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Feb 2 at 11:01 comment added Ember Edison Why is $\textbf{Ord}^\sharp$ required for $Σ^1_3$ generic absoluteness to hold, but not $\mathbb{R}^\sharp$? How should I understand the conclusion in link.springer.com/article/10.1007/BF02897063 ?
Sep 1, 2015 at 2:42 comment added Andrés E. Caicedo @William Yes.${}$
Sep 1, 2015 at 2:42 comment added William @AndresCaicedo In your paper "Projective Well-Ordering of the Reals" Theorem 3 seems to be what you are referring to. But the definition of generic absoluteness is stronger something like a two-step generic absoluteness. Also what is the sharp of an arbitary set. Should the existence of $A^\sharp$ be equivalent to there is an elementary embedding of $L[A]$ to $L[A]$? Can it be some kind of mouse?
Sep 1, 2015 at 2:31 comment added William @AndresCaicedo Is sharps really necessary? Bagaria and Friedman showed that it consistent with $\mathsf{ZFC}$ that (lightface) $\Sigma_3^1$-generic absoluteness holds, for example in the $\text{Coll}(\omega, <\lambda)$ extension where $\lambda$ is a $\Sigma_2$-correct cardinal (which is not a large cardinal).
Aug 28, 2015 at 7:52 comment added Asaf Karagila @Andres: One might even say that the consistency bound is... sharp :-P
Aug 26, 2015 at 21:42 comment added Andrés E. Caicedo The existence of sharps (as in Philip's second paragraph) is also necessary, so this assumption is optimal.
Aug 26, 2015 at 15:48 history answered Philip Welch CC BY-SA 3.0