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A topological space $X$ is called

$\bullet$ totally disconnected if for any distinct points $x,y\in X$ there exists a clopen set $U\subseteq X$ such that $x\in U$ and $y\notin U$;

$\bullet$ hereditarily disconnected if every connected subspace of $X$ has cardinality $\le 1$;

$\bullet$ punctiform if every connected compact subset of $X$ has cardinality $\le 1$.

It is well-known that $$\mbox{zero-dimensional}\Rightarrow\mbox{totally disconnected}\Rightarrow\mbox{hereditarily disconnected}\Rightarrow\mbox{punctiform}$$ and none of these implications can be reversed even in the class of separable metrizable spaces.

In this paper Roman Pol constructed his famous example of a Polish totally disconnected space, which is not a countable union of zero-dimensional spaces.

Problem 1. Is there a hereditarily disconnected metrizable separable (desirably Polish) space which is not a countable union of totally disconnected spaces?

Remark. By Main Lemma in this paper, the Hilbert cube $Q=[0,1]^\omega$ is not a countable union of hereditarily disconnected spaces. On the other hand, for any Bernstein set $B$ in $Q$ both spaces $B$ and $Q\setminus B$ are punctiform and hence one of them is not a countable union of hereditarily disconnected spaces.

Problem 2. Is there a Polish punctiform space which is not the union of countably many hereditarily disconnected spaces?

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  • $\begingroup$ For Problem 1 are you asking both versions of the question (i.e., for a Polish example or a not necessarily Polish example)? $\endgroup$ Commented Jul 24, 2022 at 3:01
  • $\begingroup$ @JamesHanson Yes. I am asking about an example of a metrizable separable (desirably Polish) hereditarily disconnected space which is not the union of countably many totally disconnected subspaces. I have made this clarification in the question. $\endgroup$ Commented Jul 24, 2022 at 5:32

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