Definition. A zero-dimensional topological space $X$ is called base zero-dimensional if for any base $\mathcal B$ of the topology that consists of closed-and-open sets in $X$, any open cover $\mathcal U$ of $X$ has a disjoint refinement $\mathcal V\subset\mathcal B$.

It can be shown that

(1) each countable regular space is base zero-dimensional;

(2) the Cantor set is not base zero-dimensional.

Let $\mathfrak z$ be the smallest cardinality $|Z|$ of a subset $Z\subset\mathbb R$, which is not base zero-dimensional. It follows that $\aleph_1\le\mathfrak z\le\mathfrak c$. So, $\mathfrak z$ is a typical small uncountable cardinal.

Problem 1. Is $\mathfrak z$ equal to some known small uncountable cardinal? Is $\mathfrak z=\mathfrak c$ under MA or PFA?

Edit 1 (written following a suggestion of @user64494): I found a (relatively) simple solution to my original question (about the base zero-dimensionality of the Cantor set) and then edited my post asking the next natural question in this context (about the base zero-dimensionality of uncountable sets of the real line).

By the way, a base $\mathcal B$ witnessing that the Cantor cube $2^\omega$ is not base zero-dimensional consists of the sets $$B_s:=\{x\in 2^\omega:x{\restriction}n=s\mbox{ and }(x(n)\ne s(n)\Rightarrow x(n{+}1)=0)\}$$where $n:=\{0,\dots,n-1\}\in\omega$ and $s\in 2^{n+1}=\{0,1\}^{n+1}$.

  • $\begingroup$ @YCor, $[0,1]^{\omega}$ is usually called the Hilbert cube. $\endgroup$ Nov 25, 2018 at 7:50
  • $\begingroup$ ah OK. I had never read "Cantor cube" anyway. $\endgroup$
    – YCor
    Nov 25, 2018 at 8:36
  • $\begingroup$ What if you take a subset of $2^\omega$, meeting each $B_s$ in a subset of cardinal $\aleph_1$? $\endgroup$
    – YCor
    Nov 25, 2018 at 23:41
  • $\begingroup$ @YCor It seems that base zero-dimemsional sets are related to strong measure zero sets, so situtation becames to be interesting. $\endgroup$ Nov 26, 2018 at 10:11
  • $\begingroup$ I think the name of the tag is a mistake. People working in the area call cardinals like the ones you investigate cardinal characteristics. The name involves a field, techniques and a body of related results. Talk of "small uncountable cardinals" doesn't do that (seems like noise to me). $\endgroup$ Mar 13, 2019 at 20:16

1 Answer 1


After an e-mail communication with Lubomyr Zdomskyy, we came to the conclusion that the base zero-dimensionality is equivalent to the Rothberger property.

So, $\mathfrak z=\mathrm{cov}(\mathcal M)$ by a result of Fremlin and Miller.


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