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I want a model of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ in which every set of reals in $L(\mathbb{R})$ has the perfect set property. In terms of consistency strength, it is known that I need at least an inaccessible: if $\mathrm{PSP}(L(\mathbb{R}))$, then $\omega_1$ is inaccessible in $L$. I haven't been able to find any other lower bounds on the consistency strength of $\mathrm{MA}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$.

The best upper bound I can find is the fact, due to Woodin, that a measurable cardinal above infinitely many Woodin cardinals outright implies $\mathrm{Det}(L(\mathbb{R}))$. So, starting with these large cardinals in the ground, I can get what I want by forcing MA (using a "small" forcing).

My question is, do I really need such strong hypotheses? The ideal answer would be "this is known; the answer can be found in...". One the other hand, if you can tell me with confidence that it's an open problem, then at least I'll know that trying to solve it isn't a waste of time.

If it's easier to answer my question for projective sets, please do!

Back in 1964, Solovay proved that Levy-collapsing an inaccessible $\kappa$ to $\omega_1$ forces every set of reals definable by an omega-sequence of ordinals---this includes every set of reals in $L(\mathbb{R})$---to have the perfect set property. The catch is that the Solovay model also satisfies CH.

There's a 1989 JSL paper by Judah and Shelah ( that looks at the consistency strength of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ (and similar forcing axioms) in conjunction with various regularity properties for projective sets: Lebesgue measurability, the Baire property, and the Ramsey property. The perfect set property is (from my point of view) conspicuously absent.

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For MA restricted to nicely definable forcings, this is known: Bagaria and Judah, "Amoeba forcing, Suslin absoluteness and additivity of measure", MR1233819. I think they do not mention the perfect set property explicitly, but iteration of nice forcings does not change the first order theory of the reals. (See also Goldstern+Judah, MR1194971, Israel Journal 1992.) – Goldstern Jul 22 '11 at 13:13
David: Now that the answer is solved, let me point out a small remark regarding your second paragraph: The consistency strength of Det$(L(\mathbb R))$ is precisely $\omega$ Woodin cardinals. You do not need more to obtain models where Martin's axiom holds and $L(\mathbb R)$ still satisfies determinacy. For example, you can force over $L(\mathbb R)$ itself using Woodin's ${\mathbb P}_{max}$ forcing to achieve this. – Andrés E. Caicedo Jul 26 '11 at 21:46
up vote 4 down vote accepted

$\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$ is equiconsistent with a Mahlo cardinal.

Before Goldstern's comment, I had assumed the perfect set property was important enough to authors to mention if their theorems covered it. With that assumption falsified, I read more carefully, and determined that the Judah-Shelah paper I mentioned in the question had the answer all along. The proof of Lemma 1.1 shows that any forcing extension satisfying the hypotheses of the lemma actually has the same $L(\mathbb{R})$ as some Solovay model. Moreover, the proof of Theorem 3.1 actually builds from a Mahlo cardinal a forcing extension that satisfies $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ and the hypotheses of Lemma 1.1. Therefore, a Mahlo cardinal suffices. A Mahlo cardinal is necessary because $\mathrm{PSP}(L(\mathbb{R}))$ is well-known to imply that $\omega_1^{L[x]}<\omega_1$ for all reals $x$, which in turn implies, by (ii)$\Rightarrow$(i) of 3.1, that if $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ also holds, then a Mahlo cardinal is consistent with ZFC (specifically, $\omega_1$ is Mahlo in $L$ by the proof of (ii)$\Rightarrow$(i)).

(The hypotheses of Lemma 1.1 are that $\kappa$ is inaccessible, $\mathbb{P}$ satisfies the $\kappa$-chain condition, $\mathbb{P}$ forces $\kappa=\omega_1$, and, for every subset $Q$ of $\mathbb{P}$ of size less than $\kappa$, $Q$ extends to $P'\subseteq \mathbb{P}$ such that $|P'|<\kappa$ and the inclusion map completely embeds $P'$ into $\mathbb{P}$.)

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