5
$\begingroup$

Let $M$ be a countable transitive model of (enough of) ZFC. Mostowski's Absoluteness Theorem says that $\Pi^1_1$ statements are absolute between $M$ and larger models, in particular, between $M$ and the universe $V$.

For general $M$, this cannot be extended to $\Sigma^1_2$ statements, see Andrés Caicedo's answer here: Failure of Shoenfield's Absoluteness.

My question is: Can we have absoluteness of $\Sigma^1_2$ (and beyond) statements between $M$ and $V$ for some countable transitive models $M$? (Possibly at the expense of some local or global large cardinal assumptions.)

Or, is it the case that for any countable transitive model $M$, there is a $\Sigma^1_2$ (lightface) statement true in $V$ which fails in $M$?

$\endgroup$
  • 1
    $\begingroup$ If $V$ satisfies $V=L$, but $M$ is a model of, say Projective Determinacy, then there's really not that much hope for absoluteness. I think. $\endgroup$ – Asaf Karagila Sep 9 '18 at 20:17
  • 1
    $\begingroup$ You have several correctness results along these lines from large cardinal assumptions. See for instance section 7.2 in Steel's Handbook article. $\endgroup$ – Andrés E. Caicedo Sep 9 '18 at 21:15
8
$\begingroup$

First, it is consistent that there is a ctm but no $\Sigma^1_2$-correct ctm, for example if $V$ is the minimal model with a ctm. If there is a model of ZFC $M$ containing all the reals (e.g., if there is an inaccessible), then there is a projectively correct ctm: take the transitive collapse $H$ of a countable elementary substructure of $M$. Since reals don't move in the collapse, $H$ is projectively correct with real parameters. (By this argument there is a projectively correct model of ZFC if and only if there is a projectively correct ctm.)

As Andrés points out, one can also obtain correct models from determinacy hypotheses, though these hypotheses are much stronger than the existence of such models. E.g., if $\Pi^1_1$-determinacy holds then there is a $\Sigma^1_2$-correct ctm (e.g., $L_\kappa$ where $\kappa$ is the least Silver indiscernible).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.