Model of ZFC Where Every Uncountable Co-Analytic Set of Reals Has a Perfect Subset I’ve been doing some reading on classical descriptive set theory for fun, and I’m trying to find a reference for the following claim:
Theorem: There is a model of ZFC Set Theory in which for every uncountable co-analytic subset $A \subseteq \mathbb{R}$, there is a non-empty perfect set $P \subseteq A$.
In 1970 of course, Solovay used forcing to construct a model of ZF (plus an inaccessible cardinal) in which every uncountable set of reals has a perfect subset. But this model does not have choice.
I have been told some of Woodin’s work should imply there is such a model, but I’m not well versed enough in Set Theory to be able to effectively comb through his work. If anyone has a reference or a place to look I would appreciate the help!
 A: I don't know if this is explicitly written down anywhere, maybe only mentioned in passing or implicit in the usual exposition of the construction of the Solovay model. Here's one way to do it:
First one proves an alternative characterization of the perfect set property for co-analytic sets (in symbols, $\mathsf{PSP}(\mathbf{\Pi^1_1})$):
Theorem. The following are equivalent:

*

*$\mathsf{PSP}(\mathbf{\Pi^1_1})$

*$\mathsf{PSP}(\mathbf{\Sigma^1_2})$

*for every real $x$, the set of real numbers in $L[x]$ is countable.

This characterization can be found in many standard texts (it's Theorem 25.38 in Jech, for instance). It involves careful descriptive set theoretic analysis of how these sets of reals are represented by special kinds of trees.
Having this in hand, one can start with an inaccessible cardinal $\kappa$ and use Levy collapse to render every ordinal below $\kappa$ countable, and $\kappa$ becomes $\omega_1$ in the extension. Working in the extension, one can show that 3 holds.
This is proven by appealing to the following property of the Levy collapse: if a real number is added by the forcing, then it is already added in an intermediate stage when $\kappa$ is still inaccessible (see Kanamori 10.21 for example). So whatever the cardinality of $\mathbb{R}\cap L[x]$ is by that stage, it is $<\kappa$ and will eventually be rendered countable by the forcing.
Note that the starting assumption of an inaccessible cardinal cannot be dispensed with, because when 3 holds, $(\omega_1)^V$ will be an inaccessible cardinal in $L$, for instance.
