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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

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    $\begingroup$ I don't know how updated this list is, but it seems to be open (question 4) $\endgroup$
    – Alessandro Codenotti
    Commented Jul 15, 2020 at 8:45
  • $\begingroup$ @AlessandroCodenotti The question arises out of a joint paper with Kihara and Ng in computability theory. So we don't know, and we know that some others don't know either, but maybe someone here knows, or has an idea how to solve this. $\endgroup$
    – Arno
    Commented Jul 15, 2020 at 17:43
  • $\begingroup$ this may be of interest mathoverflow.net/a/357441/53155 $\endgroup$
    – erz
    Commented Jul 15, 2020 at 20:07

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The Hilbert cube can be written as the union of two punctiform spaces. Just take any 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. A topological space $X$ is called hereditarily disconnected if all connected subspaces in $X$ are singletons.

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  • $\begingroup$ I suspect that your result on the hereditarily disconnected spaces is what I actually needed in the first place, thanks a lot for bringing it my attention! $\endgroup$
    – Arno
    Commented Jul 15, 2020 at 19:43
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    $\begingroup$ @Arno You are welcome. I looked at the questions of Kihara (math.berkeley.edu/~kihara/questions.html). Is his 5th question still open? $\endgroup$
    – Taras Banakh
    Commented Jul 15, 2020 at 19:44
  • $\begingroup$ To my knowledge, Question 5 is still open (and in fact is the motivation for this question). The fascinating thing about it is that it is equivalent to an open problem in computability theory concerning enumeration degrees. The latter question is Question 5 here arxiv.org/abs/1904.04107 (Page 99) $\endgroup$
    – Arno
    Commented Jul 16, 2020 at 10:40
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    $\begingroup$ @Arno Thank you for the link to your paper. What is known about metrizable subspaces of $(\omega_{cof})^\omega$? All of them are punctiform. Does $(\omega_{cof})^\omega$ contain a non-trivial metrizable connected subspace? Is the Erdos space (it is totally disconnected but not zero-dimensional) embeddable into $(\omega_{cof})^\omega$? Is each metrizable subspace of $(\omega_{fin})^\omega$ zero-dimensional? $\endgroup$
    – Taras Banakh
    Commented Jul 16, 2020 at 10:48
  • $\begingroup$ We still need to carefully check the argument, but the Erdos space should embed into $(\omega_{\mathrm{cof}})^\omega$. $\endgroup$
    – Arno
    Commented Jul 17, 2020 at 11:40

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