# Getting a model of $\mathsf{ZFC}$ that fails to nicely cover an inner model

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 $$\mathsf{SCH}$$.

• Do you want the inner model to be correct about cardinals? Nov 23, 2020 at 17:00
• @MonroeEskew: No, it doesn't need to be. Nov 23, 2020 at 18:10
• I mean, in general the failure of squares require large cardinals... Nov 23, 2020 at 18:39
• @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 $\mathsf{SCH}$.) Nov 23, 2020 at 19:09
• 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 $L[X].$ Nov 24, 2020 at 9:07

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 $$(\dagger)$$.