Recall that a (separable) metric space is called *punctiform*, if all its compact subspaces are zero-dimensional. While "natural" spaces would seem to be punctiform if they already themselves zero-dimensional, there are even infinite dimensional punctiform spaces. The constructions I have seen however are still yielding spaces that "feel sparse" to me.


The Hilbert cube $[0,1]^\omega$ is large in the sense that it is not a countable union of zero-dimensional spaces. What I am now wondering is whether we can write the Hilbert cube as a countable union of punctiform spaces. Note that I do not want to impose any complexity constraints on the pieces.


If the answer should be "yes", I'd be very interested in understanding the structure of the punctiform spaces involved.


Had the answer been "no", this would have answered an open problem in computability theory, see Question 5 on Page 99 (v1) here: <https://arxiv.org/abs/1904.04107>



Is the Hilbert cube the countable union of punctiform spaces?
The Hilbert cube can be written as the union of two punctiform spaces. Just take any [Bernstein set](https://en.wikipedia.org/wiki/Bernstein_set) $X\subset[0,1]^\omega$ and observe that compact subsets in $X$ and $Y=[0,1]^\omega\setminus X$ are at most countable. So, $X$ and $Y$ are punctiform spaces and $X\cup Y=[0,1]^\omega$.


A less trivial fact says that the Hilbert cube cannot be written as the countable union of hereditarily disconnected sets; for the proof of this fact, see Main Lemma in [this paper](https://www.impan.pl/download/pdf/fm167-2-1). A topological space $X$ is called *hereditarily disconnected* if all connected subspaces in $X$ are singletons.


