# Bernays' Reflection Principle Holding in Ranks?

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?

• 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. Oct 21, 2017 at 22:46
• Oh, yes. Sorry. I should have said simply Predicate symbol, in fact. Oct 21, 2017 at 22:48
• 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. Oct 21, 2017 at 22:55
• True; I will get rid of the first definition then. Oct 21, 2017 at 22:59
• 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. Oct 21, 2017 at 23:10

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.

• 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.) Jan 16, 2018 at 20:25