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. $\endgroup$ Dec 13, 2018 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, 2018 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. $\endgroup$ Dec 13, 2018 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, 2018 at 12:13
  • 1
    $\begingroup$ Wait a moment. I assumed you were using an $KM$ style theory. $V$ is just a constant symbol here. That makes sense. $\endgroup$
    – Master
    Jul 14, 2019 at 21:48

1 Answer 1


I claim that this is equiconsistent with "$ORD$ is Mahlo" (This is not the same as $Ord$ being actually Mahlo, as I have explained elsewhere). It turns out "$ORD$ is Mahlo" is a natural limit point for these kind of Ackermann/$KM$ based theories. Let $T$ be your theory.

First, the easy part. The consistency strength of "$ORD$ is Mahlo" $\ge$ the consistency strength of $T$. Let $C=\{\alpha|V_\alpha\prec W\}$, where $W=\{x|x=x\}$. Note that if $\alpha\in C$ and $\phi(\alpha)$, then $\exists\beta(\beta\gt\alpha\land\phi(\beta))$. The reason for this is that else $\psi(\alpha)\leftrightarrow\beta\text{ is the largest } \beta\text{ such that }\phi(\beta)$ would be a definition of $\alpha$, and so $V_\alpha\vDash\exists x(\psi(x))$, which is a contradiction. Now, if $\phi(\alpha)$ with $\alpha\in C$, then $\phi(\alpha)\land\alpha\text{ is a cardinal}$ and so $V_\alpha\vDash\forall\eta(\exists\beta\gt\eta(\phi(\beta\land\beta\text{ is a cardinal}))$. Therefore, whenever $\kappa$ is inaccessible reflecting $V_\kappa\vDash T$, and the existsence of an inaccessible reflecting cardinal is equiconsistent with "$ORD$ is Mahlo."

Second, the hard part. The consistency strength of $T\ge$ the consistency strength of "$ORD$ is Mahlo." Let $C=\{\alpha|\phi(\alpha,p)\}$, and let $C^V=\{\alpha|\phi^V(\alpha,p)\}\cap Ord$ be club in $Ord$. Then you can see $\{\alpha|\phi^V(\alpha,p)\}$ is club (In the real class of ordinals); in particular $\phi^V(Ord,p)\land Ord\text{ is regular}$. Then we can find a non-empty class of regular $\kappa\in C^V$, and so $\phi^V$ when $ZFC+ORD\text{ is Mahlo}\vdash\phi\rightarrow\phi^V$. Therefore, if $M\vDash T$ then $V^M\vDash ZFC+ORD\text{ is Mahlo}$


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