Timeline for Partitions of the real line into Borel subsets
Current License: CC BY-SA 4.0
11 events
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| May 13, 2021 at 11:26 | comment | added | Will Brian | 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 $\omega$. | |
| May 13, 2021 at 5:36 | comment | added | Taras Banakh | It seems that I found an answer to my last questions in Corollary 3.16 of your paper withh Miller. This corollary (or its minor modification) implies that it is consistent that the continuum is arbitrarily large and a cardinal $\kappa$ is $\le\aleph_1$ if $\kappa<\mathfrak c$ and $\kappa^\omega$ admits a condensation onto a Polish space. Right? | |
| May 13, 2021 at 5:17 | comment | added | Taras Banakh | After reading you preprint (about $0^\sharp$), I realized that the consistency of the statement "the continuum can be arbitrarily large and for no cardinal $\kappa$ with $\aleph_1<\kappa<\mathfrak c$ the space $\kappa^\omega$ condenses onto $2^\omega$" follows from the consistency of the statement: "the continuum can be arbitrarily large and for every $\kappa<\mathfrak c$ we have $\mathrm{cf}([\kappa]^\omega,\subseteq)<\mathfrak c$ and every cover of $2^\omega$ by $<\mathfrak c$ Borel sets has a subcover of cardinality $\le\aleph_1$". Do you know whether the last statement is consistent? | |
| May 13, 2021 at 4:54 | comment | added | Taras Banakh | That you for the prompt answer. Maybe let us start with the question: it is consistent that $\aleph_\omega<\mathfrak c$ and $\aleph_\omega^\omega$ cannot be condensed onto a Polish space? | |
| May 12, 2021 at 22:23 | comment | added | Will Brian | I think you're right. For $\kappa < \aleph_\omega$, the problems are equivalent, but possibly not for $\kappa \geq \aleph_\omega$. Recently, I've extended some of the stuff Arnie and I did to cardinals past $\aleph_\omega$, using the non-existence of $0^\dagger$. (See arxiv.org/pdf/2101.10088.pdf.) It might be the case that "$0^\dagger$ does not exist" makes these problems equivalent for $\kappa \geq \aleph_\omega$ with $\mathrm{cf}(\kappa) > \omega$. But I'm not sure without thinking more about it. | |
| May 12, 2021 at 21:45 | comment | added | Taras Banakh | 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 $\kappa\ge\aleph_\omega$). | |
| May 12, 2021 at 9:54 | comment | added | Will Brian | 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. | |
| May 12, 2021 at 9:51 | comment | added | Will Brian | 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. | |
| May 12, 2021 at 8:39 | comment | added | Taras Banakh | 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? | |
| May 12, 2021 at 3:19 | vote | accept | Taras Banakh | ||
| May 11, 2021 at 23:02 | history | answered | Will Brian | CC BY-SA 4.0 |