Let $R_\alpha$ be the $\alpha$-th infinite regular ordinal. This question assumes AC, so the following is true:

$$R_\alpha=\left\{ \begin{array}{ll} \omega_\alpha & \alpha=\kappa+n\;\mathrm{where}\;n<\omega\land\kappa\;\mathrm{is}\;\mathrm{wk. inaccessible}\;\mathrm{or}\;0 \\ \omega_{\alpha+1} & \mathrm{otherwise} \end{array} \right.$$

After this point, one could define an $\alpha$-Supercorrect cardinal $\kappa$ as follows:


Here are some remarks:

  • The $0$-Supercorrect cardinals are precisely the Correct cardinals (which are equiconsistent with ZFC).
  • In the rank of any $\alpha$-Supercorrect cardinal, there are $\beta$-Supercorrect cardinals for every $\beta<\alpha$. (Thus, the $\alpha$-Supercorrect cardinals are stronger in consistency strength the greater the $\alpha$)
  • Every $\alpha$-Supercorrect cardinal is $\beta$-Supercorrect for every $\beta<\alpha$
  • Every $1$-Supercorrect $\kappa$ is Worldly, Correct, and $V_\kappa$ satisfies the existence of a Correct cardinal.
  • In the rank of a $\kappa$ which is $\kappa$-Supercorrect, there is a proper class of $\beta$-Supercorrects for every $\beta$. Of course, this means that the consistency strength of a $\kappa$ which is $\kappa$-Supercorrect is stronger than that of "For every $\beta$ there is a proper class of $\beta$-Supercorrect cardinals".
  • For every first-order theory $T$, a $1$-Supercorrect cardinal has $V_\kappa\models T\Leftrightarrow V\models T$.

Are any of these inconsistent? Are all other than $0$-Supercorrect inconsistent? What else would be true if consistent?


Interesting result found: it is inconsistent for there to be a cardinal $\kappa$ which is $\alpha$-Supercorrect for every $\alpha$. If there were, it turns out that there would be another such cardinal in $V_\kappa$.

$L_{\infty,\infty}$ can express all of its own schemas as singular formulae. Therefore, one could let the schema $V_\kappa\prec_{L_{\infty,\infty}}V$ be the union of "$V_\kappa\models\phi\Leftrightarrow\phi$" for every $L_{\infty,\infty}$ $\phi$. Since this is expressable as a singular formula in $L_{\infty,\infty}$, we could call this singular formula $\psi(\kappa)$. If there is such a $\kappa$, $V\models\exists\kappa(\psi(\kappa))$. Therefore, $V_\kappa\models\exists\kappa(\psi(\kappa))$.


1 Answer 1


$\alpha$-supercorrect cardinals are consistent, for any $\alpha$.

Work in a mild second-order set theory, namely Gödel–Bernays set theory along with the assertion that there is a $\mathcal L_{R_\alpha,R_\alpha}$-truth predicate. (This is defined similarly to the usual Tarskian definition of a truth predicate, with the obvious modifications for larger conjunctions/disjunctions/quantifier blocks.) Call this truth predicate $T$. We can then reflect using $T$ to get unboundedly many cardinals $\gamma$ so that $(V_\gamma, T \cap V_\gamma) \prec_{\Sigma_k} (V,T)$, where $k$ is some natural number. To be clear, I mean $\Sigma_k$-elementarity in the usual first-order logic. But we can express "$T$ is an $\mathcal L_{R_\alpha,R_\alpha}$-truth predicate" as a single first-order formula. So by picking $k$ large enough we get that $T \cap V_\gamma$ is an $\mathcal L_{R_\alpha,R_\alpha}$-truth predicate for $V_\gamma$. But since $\varphi(\bar a) \in T \cap V_\gamma$ iff $\varphi(\bar a) \in T$, we get that $V_\gamma \prec_{\mathcal L_{R_\alpha,R_\alpha}} V$. So $\gamma$ is $\alpha$-supercorrect.

So what does it take to get $\mathcal L_{R_\alpha,R_\alpha}$-truth predicates? Clearly, we get $\mathcal L_{\kappa,\kappa}$-truth predicates for all $\kappa$ if we have an $\mathcal L_{\mathrm{Ord},\mathrm{Ord}}$-truth predicate, i.e. a truth predicate for the language where we allow arbitrary set-sized conjunctions/disjunctions/quantifier blocks. In a recent paper, Victoria Gitman, Joel David Hamkins, Peter Holy, Philipp Schlicht, and myself showed that, over Gödel–Bernays set theory, the existence of an $\mathcal L_{\mathrm{Ord},\mathrm{Ord}}$-truth predicate follows from what we call $\mathsf{ETR}_\mathrm{Ord}$. This is the principle of Elemetary Transfinite Recursion for recursions of length $\le \mathrm{Ord}$, i.e. that any recursion of a first-order property along $\mathrm{Ord}$ has a solution. Indeed, we show that $\mathsf{ETR}_\mathrm{Ord}$ is equivalent over Gödel–Bernays set theory to the existence of $\mathcal L_{\mathrm{Ord},\mathrm{Ord}}$-truth predicates in the language $\mathcal L_\in(A)$ for any class parameter $A$.

V. Gitman, J. D. Hamkins, P. Holy, P. Schlicht, and K. Williams, “The exact strength of the class forcing theorem,” ArXiv e-print, 2017. (manuscript under review)

$\mathsf{ETR}_\mathrm{Ord}$ has mild consistency strength, below $\mathsf{ZFC}$ plus one inaccessible. So $\alpha$-supercorrect cardinals sit below an inaccessible in consistency strength.

  • $\begingroup$ Interestingly, a $\kappa$ which is $\kappa$-supercorrect (hypercorrect) seems to evade this proof. I wonder if perhaps they are stronger consistency-wise than ZFC. $\endgroup$ Commented Sep 1, 2018 at 2:25

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