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One of the most well-known problems in set-theoretic topology is Arhangel'skii's question of whether there exists a Lindelöf Hausdorff space with "points $G_\delta$" (meaning, every point is the intersection of countably many open sets) and cardinality larger than the continuum. That is of course motivated by Arhangel'skii's Theorem stating that every Lindelof first countable Hausdorff space has cardinality at most continuum.

In 1978 Shelah was the first to construct, consistently with CH, an example of a Lindelöf space with points $G_\delta$ and cardinality $(2^\omega)^+=\omega_2$. Many more consistent examples followed (by Isaac Gorelic, Alan Dow and Toshimichi Usuba, among others), but Arhangel'skii's problem remains open in ZFC.

A space of countable extent is a space where every closed discrete set is countable. Every Lindelof space clearly has countable extent, but the converse is not true, as shown, for example, by the subspace of $\omega_2$ consisting of all ordinals of uncountable cofinality, with the topology inherited from the order topology on $\omega_2$.

QUESTION: Is there a ZFC example of a Hausdorff space with countable extent, points $G_\delta$, and cardinality larger than the continuum?

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    $\begingroup$ If $\kappa$ is an ordinal of uncountable cofinality, with the order topology, and $X$ is the subspace obtained by removing all points of uncountable cofinality, isn't the first countable space $X$ countably compact and so has countable extent? In particular, $\kappa$ can be arbitrarily large. $\endgroup$
    – Anonymous
    Commented Oct 8, 2023 at 1:49
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    $\begingroup$ Yes, you are right, and as a matter of fact if $\kappa$ is any cardinal (no need to assume that $\kappa$ has uncountable cofinality), the space of all ordinals of countable cofinality which are less than $\kappa$ has countable extent. $\endgroup$
    – Santi Spadaro
    Commented Oct 8, 2023 at 16:52
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    $\begingroup$ Of course, uncountable cofinality is not necessary to get countable extent, but I thought that being countably compact would be more dramatic. $\endgroup$
    – Anonymous
    Commented Oct 8, 2023 at 21:22

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