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## reference request – How nice can sets of reals be under $\mathsf{ZF} + \mathsf{BPI}$?

It’s well known that the full axiom of choice is not needed to prove the existence of non-measurable subsets of $\mathbb{R}$. In particular, the Boolean prime ideal theorem ($\mathsf{BPI}$) is sufficient, since non-principle ultrafilters on $\omega$coded as a subset of $2^\omega$, are not Lebesgue measurable. They also necessarily fail to be Baire measurable.

That said, they *can* satisfy a weaker regularity condition. I’ve convinced myself that under $\mathsf{ZFC}+\mathsf{CH}$ if $\mathcal{F}_0 \subset 2^\omega$ is a proper filter generated by an $F_\sigma$ subset of $2^\omega$then $\mathcal{F}_0$ can be extended to an ultrafilter $\mathcal{F}$ with the property that for every Borel set $B \subseteq 2^\omega$both $B \cap \mathcal{F}$ other $B \set minus \mathcal{F}$ have the perfect set property (ie, are either finite, countable, or have a non-empty perfect subset). I’m not sure about filters generated by Borel sets in general, and I don’t know what $\mathsf{ZFC}$ entails regarding the question of whether every ultrafilter on $\omega$ has this property.

This has led me to the following question.

question.is $\mathsf{ZF}+\mathsf{BPI}+\text{“Every set of reals has the perfect set property.”}$ consistent?

Looking through *Consequences of the Axiom of Choice* by Howard and Rubin, it seems like this was open 24 years ago at the very least.

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