Tue, 7 Feb 2023, 1:25 pm MST

I will sketch a proof that, assuming $0^{\dagger}$ does not exist, if there is a partition of $\mathbb R$ into $\aleph_{\omega}$ Borel sets, then there is also a partition of $\mathbb R$ into $\aleph_{\omega+1}$ Borel sets. (And the same is true for any singular cardinal of countable cofinality in place of $\aleph_{\omega}$.) This contrasts starkly with the situation for cardinals with uncountable cofinality and their successors, where the spectrum of possible sizes of partitions of $\mathbb R$ into Borel sets can be made completely arbitrary via forcing.

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