# Is there a first-countable space containing a closed discrete subset which is not $G_\delta$?

Being motivated by this problem, I am searching for an example of a first-countable regular topological space $X$ containing a closed discrete subset $D$, which is not $G_\delta$ in $X$.

It is easy to show that such set $D$ cannot be countable. Also the space $X$ cannot be Moore.

On the other hand, there exists a simple example of a non-regular second-countable Hausdorff space, containing a countable closed discrete subsets which is not $G_\delta$.

EDIT: fixed some error in the proof that the diagonal is not a $G_\delta$, and added details. I hope that the proof is now correct.

I think that the following works. First I'll describe a first countable non-regular example, and then explain how it can be modified to obtain a regular one.

Endow $\omega_1$ with the usual order topology. Take the space $\omega_1\times\omega_1$ with the product topology and refine it by declaring open the sets of the form $$U = \{\langle\alpha,\alpha\rangle\}\cup (V-\Delta),$$ where $V\ni \langle\alpha,\alpha\rangle$ is open in $\omega_1\times\omega_1$ in the usual sense and $\Delta$ is the diagonal, which is thus closed discrete in this topology. Any open set containing $\Delta$ is also open in the product topology, and it is well known that it must then contain $[\alpha,\omega_1) \times [\alpha,\omega_1)$ for some $\alpha$. Hence, any countable intersection of such sets contains $[\beta,\omega_1) \times [\beta,\omega_1)$ for some $\beta$.

This topology is not regular, but we can modify it this way. Only the topology on $\Delta$ will be changed, and since successor ordinals are isolated in $\omega_1$, we only need to worry about limit ordinals. For each such limit $\alpha\in\omega_1$, fix a sequence of successor ordinals $\alpha_n\nearrow\alpha$. Then a neighborhood of $\langle\alpha,\alpha\rangle$ is given by $$U_{\alpha,k,m} = \{\langle\alpha,\alpha\rangle\} \, \cup \, \bigcup_{n\ge m} [\alpha_{n},\alpha_{n+1}]\times (\alpha_{n+k},\alpha] \, \cup \, \bigcup_{n\ge m} (\alpha_{n+k},\alpha]\times [\alpha_{n},\alpha_{n+1}].$$ That is, we take "triangles" (more akin to step pyramids, actually) pointing at $\langle\alpha,\alpha\rangle$ from below and from the side. $\Delta$ is obviously closed discrete since $U_{\alpha,k,m}\cap\Delta = \{\langle\alpha,\alpha\rangle\}$, and the following holds:

Lemma If $U\supset\Delta$ is open, then there is some $\beta$ such that $U$ contains the terminal part of $\omega_1\times \{\alpha\}$ whenever $\alpha>\beta$.

By terminal part, I mean $[\gamma,\omega_1)\times \{\alpha\}$ for some $\gamma$. The proof is by using twice Fodor's Lemma. First, by definition for each $\alpha$ there is $\beta(\alpha)$ such that $U$ contains $\{\alpha\}\times[\beta(\alpha),\alpha]$. Hence by Fodor there is some $\beta$ such that $U$ contains $\{\alpha\}\times[\beta,\alpha]$ for $\alpha$ in a stationary subset $E\subset\omega_1$. If $\alpha>\beta$, $\omega_1\times \{\alpha\}\cap U$ is stationary, so another use of Fodor gives the result.

Corollary The diagonal $\Delta$ is not a $G_\delta$.

Indeed, a countable collection of open sets containing $\Delta$ will contain the terminal part of $\omega_1\times \{\alpha\}$ if $\alpha$ is big enough.

I hope I did not overlook something, but since this is very similar to the method of "Prüferizing" a surface, which dates back to Rado (1925) and is described in D. Gauld's book on non-metrisable manifolds or in this preprint (free access), I believe that everything works. I am actually fairly convinced that by Prüferizing the diagonal of the square of the longray we can obtain an example which is a topological surface. I have some vague remembrance of an example of this type in a paper of Nyikos, by the way, though I don't remember which.

• @Matthieu-Baillif Thank you for your answer. Unfortunately, I cannot fill all details. Namely, I cannot show that the diagonal is not of type $G_\delta$. Indeed, using the pressing-down lemma, it is possible to prove that each open neighborhood $U$ of the diagonal contains some vertical and horisontal strips. But the intersection of countably many such strips can be empty. How to proceed further? – Taras Banakh Feb 25 '18 at 9:06
• @Taras Banakh You are right, there is some gap in my arguments, because actually an open set containing the diagonal might fail to be open in the usual topology. I believe that it will work in the end, but I need to think more about it. – Mathieu Baillif Feb 25 '18 at 9:16
• @Matthieu Baillif In your example the diagonal is not $\bar G_\delta$-set, i.e., is not a countable union of closures of neighborhoods of $\Delta$. – Taras Banakh Feb 25 '18 at 13:50
• Maybe I again overlooked something, but I made some edits that I hope solve the problem. Sorry for the messy answer. – Mathieu Baillif Feb 25 '18 at 13:55
• Now everything seems to be correct. Thank you for the great answer! – Taras Banakh Feb 25 '18 at 20:02