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A Tychonoff space $X$ is called strongly zero-dimensional if each functionally closed subset $F$ of $X$ is a $C$-set, which means that $F$ is the intersection of a sequences of clopen sets in $X$.

A Tychonoff space $X$ is called almost strongly zero-dimensional if each functionally closed subset of $X$ is the union of a sequence of $C$-sets.

Question. Does there exists a (metrizable separable) Tychonoff space which is almost strongly zero-dimensional but not strongly zero-dimensional?

This problem was posed on 30.11.2019 by Olena Karlova (from Chernivtsi) on page 35 of Volume 3 of the Lviv Scottish Book.

Prize. A portrait of a mathematician who will solve this problem :)

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    $\begingroup$ The definition of strong zero-dimensionality in most books is stronger: disjoint functionally closed sets can be separated by clopen sets. $\endgroup$ – KP Hart Dec 6 '19 at 9:16
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    $\begingroup$ @KPHart You are right, but I just have preserved the terminology of the author of the problem. In fact, the question is non-trivial already at the level of metrizable separable spaces where the strong zero-dimensionality (in all senses) is equivalent to the standard zero-dimensionality. $\endgroup$ – Lviv Scottish Book Dec 6 '19 at 10:25
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    $\begingroup$ The ``metrizable separable" case of this problem is equivalent to this MO-problem (mathoverflow.net/questions/240215/…), which has an affirmative answer if the space $X$ is analytic, i.e., a continuous image of a Polish space. Therefore, we have a partial answer for analytic spaces: each almost strongly zero-dimensional analytic metrizable separable space is strongly zero-dimensional. $\endgroup$ – Taras Banakh Dec 6 '19 at 10:30
  • $\begingroup$ @KPHart "Your" definition and "my" definition of strongly zero-dimensional space are equivalent, because every two disjoint C-sets can be separated by clopen sets $\endgroup$ – MasleniZZa Dec 7 '19 at 8:53
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    $\begingroup$ @D.S.Lipham Probably, you are right: the metrizability does not follow, but maybe the analyticity can be still applied? $\endgroup$ – Taras Banakh Dec 12 '19 at 4:41
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Not an answer, but this may be helpful:

Theorem 1. If $X$ is a Lindelöf Tychonoff almost strongly zero-dimensional space, then the following are equivalent:

(i) $X$ is strongly zero-dimensional;

(ii) $X$ is almost zero-dimensional, that is, $X$ has a neighborhood basis of C-sets.

Proof. (i)$\Rightarrow$(ii) is trivial, and the converse follows from Theorem 4.3 in this paper (we assume separable metrizable there, but Lindelöf should be enough). $\square$

In light of Taras Banakh's comment above, for separable metrizable spaces I believe the question is: If $X$ is separable metrizable and $f:X\to Y$ is a continuous bijection onto a zero-dimensional space $Y$ which maps open sets to $G_\delta$-sets, then is $X$ almost zero-dimensional?

Theorem 2. Every almost strongly zero-dimensional homogeneous Polish space $X$ is (strongly) zero-dimensional.

Proof. If $U$ is any open subset of $X$, then $U$ is a $\sigma$C-set, so by the Baire property there is a C-set $F\subseteq U$ which contains a non-empty open set. Continuing this process we construct C-sets $F_n$ such that $F_{n+1}\subseteq F^\mathrm{o}_n$ and $\text{diam}(F_n)\leq 1/n$ in a complete metric. Then there exists $x\in \bigcap F_n$, and $x$ has a neighborhood basis of C-sets. By homogeneity, $X$ is almost zero-dimensional, so by Theorem 1 $X$ is strongly zero-dimensional. $\square$

More generally it is true that each almost strongly zero-dimensional Polish space is zero-dimensional at a dense $G_\delta$-set of points.

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