Problem 1. Is it true that for every cardinal $\kappa\le\mathfrak c$ there exists a partition $(B_\alpha)_{\alpha\in\kappa}$ of the real line into $\kappa$ pairwise disjoint non-empty Borel subsets?

Remark. The answer to this problem is affirmative if $\mathfrak c\le \aleph_2$.

If the answer to Problem 1 is negative, then what about

Problem 2. Let $\mathcal P$ be a partition of the real line into Borel subsets. Is $|\mathcal P|=\mathfrak c$ or $|\mathcal P|\le\aleph_1$ in ZFC?


The answer to both problems is no!

If the Cohen forcing is used to add lots of reals to a countable transitive model of GCH, then in the resulting extension, any partition of the real line into Borel sets has size $\leq \aleph_1$ or $\mathfrak{c}$. (But the continuum can be anything with uncountable cofinality.) This is a theorem of Arnie Miller, Theorem 3.7 in

A. W. Miller, "Infinite combinatorics and definability," Annals of Pure and Applied Mathematical Logic 41 (1989), pp. 179-203.

On the other hand, it is consistent with any permissible value of $\mathfrak{c}$ that there is a partition of the real line into $\kappa$ closed sets, for every uncountable $\kappa \leq \mathfrak{c}$. This is a joint theorem of myself and Arnie Miller, Theorem 3.11 in

W. Brian and A. W. Miller, "Partitions of $2^\omega$ and completely ultrametrizable spaces," Topology and Its Applications 184 (2015), pp. 61-71 (link).

  • $\begingroup$ Thank you for the prompt answer. I have a question concerning your model from 2015 paper with Miller. Can we additionally have the equality $\mathrm{cov}(\mathcal N)=\mathfrak c$ holds in this model (or its suitable modification)? Here $\mathcal N$ is the ideal of Lebesgue null sets in the real line. More precisely, is it consistent that $\mathrm{cov}(\mathcal N)=\mathfrak c$ is arbitrarily large and for every $\kappa<\mathfrak c$ there exists a partition of the real line into $\kappa$ Borel subsets? $\endgroup$ – Taras Banakh May 12 at 8:39
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    $\begingroup$ The model from our paper has $\mathrm{cov}(\mathcal N) = \aleph_1$. The forcing used is an iteration of length $\omega_1$, and it adds Cohen reals at every stage. This means that no matter what the ground model is, we'll end up with $\mathrm{cov}(\mathcal N) = \aleph_1$. Since size-$\aleph_1$ partitions are guaranteed in the extension anyway, we could change the length of the iteration to $\omega_2$ and (maybe changing a few other things too) end up with $\mathrm{cov}(\mathcal N) = \aleph_2$. But a new technique would be needed to do any better than that. $\endgroup$ – Will Brian May 12 at 9:51
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    $\begingroup$ You can get $\mathrm{cov}(\mathcal N) = \mathfrak{c} = \aleph_3$, together with a partition of $\mathbb R$ into $\aleph_2$ Borel sets, as follows. First force $MA+\mathfrak{c} = \aleph_2$, and then add $\aleph_3$ random reals. Adding the random reals makes $\mathrm{cov}(\mathcal N) = \mathfrak{c} = \aleph_3$, and by an argument due to Stern/Kunen, adding any number of random reals to a model of $MA+\mathfrak{c} = \aleph_2$ will leave you with a partition of $\mathbb R$ into $\aleph_2$ closed sets. $\endgroup$ – Will Brian May 12 at 9:54
  • $\begingroup$ Will, I am reading your paper with Miller and have a question: is it consistent that the continuum is arbitrarily large and for no cardinal $\kappa$ with $\aleph_1<\kappa<\mathfrak c$ the countable power $\kappa^\omega$ condenses onto $2^\omega$? As far as I understand this problem does not reduce to partitions of $2^\omega$ into $\kappa$ Borel subsets (at least for $\kappa\ge\aleph_\omega$). $\endgroup$ – Taras Banakh May 12 at 21:45
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    $\begingroup$ Yes, that's right. The model described in that corollary has no condensation from $\kappa^\omega$ to any Polish space when $\aleph_1 < \kappa < \mathfrak{c}$. Also, using the results from the newer paper, it seems that if $\mathrm{cf}(\kappa) > \omega$, then $\omega^\omega$ is a condensation of $\kappa^\omega$ if and only if there is a partition of $[0,1]$ into $\kappa$ Borel sets. I don't see how to extend this to include singular cardinals with cofinality $\omega$. $\endgroup$ – Will Brian May 13 at 11:26

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