# Is there an almost strongly zero-dimensional space which is not strongly zero-dimensional

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 :)

• The definition of strong zero-dimensionality in most books is stronger: disjoint functionally closed sets can be separated by clopen sets. – KP Hart Dec 6 '19 at 9:16
• @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. – Lviv Scottish Book Dec 6 '19 at 10:25
• 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. – Taras Banakh Dec 6 '19 at 10:30
• @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 – MasleniZZa Dec 7 '19 at 8:53
• @D.S.Lipham Probably, you are right: the metrizability does not follow, but maybe the analyticity can be still applied? – Taras Banakh Dec 12 '19 at 4:41

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.