7
$\begingroup$

Last night I was thinking about some related statements which follow from ZF+DC, but it actually seems they only need DC to hold in some outer model of the universe. In particular, let $M \models ZF.$ Consider the following claims (all relativized to $M$):

  1. For any sentence $\sigma$ such that $M \models \sigma,$ there is a countable transitive model $M_0 \in M$ such that $M_0 \models \sigma.$
  2. $L^M$ and $M$ agree on $\Sigma_1$ sentences (Levy's version of Shoenfield absoluteness).
  3. $HC^M \prec_1 M.$

I believe these follow from there being some outer model $N \supset M$ where $N \models ZF + DC,$ or even $M$ existing in some ambient universe $V$ such that $V \models ZF+DC \wedge ``M \text{ is transitive}" \wedge \text{ } \omega_1 \subset M.$ E.g., to prove (1), use DC in $V$ to construct a countable $S \subset M$ which collapses to $M_0'$ such that $M_0' \models \sigma.$ The claim that such a ctm exists is $\Sigma_2^1$ if I'm not mistaken, so Shoenfield absoluteness implies $M$ also has such a ctm.

So I'm wondering if these claims can be proven directly in ZF; I've heard (2) can be, but I've never seen claim (1) proven without DC. Is there a way to formalize "using choice in an ambient universe" within a model? I know there's a theorem of Woodin that says collapsing a supercompact cardinal in a model of ZF forces DC to hold, but that seems overkill.

$\endgroup$
  • $\begingroup$ The second claim is easy to verify: $\Delta_0$ statement are absolute, so $\Sigma_1$ is upwards absolute. $\endgroup$ – Asaf Karagila May 13 '17 at 17:08
  • $\begingroup$ You seem to be missing the hypothesis that $M$ is transitive, which I believe you have in mind. Or do you mean instead to assert merely that $M$ thinks $M_0$ is transitive? $\endgroup$ – Joel David Hamkins May 13 '17 at 17:15
  • $\begingroup$ Asaf, the point is to prove downwards absoluteness. Joel, the latter. Each of these three claims are all relativized to $M,$ under the hopes that they can be proven from ZF alone. $\endgroup$ – Elliot Glazer May 13 '17 at 17:20
  • $\begingroup$ Well. $\Pi_1$ statements are downwards absolute. If $M\models\lnot\exists x\varphi(x,y)$ for some $y\in L$, then $L$ ought to satisfy the same. $\endgroup$ – Asaf Karagila May 13 '17 at 17:36
  • 1
    $\begingroup$ It is a theorem that if $V \models ZF+DC,$ then $\Sigma_1$ sentences are downward absolute to $L.$ This is because $ZF+DC \vdash HC \prec_1 V,$ so any $\Sigma_1$ sentence is equivalent to a $\Sigma_1^{HC}$ sentence, which is equivalent to a $\Sigma_2^1$ sentence, which is downward absolute to $L$ by Shoenfield absoluteness. The point of this question is determining whether we can get rid of the use of DC in these arguments. $\endgroup$ – Elliot Glazer May 13 '17 at 19:43
8
$\begingroup$

Yes, all three of these statements can be proved in ZF, without any DC assumption.

For statement 1, assume $M\models\newcommand\ZF{\text{ZF}}\ZF+\sigma$. By the reflection theorem, there is some ordinal $\theta$ with $(V_\theta)^M\models\sigma$. One doesn't need DC to prove the reflection theorem, since the argument is about finding an ordinal that is closed under the ranks of witnesses, rather than being able to pick out particular witnesses.

So in $M$, we have a transitive set, $(V_\theta)^M$, which is a model of $\sigma$. Let $M[G]$ be a forcing extension of $M$ in which $|V_\theta|^M$ is countable. In the forcing extension $M[G]$, there is a countable transitive model of $\sigma$, namely, the set $(V_\theta)^M$, which is countable in $M[G]$. But the assertion "there is a countable transitive model of $\sigma$" is a $\Sigma^1_2$ assertion, and so it is absolute to $M$. So $M$ has a countable transitive model of $\sigma$, as desired.

A similar argument works for statement 2. If $M\models \sigma$ and $\sigma$ is $\Sigma_1$, then by the above, there is a countable transitive model of $\sigma$ in $M$. By Shoenfield absoluteness again, there is a countable transitive model of $\sigma$ in $L$. But if $\sigma$ is $\Sigma_1$, then $\sigma$ is upward absolute to $L$, and so $L\models\sigma$.

An essentially similar argument works for statement $3$. Namely, if $M\models\sigma$ and $\sigma$ is $\Sigma_1$, then there is countable transitive model of $\sigma$, and this countable transitive model is contained in $H_{\omega_1}$. So by upward absoluteness of $\Sigma_1$ assertions, it is true in $H_{\omega_1}$, as desired.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Nice. Real nice! $\endgroup$ – Asaf Karagila May 14 '17 at 13:38
  • $\begingroup$ Wow, it's surprising to me that statement 1, which is a theorem in pretty much every textbook, is always stated with an unnecessary assumption. Maybe that's just for simplicity of proof, since this theorem is usually earlier than the sections on forcing and well-founded trees. $\endgroup$ – Elliot Glazer May 14 '17 at 13:40
  • $\begingroup$ Just a nitpick, you only proved $HC$ is correct about $\Sigma_1$ sentences, not $\Sigma_1$ formulae. Not that it really matters, since Shoenfield absoluteness can be relativized to a real (and therefore any element in $HC,$ I think), so the same proof carries through. $\endgroup$ – Elliot Glazer May 14 '17 at 13:43
  • $\begingroup$ @Elliot: It's not at all surprising. The machinery needed here is far more complicated than the one needed for (1). You don't need Shoenfield, forcing or other various nontrivial facts presented here in order to prove (1) in ZF+DC. You just need Reflection, downward LS, and Mostowski collapses. This is why you can teach (1) in a basic course of axiomatic set theory, and maybe the choice-free proof in a third course after taking forcing and basic descriptive set theory as well. $\endgroup$ – Asaf Karagila May 14 '17 at 13:49
  • $\begingroup$ @ElliotGlazer Yes, one can relativize the proof to any parameter. $\endgroup$ – Joel David Hamkins May 14 '17 at 13:52

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.