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 !

  • 1
    $\begingroup$ $\Sigma^1_2$, not $\Sigma^2_1$. $\endgroup$ Nov 26, 2013 at 18:09
  • 2
    $\begingroup$ 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). $\endgroup$ Nov 26, 2013 at 18:11
  • 5
    $\begingroup$ Finally. Now all the set theory Ph.D. students are using MathOverflow. :-) $\endgroup$
    – Asaf Karagila
    Nov 26, 2013 at 18:17
  • $\begingroup$ Very closely related - mathoverflow.net/questions/71965/… $\endgroup$ Nov 26, 2013 at 20:34
  • 4
    $\begingroup$ Which hypothesis did you want to fail? That $M \subseteq N$? That $\omega_1^M \subseteq N$? That the formula is $\Sigma^1_2$? $\endgroup$ Nov 27, 2013 at 12:56

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Nov 27, 2013 at 13:46
  • $\begingroup$ 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 ? $\endgroup$ Nov 27, 2013 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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