# What's the consistency status/strength of this limitation principle?

$$\DeclareMathOperator\iCard{iCard}$$In a prior posting If we limit matters what ZFC can prove, would that be consistent? to MO, I tried to capture the informal principle of whatever ZFC proves, it is, which I've done erroneously. Here, I present a version along the same general lines which seems informally like saying whatever ZFC proves, then it's consistent to limit matters as such.

Formal capture:

Define: $$\text{ ZFC cannot decide on fulfillment of }\phi \text { by infinite cardinals } \\ \iff \\ \bigl{[} \exists M: M \text { is CTM(ZFC}+ \forall \text { infinite cardinal } \kappa : \phi ) \land \\\forall \kappa \in \iCard^M \exists N: N \text { is CTM(ZFC) } \land \iCard^N=\iCard^M \land N \models \neg \phi(\kappa) \bigr{]}$$

Where: $$\iCard^S=\{\kappa : S \models \exists \text { infinite } x \, (\kappa=\lvert x\rvert)\}$$; $$\text { CTM}$$ stands for "Countable Transitive Model".

In English: We say that ZFC cannot decide on fulfillment of a formula $$\phi$$ (in one free variable) by the infinite cardinals, if and only if, there exists a countable transitive model $$M$$ of "ZFC + all infinite cardinals satisfy $$\phi$$", and such that for every infinite $$M$$-cardinal $$\kappa$$ there is a countable transitive model $$N$$ of ZFC whose infinite cardinals are exactly the same infinite cardinals of $$M$$, and which satisfies negation of $$\phi$$ by $$\kappa$$.

Axiom schema of limitation: if $$\phi$$ is a formula in which only symbol "$$\kappa$$" occurs free, then: $$\text{ ZFC cannot decide on fulfillment of }\phi \text { by infinite cardinals } \\\implies \\ \exists \mathcal M : \mathcal M \models (\text{ZFC} + \forall \text{ infinite cardinal } \kappa: \neg \phi(\kappa)).$$

I do realize that there are deep concerns with such attempts (see comment), yet I need to elaborate that such an axiom is very strong. It takes a relatively weak system and pump matters to very strong altitudes. As an example if we let $$\phi$$ in the above axiom to be $$\operatorname {CH}$$ which stands for fulfilment of the continuum hypothesis (i.e. $$\operatorname {CH}(\kappa) \equiv_\text{def} \neg \exists \zeta: \kappa < \zeta < 2^\kappa$$), then it's known that ZFC cannot decide on fulfillment of the continuum hypothesis by infinite cardinals (Cohen), which is here of the strength of $$\text{ZFC} +\exists M: M \text { is CTM(ZFC)}$$, yet this principle would blow matters up to the level of consistency of Global failure of the continuum hypothesis.

Is there a clear inconsistency with this principle?

If not, then is there a proof of its consistency, and at which level?

• Your link to "a prior posting" pointed to the comment to which you explicitly link later, but it seemed more likely to be intended to point to the answer itself. I edited accordingly. Apr 28, 2022 at 18:35
• Thank you. Yes, that's the intended reference. Apr 29, 2022 at 4:51

This principle is inconsistent: consider the formula $$\theta(x)$$ = "$$x^+$$ is the smallest infinite cardinal at which $$\mathsf{CH}$$ fails." The formula $$\theta$$ cannot hold on more than one infinite cardinal, let alone on all infinite cardinals, yet your principle (applied to $$\phi:=\neg\theta$$) would require this.
(I originally omitted the "$$+$$"-superscript; that's actually a mistake, since there are constraints on when $$\mathsf{CH}$$ can fail first. Looking at a successor simplifies things.)
Note that this argument is rather flexible: for example, it also shows that we can't "go from internally countable failures to global failures" by considering $$\psi(x)=$$ "$$x^+$$ is one of the first $$\omega_1$$-many failures of $$\mathsf{CH}$$."
• We can even have a counter-example with $\neg \phi$ being a proper class! Take $\theta$ to be $\operatorname {CH}(\kappa) \land \neg \operatorname {CH}(\kappa^+)$, so our $\phi$ is $\neg \theta$. Apr 29, 2022 at 7:55