Consider the following statement:

$(\dagger)$ $\ $ There is an inner model $M$ such that $M \models \mathsf{GCH}+\square$ and for every countable $X \subseteq \mathrm{Ord}$, there is a countable $Y \in M$ such that $X \subseteq Y$.

When I say $M$ is an ``inner model'' I mean that $M$ is a class (definable with parameters) such that $M \supseteq \mathrm{Ord}$ and $\langle M,\in \rangle \models \mathsf{ZFC}$.

Question: Does the negation of $(\dagger)$ have any large cardinal strength? That is, does the consistency of $\mathsf{ZFC}+\neg(\dagger)$ imply the consistency of large cardinals? And if so, what kind of large cardinals are required for $\neg(\dagger)$?

Here are a few observations:

$\bullet \ $ The statement of $(\dagger)$ seems close in spirit to the statement of Jensen's Covering Lemma. If we were to replace "countable" with "of size $\kappa$" for any uncountable $\kappa$, then this modified version of $(\dagger)$ would follow from the Covering Lemma by taking $M = \mathrm{L}$, and therefore its negation would imply the existence of $0^\sharp$.

$\bullet \ $ However, the Covering Lemma does not imply $(\dagger)$. Furthermore, if we modify $(\dagger)$ by insisting on $M = \mathrm{L}$, then we don't need the failure of the Covering Lemma, or any large cardinal strength at all, to get this modified version of $(\dagger)$ to fail. This is because of Namba forcing. If we start with $\mathrm{L}$ and add a Namba-generic filter $G$, then $\mathrm{L}[G]$ will fail to satisfy ``$(\dagger)$ with $M = \mathrm{L}$.'' However, (I'm fairly certain that) $\mathrm{L}[G]$ is itself still a model of $\mathsf{GCH}+\square$, which means that $\mathrm{L}[G]$ satisfies $(\dagger)$, simply by taking $M = \mathrm{L}[G]$. Therefore Namba forcing does not make $(\dagger)$ fail.

$\bullet \ $ The Singular Cardinals Hypothesis follows from $(\dagger)$. Hence one can force a failure of $(\dagger)$ by forcing $\neg\mathsf{SCH}$, which requires a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$. So I know that $\neg(\dagger)$ is consistent relative to large cardinals -- I just don't know whether any large cardinals are actually necessary.

$\bullet \ $ If $(\dagger)$ holds in some ground model $V$, then it continues to hold in any $\omega$-distributive forcing extension of $V$. If $(\dagger)$ and Jensen's Covering Lemma both hold in $V$, then both continue to hold in any cardinal-preserving forcing extension of $V$.

$\bullet \ $ I suppose $(\dagger)$ isn't expressible as a first-order statement in the language of set theory, but it is expressible as a scheme in the metatheory.

My motivation for asking this question is that I've proved a topological theorem using $(\dagger)$ as a hypothesis. I'd like to know my hypothesis can be negated without assuming something with large cardinal strength, like the failure of $\mathsf{SCH}$.

  • $\begingroup$ Do you want the inner model to be correct about cardinals? $\endgroup$ Nov 23, 2020 at 17:00
  • $\begingroup$ @MonroeEskew: No, it doesn't need to be. $\endgroup$
    – Will Brian
    Nov 23, 2020 at 18:10
  • $\begingroup$ I mean, in general the failure of squares require large cardinals... $\endgroup$
    – Asaf Karagila
    Nov 23, 2020 at 18:39
  • $\begingroup$ @AsafKaragila: True, but the question isn't just about square failing. In fact, I'm not sure that removing square from the question would make it any easier. Finding an inner model of $\mathsf{GCH}$ with this strong covering property is already nontrivial. (We know it's nontrivial because even if we delete any mention of $\square$ from $(\dagger)$, it still implies $\mathsf{SCH}$.) $\endgroup$
    – Will Brian
    Nov 23, 2020 at 19:09
  • 3
    $\begingroup$ What about the following idea (in the absence of square at least): if $0^\sharp$ does not exist, let $X \subseteq \aleph_2$ be such that $V$ and $L[X]$ have the same $\aleph_1$ and $\aleph_2$. By the Jensen covering lemma and correctness of $\aleph_1$ and $\aleph_2$, every countable set of ordinals is covered by one in $L[X].$ $\endgroup$ Nov 24, 2020 at 9:07

1 Answer 1


The consistency strength of the failure of $(\dagger)$ is an inaccessible cardinal.

Building on the comment of Mohammad, if $\omega_2^V$ is a successor cardinal in $L$ then there is a set $X \subseteq \aleph_1^V$ such that $L[X]$ computes $\aleph_1, \aleph_2$ correctly, which (assuming $0^\#$ does not exist) is enough, since this model would satisfy $\mathrm{GCH}$.

On the other direction, in the paper "Inner Models from Extended Logics" by Kennedy, Magidor and Vaananen, Theorem 6.6 they show that starting with $V=L$ and an inaccessible cardinal $\kappa$, there is a (modification) of revised-countable-support iteration of a variant of the Namba forcing, such that $V[G] \models \kappa = \aleph_2 = 2^{\aleph_0}$ and $V[G] = (C^*)^{V[G]}$, where the model $C^*$ is $L[A]$, for $A$ the class of all ordinals of countable cofinality.

Since every model witnessing $(\dagger)$ would be able to compute the class $A$ and thus would contain $L[A]$, this model would witness the failure of $(\dagger)$.

  • 2
    $\begingroup$ Well, that's disappointing. $\endgroup$
    – Asaf Karagila
    Nov 24, 2020 at 21:02

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.