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If there are infinitely many Woodins, then every projective set is determined, whence every projective set has the perfect set property (PSP). Since determinacy is such a stronger property than the PSP, it seems this should be overkill, and we should be able to prove that every projective set has the PSP from weaker assumptions. Indeed, in terms of consistency strength (rather than outright implication) we do know the gap: projective determinacy needs the Woodins while the PSP for projective sets only needs one inaccessible.

It's clear that no large cardinal consistent with $V = L$ suffices. And Solovay's work mentioned in this related question and this other related question gives an epsilon more precise lower bound. But there's a pretty big gap from that to infinitely many Woodins.

Are more precise bounds known?

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    $\begingroup$ Schindler and Wilson proved that the consistency strength of $\mathsf{ZFC}$ with the perfect set property for universally Baire sets is equal to $\mathsf{ZFC}$ with a large cardinal called virtually Shelah cardinal, whose consistency strength is below $0^\sharp$. It does not answer your question, but might be relevant. $\endgroup$
    – Hanul Jeon
    Jan 27, 2023 at 21:48
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    $\begingroup$ Finitely many Woodins is not enough, because in $M_n$ (the minimal iterable proper class model with $n<\omega$ Woodin cardinals) it fails. $\endgroup$
    – Farmer S
    Jan 27, 2023 at 23:16
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    $\begingroup$ But "infinitely many Woodins" is certainly overkill, even for projective determinacy. By well known results, for that it suffices to have that for every $n<\omega$, $M_n^{\#}$ exists and is $\omega_1$-iterable. $\endgroup$
    – Farmer S
    Jan 27, 2023 at 23:22
  • $\begingroup$ @Farmer: So, more or less Con(ZFC+PD), right? $\endgroup$
    – Asaf Karagila ♦
    Jan 29, 2023 at 1:37
  • $\begingroup$ @FarmerS Thanks for the answer! If you want to write it as an answer instead of a comment I can accept it. $\endgroup$
    – Kameryn Williams
    Jan 30, 2023 at 15:24

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