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In the paper "On the consistency strength of projective uniformization" Woodin proves a lemma "Assume $M$ is a model of ZFC that is $\Sigma^{1}_{3}$-absolute. Then $M\vDash\forall x\in\mathbb{R}\,[x^{\sharp} \mathrm{\ exists}]$." He then goes on to say after the proof "It in fact now follows by a theorem of Martin-Solovay [6] that $\Sigma^{1}_{3}$-absoluteness is equivalent to the existence of $S^{\sharp}$ for every set $S$."

[6] Martin, D. A. and Solovay, R. M., A basis theorem for $\Sigma^{1}_{3}$ sets of reals, Ann. of Math. 89 (1969), 138-160.

When I read this article of Martin and Solovay I have trouble seeing the connection with the assertion that $S^{\sharp}$ exists for all sets $S$. I was wondering if anyone could clarify this for me.

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up vote 6 down vote accepted

Rupert, the Martin-Solovay paper shows the absoluteness result follows form the existence of measurable cardinals. The measurables are used in the construction of certain trees (now called Martin-Solovay trees), where some ordinals are chosen by means of the measure to serve as witnesses (of ill-foundedness of some branches, say). There are standard arguments to tighten up this approach so indiscernibles (coming from sharps) suffice for this. (This is closely related to how $\Pi^1_1$-determinacy can be established form sharps rather than measurables.)

I suspect Martin's draft of a book on determinacy has the details, but I do not know whether you have access to it (and I do not have my copy handy at the moment to double-check). In any case, Ralf Schindler and I have a joint paper where we explain in detail how $\Sigma^1_3$-absoluteness is equivalent to the existence of sharps. The paper is "Projective well-orderings of the reals", Arch. Math. Logic (2006) 45:783–793. It is available at my webpage, see Theorem 3.

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