Is this theory equivalent to TG set theory

large-cardinalsset-theory

Is the following theory equivalent to Tarski-Grothendieck set theory?

The language is first order logic with equality, and membership, with axiom schemata:

  1. Specification: $\forall A \exists! x \, \forall y \, (y \in x \iff y \in A \land \phi )$

  2. Reflection: $(\varphi \implies \exists \alpha: \operatorname {superinclusive}(V_\alpha) \land \varphi^{V_\alpha} )$

$\operatorname {superinclusive}(X) \equiv_{df} \forall Y \subseteq X \, (|Y|<|X| \implies Y \in X)$

$\varphi$ is a first order formula in the language of set theory + defined predicates and functions, that doesn't use the symbol $\alpha$. And $\varphi^X$ is the formula obtained by merely bounding all quantifiers in $\varphi$ by $\in X$. The formula $\phi$ (of specification) is any formula that doesn't have the symbol $x$ occurring free.

Best Answer

This theory proves ORD is $\Sigma_2$-Mahlo, that is, that every $\Sigma_2$-definable closed unbounded class contains an inaccessible cardinal. If $\phi$ is a $\Sigma_2$ formula that defines a closed unbounded class, we can apply the reflection schema to get an inaccessible cardinal $\alpha$ such that $V_\alpha \vDash \text{the class defined by $\phi$ is closed and unbounded}$. Since $\phi$ is $\Sigma_2$ and $\alpha$ is inaccessible and thus a beth fixed point, $\phi$ is upward absolute from $V_\alpha$, so $\alpha$ is a limit of the class defined by $\phi$ and thus, since that class is closed, $\alpha$ is in that class.

For example, this proves that there are unboundedly many inaccessible limits of inaccessible cardinals, so this theory is stronger than Tarski-Grothendieck set theory. I don't know if this theory proves the full ORD is Mahlo schema.