I was reading about the $\Omega$-conjecture and have thought of a refutation of it, which seems too simple to not have been noticed since the $\Omega$-conjecture has been around, so i'm skeptical and want to see whether anyone can spot a flaw in it.

I will assume there is a proper class of hyper-huge cardinals. This assumption implies a proper class of Woodin cardinals and by Usuba implies the existence of the bedrock model $W$, and since $V$ is a set-generic extension of $W$ and our assumption (as well as the $\Omega$-conjecture) is invariant throughout the set-generic multiverse, we can work in $W$. Suppose for a contradiction that the $\Omega$-conjecture holds. Since $V = W$ is $\Sigma_2$ definable in any set-forcing $V[G]$, we can uniformly evaluate $\Sigma_2$-truths of $V$ in any set-forcing of $V$ by other recursively given $\Sigma_2$ sentences. Since the $\Sigma_2$-laws of the set-generic multiverse are definable in $H(\delta_0^+)$ where $\delta_0$ is the least Woodin cardinal (by the $\Omega$-conjecture), and they can be used to compute all the $\Sigma_2$ truths of $V$ (including the theory of $H(I_0^{+})$ and beyond), this violates Tarski's undefinability of truth. Thus the $\Omega$-conjecture must fail assuming our large cardinal hypothesis. $\square$

Is this argument really valid? (I'm worried 2016 will end by refuting $V = \text{Ultimate }L$)

  • 4
    $\begingroup$ Uncountable_turtle, could you clarify how you are violating Tarski? After all, there is no problem with having a $\Sigma_2$ truth predicate of $V$--this doesn't violate Tarski. Also, could you clarify the role of parameters when you refer to "$\Sigma_2$-truths". $\endgroup$ Dec 31, 2016 at 12:50
  • 4
    $\begingroup$ Also, why do you worry that the year will end with a refutation of $V=\text{Ultimate }L$? That's a good thing, not a bad thing! :) $\endgroup$
    – Asaf Karagila
    Dec 31, 2016 at 12:58
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because it asks to verify a proof / disproof. $\endgroup$
    – Stefan Kohl
    Dec 31, 2016 at 17:22
  • 11
    $\begingroup$ @StefanKohl Please don't close the question; it is a fine question, and definitely on topic! I for one would like to get to the bottom of it. $\endgroup$ Dec 31, 2016 at 17:28
  • 8
    $\begingroup$ @StefanKohl I know that you already retracted your close vote, so no harm no foul, but the usual application of "we don't vet proofs here" is to crackpot posts, which doesn't apply here. This seems like a pretty serious and high-level discussion. $\endgroup$
    – Todd Trimble
    Jan 1, 2017 at 19:19

1 Answer 1


Woodin's theorem says that assuming the $\Omega$ Conjecture and the existence of a proper class of Woodin cardinals, the set $\mathcal V_\Omega$ of $\Pi_2$ sentences that hold in every universe of the generic multiverse is lightface definable over $H_{\delta^+}$ where $\delta$ is the least Woodin cardinal. You claim there is a Turing reduction from the $\Pi_2$ theory of the bedrock to $\mathcal V_\Omega$, obtaining in this way a $\Sigma_2$ definition of $\Pi_2$ truth (assuming the Ground Axiom), a contradiction. The proposed reduction sends a formula $\phi$ to the sentence $f(\phi)$ expressing "The bedrock satisfies $\phi$." Tell me if I'm misunderstanding you.

One problem: it isn't clear that $f(\phi)$ is $\Pi_2$. Note that if $W$ is a $\Sigma_2$ or $\Pi_2$ or even $\Delta_2$ inner model and $\phi$ is a $\Pi_2$ sentence, the sentence $W\vDash \phi$ is not obviously $\Pi_2$. We can write $\phi$ as $$\forall x \ (x\notin W \vee \exists y\ (y\in W\wedge \psi(x,y))$$ for some $\Delta_0$ formula $\psi$. This seems to be no simpler than $\Pi_3$ even if $W$ is $\Delta_2$.

For example, we claim that the statement $$\Psi\equiv \text{CH}\text{ fails in the bedrock}$$ is not equivalent over ZFC + the Bedrock Axiom to a $\Pi_2$ formula, even though $\neg\text{CH}$ is $\Delta_2$. (The Bedrock Axiom just asserts that the generic multiverse has a bedrock.) Suppose towards a contradiction that this statement is equivalent to a $\Pi_2$ formula, which we may assume is of the form $\forall \alpha\ (V_\alpha\vDash \psi)$. Fix a model $$M\vDash\text{the Bedrock Axiom}+\neg\text{CH} + \neg\Psi$$ (so the bedrock of the generic multiverse of $M$ satisfies $\text{CH}$). Since $M\vDash \neg \Psi$, there is some ordinal $\alpha$ of $M$ such that $M_\alpha\vDash \neg \psi$. Now pass to a class forcing extension $M[H]$ satisfying the Ground Axiom and such that $M[H]_{\alpha} = M_{\alpha}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$. (See Reitz's thesis, Theorem 12.) We have $M[H]\vDash \neg\text{CH}$ (since $M\vDash \neg \text{CH}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$). On the other hand since $M[H]_\alpha = M_\alpha\vDash \neg \psi$, $M[H]\vDash \exists \alpha\ (V_\alpha\vDash \neg \psi)$. Thus $M[H]\vDash \neg \Psi,$ so the bedrock of $M[H]$ satisfies $\text{CH}$, which contradicts the fact that $M[H]$ is its own bedrock.


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.