I wanted to see for which ranks Bernays' Reflection Principle holds; that is, for every class and every property (allowing quantification over all classes) which is true about that class, there is a transitive set $u$ for which $\mathcal{P}(u)$ satisfies that property about that class's intersection with $u$. Like Vopenka did with Vopenka cardinals, I allow "arbitrary $A\subseteq V_\kappa$" to be classes in $V_\kappa$.

It turns out that once analyzed, these ranks bear a striking similarity to those ranks of Indescribable cardinals. I formalized this definition in ZFC as follows:

Let $\kappa$ be $\Pi_m$-Bernays when for every first-order $\Pi_m$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is an unary predicate symbol:

$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\langle \mathcal{P}(u);\in,A\cap u\rangle\models\phi))$$

A similar definition in ZFC would be that for every first-order $\Pi_m$ unary formula $\phi$ in the language $\mathcal{L}_\in$:

$$\forall A\subseteq V_\kappa(\phi^{V_{\kappa+1}}(A)\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\phi^{\mathcal{P}(u)}(A\cap u)))$$

Of course, the existence of a $\Pi_{<\omega}$-Bernays cardinal is equivalent to the existence of a cardinal rank which satisfies Bernays' Reflection Principle. Thus, $\Pi_{<\omega}$-Bernays cardinals, in consistency strength, are somewhere above Bernays' Reflection Principle, which oddly enough implies the existence of an Inaccessible.

These cardinals are therefore consistency-wise stronger than that of Inaccessible cardinals.

Every cardinal is $\Pi_0$-Bernays. This is pretty simple once one considers that all of the unary first-order formulas in the language $\mathcal{L}_\in$ are either true for every set or true for no set.

This brings me to my question:

Are these cardinals inconsistent with ZFC? Are they any weaker or stronger than Indescribability?

  • $\begingroup$ You write "$c$ is a constant symbol". Do you really mean "$c$ is a relation symbol"? Because you want to interpret $c$ by $A$, which is a relation on the universe $V_\kappa$, not an element. $\endgroup$
    – Goldstern
    Oct 21, 2017 at 22:46
  • $\begingroup$ Oh, yes. Sorry. I should have said simply Predicate symbol, in fact. $\endgroup$ Oct 21, 2017 at 22:48
  • $\begingroup$ But if you let $\varphi$ be the formula $\forall x\ P(x)$, then (using $A:=V_\kappa$) $(V_\kappa,\in,V_\kappa)$ satisfies $\varphi$, but $(\mathcal P(u),\in,u)$ does not. $\endgroup$
    – Goldstern
    Oct 21, 2017 at 22:55
  • $\begingroup$ True; I will get rid of the first definition then. $\endgroup$ Oct 21, 2017 at 22:59
  • $\begingroup$ Oh, I found the reason my original definition was wrong. I should have defined it using $\langle V_{\kappa+1};\in,A\rangle$, as to accommodate for classes. $\endgroup$ Oct 21, 2017 at 23:10

2 Answers 2


I would like to add that, for $n>1$, the $\Pi_n-$Bernays cardinals are precisely the $\Pi_n^1-$indescribable cardinals. We can do this really simply by giving a $\Pi_2$ definition of the Axiom of limitation of size:

$$ALZ\leftrightarrow\forall x(\exists W\ni x\leftrightarrow\exists F,y(F\text{ is a bijection } F:x\rightarrow y\land\exists U\ni y))$$

Then $2^u\vDash ALZ$ precisely when $2^u$ satisfies limitation of size. Let $AU$ be the axiom of union (Which is also $\Pi_2$). Then whenever $u$ is infinite, transitive, and $2^u\vDash ALZ\land AU$, then $u$ is a Grothendrick universe and therefore of the form $V_\lambda$ for some inaccessible $\lambda$.

Then, for any $\Pi_2^1$ formula $\phi$ in the language of $\{\in,S\}$ such that $(V_{\kappa+1},\in,S)\vDash\phi$, to get some $(V_{\lambda+1},\in,S\cap V_\lambda)\vDash\phi$, let $\psi$ be the formula $ALZ\land AU\land\phi$, and take some transitive $u$ such that $(2^u,\in,S\cap u)\vDash\psi$. It is easy to see that $\kappa$ is inaccessible.


Every $\Pi_n^1$-Indescribable cardinal is $\Pi_n$-Bernays, and thus the $\Pi_{n+1}^1$-Indescribable cardinals are strictly of greater consistency strength than that of the $\Pi_n$-Bernay cardinals, and also the $1$-Indescribable cardinals are of strictly greater consistency strength than that of Bernay's principle itself.

Here's the proof:

A little Lemma before the proof is to show that if $\exists\alpha<\kappa(\langle V_{\alpha+1};\in,A\cap V_\alpha\rangle\models\phi)$ then $\exists u\in V_\kappa(\langle\mathcal{P}(u);\in,A\cap u\rangle\models\phi)$. Specifically, this becomes quite obvious when one realizes that $V_\alpha$ is such a $u$ described when the first condition holds for $\alpha$.

A cardinal $\kappa$ is $\Pi_n^1$-Indescribable iff for every first-order $\Pi_n$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is an unary predicate symbol:

$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_{\alpha+1};\in,A\cap V_\alpha\rangle\models\phi))$$

By the previous lemma, this implies that for every first-order $\Pi_n$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is a unary predicate symbol:

$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists u\in V_\kappa(\langle\mathcal{P}(u);\in,A\cap u\rangle\models\phi))$$

Of course, this is the first definition I gave for a $\Pi_n$-Bernays cardinal. Thus, every $\Pi_n^1$-Indescribable cardinal is $\Pi_n$-Bernays.

  • $\begingroup$ It seems to me that $\Pi^1_n$-Bernays and $\Pi^1_n$-indescribable ought to be equivalent once $n$ is big enough (maybe 3 or so). The idea is that one could build into the formula $\phi$ enough requirements to ensure that any $u$ as in the definition of $\Pi^1_n$-Bernays is necessarily of the form $V_\alpha$. (There might even be something like this in Bernays's paper.) $\endgroup$ Jan 16, 2018 at 20:25

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