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)?

  • $\begingroup$ I am not sure about your first sentence. If $V=L$ (or something quite small over $L$), then the constructible reals are not countable... $\endgroup$
    – Asaf Karagila
    Sep 26, 2012 at 23:20
  • $\begingroup$ 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. $\endgroup$ Sep 26, 2012 at 23:50
  • $\begingroup$ 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 $\omega_1$) $\endgroup$ Sep 26, 2012 at 23:56
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    $\begingroup$ 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) $\endgroup$ Sep 27, 2012 at 0:16

1 Answer 1


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.

What the actual optimal statement is in terms of large cardinal strength is hard to tell at the moment, as inner model theory does not reach that high. We expect it to be somewhere around the sharp for a mouse with a measurable Woodin.

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.


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