# Sets of reals and absoluteness

Schoenfield's absoluteness states that if $\phi$ is $\Sigma^1_2$ then $V\models \phi$ iff $L\models \phi$. The set of reals in $L$ is $\Sigma^1_2$ and it is the largest countable $\Sigma^1_2$ set of reals if $\omega_1 ^L < \omega_1$.

If $\phi$ is $\Sigma^1_4$ then $V\models \phi$ iff $\mathcal M_2 \models \phi$, where $\mathcal M_2$ is the minimal proper class mouse with $2$ Woodins. The largest countable $\Sigma^1_4$ set of reals is exactly the set of reals in $\mathcal M_2$.

In general the largest countable $\Sigma^1_{2n+1}$ set of reals is exactly the set of reals in the minimal proper class mouse with $n$ Woodins $\mathcal M_n$. Could you redirect me to a reference please, I would like to see a proof.

Also how far can this phenomenon be pushed in general? For example, if $\phi$ is a second order formula (say $\Sigma^2_1$), how many Woodins would we need so that $\phi$ is absolute between $V$ and the appropriate proper class mouse containing these Woodin cardinals? Would the reals of that proper class mouse necessarily be the largest countable $\Sigma^2_1$ set of reals ( if it exists, I don't know if it does)?

• I am not sure about your first sentence. If $V=L$ (or something quite small over $L$), then the constructible reals are not countable... Sep 26 '12 at 23:20
• Yes I needed the extra assumption that I added. Of course if $V=L$ then the reals in $V$ are just the reals in $L$ and also Schoenfield's absoluteness doesn't need to be stated in this case. Sep 26 '12 at 23:50
• Actually any countable $\Sigma^1_2$ set must be in $L$ since they come from the Schoenfield tree and the Schoenfield tree is definable in $L$. (if the $\omega_1$ of $L$ is smaller than the real $\omega_1$) Sep 26 '12 at 23:56
• Actually I think I found the reference I was looking for. It seems like it is in Steel, Projectively well ordered inner models, 1995. I am not sure it is treating the $\Sigma^1_2$ case (if it's true, I don't know) Sep 27 '12 at 0:16

At the projective level, there are nice level by level generalizations, and looking at Steel's paper in the Handbook should give you the proof and the pre-requisites to understand it fully. This is what is behind the relation between determinacy and large cardinals. On the other hand, $\Sigma^2_1$ is never going to be possible, at least given our current understanding of how large cardinals work, because $\mathsf{CH}$ is $\Sigma^2_1$.
On the other hand, Woodin proved around 1985 that a conditional version of $\Sigma^2_1$ absoluteness holds. In fact, it identifies $\mathsf{CH}$ as a "maximal" sentence, in the following sense:
Theorem. Assume there are a proper class of cardinals that are simultaneously measurable and Woodin. If $\phi$ is a $\Sigma^2_1$ statement (with real parameters from the ground model), then: $\phi$ is true in some set forcing extension of the universe iff $\phi$ is true in every set forcing extension that satisﬁes $\mathsf{CH}$.
For a nice recent account of the argument, see Ilijas Farah, "A proof of the $\Sigma^2_1$ absoluteness theorem", in Advances in Logic, S. Gao, S. Jackson and Y. Zhang, eds., Contemporary Mathematics, 425 (2007) American Mathematical Society, RI., 9-22.
Beyond $\Sigma^2_1$, there is much speculation. It is expected some strengthening of diamond will be maximal for $\Sigma^2_2$, and we will get a similar theorem, but beyond $\Sigma^2_2$ this starts to conflict with other conjectures.