# Failure of Shoenfield's Absoluteness

Shoenfield's absoluteness states that if $M \subseteq N$ are models of $ZF$ and $M \supseteq \omega_1^N$, then every $\Sigma^1_2$ formula with parameters in $M$ is absolute between $M$ and $N$. In particular, $\Sigma^1_2$ properties are preserved under generic extensions of the universe.

What I'm looking for are examples of failures of $\Sigma^1_2$ absoluteness between $M \subseteq N$ models of $ZF$ when $M$ does not contain all countable ordinals of $N$. I'll be even happier if the examples are in $ZF$ and as natural as possible, although any help will be appreciated of course.

Thanks !

• $\Sigma^1_2$, not $\Sigma^2_1$. Nov 26, 2013 at 18:09
• Consider the statement "there is a transitive model of set theory". This statement is $\Sigma^1_2$: There is a real that codes a model of set theory that is well-founded (that is, no sequence through its ordinals is strictly decreasing). If there is such a model, this is true in $V$ that fails in the smallest such model (which is countable). Nov 26, 2013 at 18:11
• Finally. Now all the set theory Ph.D. students are using MathOverflow. :-) Nov 26, 2013 at 18:17
• Very closely related - mathoverflow.net/questions/71965/… Nov 26, 2013 at 20:34
• Which hypothesis did you want to fail? That $M \subseteq N$? That $\omega_1^M \subseteq N$? That the formula is $\Sigma^1_2$? Nov 27, 2013 at 12:56

I'm turning my comment into an answer. With $\Sigma^1_2$ statements we can discuss well-foundedness: A real codes a well-founded model of enough set theory iff it codes a model (which is an arithmetic statement) and the model is well-founded (this you can express by saying that no sequence through the ordinals of the model is strictly decreasing). So, to say that there is such a well-founded model is $\Sigma^1_2$. As for "enough set theory", pick $T$ a finite and sufficiently strong fragment of $\mathsf{ZF}$.
By the reflection theorem, there are transitive models of $T$, so there are countable transitive ones (by Lowenheim-Skolem and Mostowski). Pick an example $M$ of smallest height. We have that, in $M$, there are no transitive set models of $T$. That is, the $\Sigma^1_2$ statement discussed in the previous paragraph is true in $V$ (and in set models of certain stronger fragments of $\mathsf{ZF}$), but fails in $M$.
If we assume that there are transitive set models of $\mathsf{ZF}$, then we can run this argument without using reflection, of course. We can even pick $M$ to be an $L_\alpha$. But the point of picking $T$ finite is so that we can formalize this: If it were the case that Shoenfield's absoluteness result goes through without the requirement on $\omega_1$, then this would be provable in $\mathsf{ZF}$, and the proof of course only uses a finite set of axioms, and would apply to finite fragments of set theory, as long as they are strong enough. We can then let $T$ be so strong that, in particular, it contains all these axioms. This, naturally, leads to a contradiction, and the conclusion is that Shoenfield's theorem indeed needs the uncountability assumption.
• Thanks ! So if I may rephrase it as a $ZF$ counterexample : Let $T \subseteq ZF$ be a finite fragment of $ZF$ that proves Shoenfield's Absoluteness. Choose $\kappa$ large enough so that $V_\kappa \models T$, and $M \subseteq V_\kappa$ transitive and of minimal height between models of $T$. Then $V_\kappa$ thinks that there is a well founded model of $T$, while $V_\kappa$ disagrees. Nov 27, 2013 at 13:46
• This is obviously an excellent answer to my question, Thanks Andres ! Still, if anyone can think of a counterexample which is more natural, in the sense that it is in $ZF$ and have nothing to do with Shoenfield's absoluteness, I'll be glad if you could post it here. Just to clarify what I mean, failure of absoluteness for $\Sigma^1_3$ formulas is demonstrated with statements about constructible elements - definitely more natural, isn't it ? Nov 27, 2013 at 13:53