-space-containing-a-closed-discrete-subset-which-is-no).



Is a locally finite union of $G_\delta$-sets a $G_\delta$-set?
I found a simple counterexample, which is however not regular.



> 
> **Example.** There exists a functionally Hausdorff second-countable space $X$ containing a closed discrete subset $D$, which is not of type $G\_\delta$ in $X$.
> 
> 
> In this space the countable the family of singleton $\mathcal F=\{\{x\}:x\in D\}$ is a locally finite family of closed $G\_\delta$-set whose union $\cup\mathcal F=D$ is not a $G\_\delta$-set in $X$.
> 
> 
> 


*Proof.* Let $X$ be the real line endowed with the topology $\tau$ consisting of sets $W\subset \mathbb R$ such that for each point $w\in W$ there exists $\varepsilon>0$ such that each point $x\in\mathbb R\setminus \mathbb Q$ with $|x-w|<\varepsilon$ belongs to $W$. It is easy to see that the topological space $X$ is second-countable and functionally Hausdorff but not regular.


The definition of the topology $\tau$ ensures that the countable set $D:=\mathbb Q$ of rational numbers is closed and discrete in $X$. We claim that $D$ is not a $G\_\delta$-set in $X$. To derive a contradiction, assume that $D=\bigcap\_{n\in\omega}W\_n$ for some decreasing sequence $(W\_n)\_{n\in\omega}$ of open sets in $X$. By the definition of the topology $\tau$, for every $n\in\omega$ and $x\in W\_n$ there exists an neighborhood $V\_{n,x}$ of $x$ in the Euclidean topology of $\mathbb R$ such that $V\_{n,x}\setminus D\subset W\_n$. Then $V\_n:=\bigcup\_{w\in W\_n}V\_{n,x}$ is an open set such that $D\subset W\_n\subset V\_n$ and $V\_n\setminus D=W\_n\setminus D$, which implies that $V\_n=D\cup (V\_n\setminus D)=D\cup(W\_n\setminus D)=W\_n$. This means that each set $W\_n$ is open in the Euclidean topology of the real line and the set of rational numbers $D=\bigcap\_{n\in\omega}W\_n$ is of type $G\_\delta$ in $\mathbb R$, which contradicts the Baire Theorem.




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**Added in Edit.** Answering [this question](https://mathoverflow.net/questions/293714/is-there-a-first-countable-space-containing-a-closed-discrete-subset-which-is-no), Mathieu Baillif constructed a first-countable zero-dimensional Hausdorff space $X$ of cardinality $|X|=\omega\_1$ containing a closed discrete subset $D$, which is not a $G\_\delta$-set in $X$. Then $\mathcal F=\{\{x\}:x\in D\}$ is a locally finite family of compact $G\_\delta$-sets
in $X$ whose union $\cup\mathcal F$ is not a $G\_\delta$ in $X$.


