# Is a locally finite union of $G_\delta$-sets a $G_\delta$-set?

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. 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.

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.

Added in a Next Edit. A regular counterexample was constructed by Mathieu Baillif in his answer to this question.

• I tried $X = \omega_1$ for a possible counterexample (since it isn't perfect), but it doesn't work. Note all bounded sets are $G_\delta$. Now, is it the case that for every $x$, there are infinitely many $G \in \mathcal{F}$ each containing some value greater than $x$? If no, then we can find $x$ such that $A = \cup \mathcal{F} = ([0,x] \cap A) \cup G_1 \cup \dots \cup G_n$ where $G_1, \dots, G_n \in \mathcal{F}$. Then $A$ is a finite union of $G_\delta$s, hence $G_\delta$. ... – Nate Eldredge Feb 23 '18 at 15:41
• If yes, then we can find distinct $G_1, G_2, \dots \in \mathcal{F}$ and an increasing sequence $x_n \in G_n$, and then $\mathcal{F}$ is not locally finite at $x = \sup x_n$. – Nate Eldredge Feb 23 '18 at 15:41
• @NateEldredge Your comments simply show that the perfectness is not a necessary condition for preservation of $G_\delta$-sets by locally finite unions. And this is good observation. – Taras Banakh Feb 23 '18 at 15:50
• @NateEldredge Since the space $\omega_1$ is pseudocompact, each locally finite family in $\omega_1$ is finite and this is true reason why your argument works. – Taras Banakh Feb 23 '18 at 20:47

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.
Added in Edit. Answering this question, 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 $$X$$.