Being motivated by [this problem](https://mathoverflow.net/questions/293682/is-a-locally-finite-union-of-g-delta-sets-a-g-delta-set/293703#293703), 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](https://mathoverflow.net/questions/293682/is-a-locally-finite-union-of-g-delta-sets-a-g-delta-set/293703#293703) of a non-regular second-countable Hausdorff space, containing a countable closed discrete subsets which is not $G\_\delta$.



Is there a first-countable space containing a closed discrete subset 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](https://en.wikipedia.org/wiki/Fodor%27s_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](http://www.springer.com/gp/book/9789812872562) or in [this preprint](https://arxiv.org/pdf/1303.6714.pdf) (free access), I believe that everything works. I am actually fairly convinced that by Prüferizing the diagonal of the square of the [longray](https://en.wikipedia.org/wiki/Long_line_(topology)) 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.


