$\DeclareMathOperator\iCard{iCard}$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?

  • $\begingroup$ 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. $\endgroup$
    – LSpice
    Apr 28, 2022 at 18:35
  • 1
    $\begingroup$ Thank you. Yes, that's the intended reference. $\endgroup$ Apr 29, 2022 at 4:51

1 Answer 1


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 $\mathsf{CH}$."

  • $\begingroup$ 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 $\neg \theta$. $\endgroup$ Apr 29, 2022 at 7:55

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