# 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 $$B \cap \mathcal{F}$$ and $$B \setminus \mathcal{F}$$ have the perfect set property (i.e., 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 do not know what $$\mathsf{ZFC}$$ entails regarding the question of whether every ultrafilter on $$\omega$$ has this property.

This has lead 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.