Is there any paper discussing the consistency strength (or possible equivalents, maybe large cardinals) of just assuming the perfect set property for certain levels of the projective hierarchy?
2 Answers
Analytic sets have the perfect set property, provable in, say, ZF+DC. This goes back to Suslin, and is discussed in Kanamori's book "The higher infinite" (Around section 12).
Large cardinals imply that projective sets have the perfect set property, but the cardinals needed (Woodin cardinals) are more than those needed to establish mere consistencyan inaccessible suffices, of course, by Solovay's result: If $\kappa$ is inaccessible, and ${\mathbb P}$ is the Levy collapse that makes countable all cardinals below $\kappa$, then in the forcing extension by ${\mathbb P}$, all sets of reals in $L({\mathbb R})$ have the classical regularity properties (Lebesgue measurability, the property of Baire, and the perfect set property). (Solovay's result is stronger than I've stated, this is discussed in detail in sections 10 and 11 of Kanamori's book.)
In fact, for sets of reals in $L({\mathbb R})$, an inaccessible is required as well. This goes back to Specker, the point being that if $\aleph_1\le{\mathfrak c}$, then there is a set of reals without the perfect set property, so $\omega_1^{L[r]}<\omega_1$ for all reals $r$, and it follows that $\omega_1$ is inaccessible in all $L[r]$.
For (boldface) $\Sigma^1_2$ sets, sharps for reals imply the result and already an inaccessible is needed in consistency strength: Solovay showed that if $A$ is $\Sigma^1_2$ in a parameter $r$, then either $A\in L[r]$ or else $A$ contains a perfect set. If $A$ is the standard wellordering of ${\mathbb R}\cap L[r]$, then $A$ is $\Sigma^1_2$ in $r$. But then, if $\omega_1^{L[r]}=\omega_1$, we have that $A$ does not have a perfect subset and is uncountable (This can be unearthed from Section 14 of Kanamori's book). It follows that already at the $\Sigma^1_2$ level we need $\omega_1$ to be inaccessible in all $L[r]$.
In summary: An inaccessible is precisely what is needed.
Bartoszynski and Judah' book "Set theory. On the structure of the real line" discusses extensively the effect of Martin's axiom (and large cardinals) on the regularity properties of the projective sets, but it focuses on the Baire property and Lebesgue measurability. David Milovich asked a closely related question not long ago, and you may want to look at the answer there.
However, I suspect the real question is what is the consistency strength of, say, "the $\Sigma^1_5$ sets have the perfect set property but the $\Sigma^1_6$ do not." I think this is wide open in general. It is also open with Lebesgue measurability or Baire property in place of the perfect set property, as far as I know.
[Added Jul. 14, 2023.] For some progress on the more ambitious version of the question mentioned in the last paragraph above, see
MR4194891. Friedman, Sy David; Schrittesser, David. "Projective measure without projective Baire", Mem. Amer. Math. Soc. 267 (2020), no. 1298, v+150 pp. ISBN: 9781470442965; 9781470463953. ArXiv.
In this work, Sy and David show that a Mahlo cardinal is an upper bound for the consistency of $\mathsf{ZFC}$ together with ``all projective sets of reals are Lebesgue measurable, but there is a $\Delta^1_3$ set without the Baire property''.

$\begingroup$ May I just ask what "Solovay's result" refers to? Thank you for the answer! $\endgroup$– ftontiCommented Aug 7, 2011 at 21:05

$\begingroup$ @ftonti: I've edited the answer to include a short description. $\endgroup$ Commented Aug 7, 2011 at 21:16
The classical argument Andres mentions can be taken one step further to show that if every uncountable $\Pi_1^1$ set contains a perfect subset then $\aleph_1$ is inaccessible in $L$. This follows from the same fact for $\Sigma^1_2$sets since, by the uniformization theorem, every $\Sigma^1_2$set is the injective projection of a $\Pi^1_1$set.

$\begingroup$ So I just want to see if I understand this correctly: the perfect set property for coanalytic sets already leads to an inaccessible? Sorry for the probably trivial restating of your answer. $\endgroup$– ftontiCommented Aug 13, 2011 at 18:36

$\begingroup$ Yesat least to consistency of an inaccessible. Namely it proves that the real $\aleph_1$ is in an inaccessible cardinal in $L$. $\endgroup$ Commented Aug 13, 2011 at 19:27

$\begingroup$ Thank you very much, then I had understood it correctly. $\endgroup$– ftontiCommented Aug 13, 2011 at 20:11