8
$\begingroup$

Is $[0,1]$ a disjoint union of $\aleph_1$ compact subsets with empty interior?

The answer is obviously yes assuming the continuum hypothesis. Also, by Baire's lemma, the answer is negative if one replaces $\aleph_1$ with $\aleph_0$.

Does anybody know if this question has a definite answer or whether this is independent of ZFC?

$\endgroup$
  • 1
    $\begingroup$ Related cool fact: if you replace "compact" with "Borel", then the answer is always yes at $\aleph_1$, but independent of ZFC for all $\kappa > \aleph_1$. $\endgroup$ – Will Brian Feb 1 '18 at 13:58
  • $\begingroup$ @WillBrian I'm not sure I understand: if $\mathbf{R}$ is a union of $\aleph_1$ Borel subsets of empty interior, it's a union of $\kappa\ge\aleph_1$ as well (just use a non-injective indexation by $\kappa$)... it's easy to make it injective if $\kappa\le c$ (use $c$ distinct compact subsets to cover a Cantor subset and work with the complement with $\aleph_1$ additional guys). Are you asking something more (pairwise non-comparable for inclusion?) $\endgroup$ – YCor Feb 1 '18 at 17:18
  • $\begingroup$ @YCor: $[0,1]$ is a disjoint union of $\aleph_1$ Borel sets with empty interior (a theorem of Hausdorff), but this is not necessarily so for $\aleph_2 \leq \kappa < \mathfrak{c}$. The Cantor set can be partitioned into $\mathfrak{c}$ sets, but if you try to put $\kappa$ of them into a partition of $[0,1]$, you're left needing to do something with the complement of those $\kappa$ sets. It might be really ugly (badly non-Borel), and you might not be able to cover it with fewer than $\mathfrak{c}$ extra Borel sets. $\endgroup$ – Will Brian Feb 1 '18 at 17:27
  • $\begingroup$ @WillBrian thanks - sorry I misread the question (I read it 30 times without seeing "disjoint")... my previous comment is thus pointless. As regards the question without the word "disjoint", it's also solved in your answer. $\endgroup$ – YCor Feb 1 '18 at 17:33
18
$\begingroup$

It is independent of ZFC. As you mention, CH implies the answer is yes. A different axiom, MA+$\neg$CH, implies the answer is no.

A little more precisely, there is a cardinal number denoted $\mathrm{cov}(\mathcal M)$, one of the so-called ``small cardinals'', that is defined to be the smallest number of meager sets needed to cover $[0,1]$. MA implies this cardinal is equal to the continuum, and if that is the case (and CH fails) then the answer to your question is no, because a partition of $[0,1]$ as you describe is also a covering of $[0,1]$ with meager sets.

But this is not the end of the story. There is a difference between covering and partitioning, and in general it seems harder to partition $[0,1]$ into copies of the Cantor space than to cover it with copies of the Cantor space. I have a paper with Arnie Miller (link) where we look at this question (and some similar ones), and show that it is consistent with the continuum being very large that for every $\kappa < \mathfrak{c}$ there is a partition of $[0,1]$ into exactly $\kappa$ copies of the Cantor space.

EDIT: I just realized that Taras Banakh has asked a very similar question here. The statement of his question actually contains an answer to your question, and you can learn a lot more about this question by looking at his question, the comments under it, and the answer I posted there.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.