If $\psi$ is a predicate that is definable in $FOL(\in,=)$ by a formula from parameters in $V$, then if $\psi$ hold of the class $\small ORD$ of all ordinals in $V$, then the class of all cardinals in $V$ satisfying $\psi$ is non empty and is isomorphic on membership with $\small ORD$

Formally this is: $\psi(\small ORD) \to \forall x (x = \{y| \ y \text{ is a cardinal } \wedge \psi(y)\} \to x \neq \emptyset \wedge x \cong \small ORD)$

Now this axiom scheme is to be added on top of axioms of the following theory formulated in first order predicate logic with extra-logical primitives of equality, membership and a single primitive constant symbol $V$ denoting the class of all sets.

The axioms are those of first order identity theory +

  1. Extensionality: $\forall x (x \in a \leftrightarrow x \in b) \to a=b$
  2. Foundation over all classes: $\exists m \in x \to \exists y \in x (y \cap x = \emptyset)$
  3. Class comprehension axiom: if $\phi(y)$ is a formula in which the symbol $``y"$ occurs free, then all closures of: $\exists x \forall y (y \in x \leftrightarrow y \in V \wedge \phi(y))$ are axioms.

Define $\{|\}: x=\{y|\phi(y)\} \iff \forall y (y \in x \leftrightarrow y \in V \wedge \phi(y))$

  1. Transitivity: $x \in V \wedge y \in x \to y \in V$

  2. Supertransitivity: $x \in V \wedge y \subseteq x \to y \in V$

  3. Pairing: $a,b \in V \to \{a,b\} \in V$

  4. Set Union: $a \in V \to \{x| \exists y \in a (x \in y)\} \in V$

  5. Power: $a \in V \to \{x| x \subseteq a\} \in V$

  6. Limitation of size: $|x| < |V| \wedge x \subset V \to x \in V$

Where $``||"$ denotes cardinality function defined in the usual manner.

Now its clear that this theory goes beyond $ZFC$, since $\small ORD$ would provably be a regular cardinal and so the set of all regular cardinals in $ORD$ must be isomorphic on membership with $ORD$, and so we must have inaccessible cardinals in $ORD$, actually it is even simpler than that, simply take the property of being "inaccessible cardinal" which is definable in the pure language of set theory, clearly $ORD$ fulfills that, so there must be an inaccessible cardinal in $ORD$. However it is not clear to me how far this theory can go to?

Question: what is the consistency strength of this theory?

  • $\begingroup$ Note that over ZFC, "unbounded in $ORD$" is the same as "$\in$-isomorphic to $ORD$." So this simplifies things substantially. $\endgroup$ – Noah Schweber Dec 13 '18 at 0:05
  • $\begingroup$ I am confused by your intuitive explanation. If something holds for all ordinals, it therefore holds for all cardinals. So the class of cardinals satisfying that property is all the cardinals, and is therefore isomorphic to the ordinals. Where do we go beyond ZFC here? $\endgroup$ – Asaf Karagila Dec 13 '18 at 7:22
  • $\begingroup$ @AsafKaragila if something holds for $ORD$, i.e. holds for the CLASS of all ordinals in $V$, this doesn't mean that it holds for every ordinal in $V$, you are confusing the class for its members. For example take the predicate $\psi$ to be $\text{ "is regular"} $, clearly $ORD \text{ is regular}$, but that doesn't entail that every ordinal in $ORD$ is regular, nor does it mean that all cardinals in $ORD$ are regular. $\endgroup$ – Zuhair Al-Johar Dec 13 '18 at 12:11
  • 2
    $\begingroup$ I don't understand $\psi(ORD)$ in your notation, then. It's not a first-order statement. What is it, then? $\endgroup$ – Asaf Karagila Dec 13 '18 at 12:13
  • $\begingroup$ $ORD$ is an object of our universe of discourse, so $\psi(ORD)$ is a first order logic formula. $\endgroup$ – Zuhair Al-Johar Dec 13 '18 at 12:15

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