-1
$\begingroup$

[EDIT] This posting had been edited to assert that we are speaking about regular mutual stages.

Let $H_{\kappa}$ be the set of all sets that are hereditarily strictly smaller in cardinality than cardinality $\kappa$. Formally

$H_\kappa = \{x: |TC(x)|<\kappa\}$

Where $TC(x)$ is the transitive closure of $x$ defined as the intersectional set of all transitive supersets of $x$.

Now a stage of the cumulative hierarchy $V_{\kappa}$ is either identical to $H_{\kappa}$ or not. For example $V_\omega = H_\omega$, and of course $V_\emptyset = H_\emptyset$. Lets call such stages as "mutual" stages.

Define: $mutual(x) \iff \exists \kappa (x=V_\kappa =H_\kappa) $

Axiom: The class of all regular mutual stages is proper.

Formally: $\forall x \exists \kappa (x \in V_\kappa \land mutual(V_\kappa) \land regular(\kappa) )$

What is the strength of adding this axiom to ZFC?

$\endgroup$
6
$\begingroup$

The axiom in your question adds no strength; it's provable in ZFC. The claim in your title, involving inaccessibility, has the same strength as the existence of an inaccessible cardinal.

$\endgroup$
  • $\begingroup$ Ok, I've edited it, the mutuals are meant to be regular. $\endgroup$ – Zuhair Al-Johar Jun 13 at 14:36
  • 5
    $\begingroup$ @ZuhairAl-Johar If $\kappa$ is infinite and regular, then $V_\kappa=H_\kappa$ iff $\kappa$ is inaccessible; so this principle is just "There is a proper class of inaccessibles." $\endgroup$ – Noah Schweber Jun 13 at 15:06
  • $\begingroup$ you mean $\kappa$ is strongly inaccessible! (we are of course not including $\omega$). $\endgroup$ – Zuhair Al-Johar Jun 13 at 15:31
  • 1
    $\begingroup$ @ZuhairAl-Johar For at least the last few decades, "inaccessible" without further qualifications has meant strongly inaccessible. If one wants to talk about weak inaccessibility, then one must explicitly say so. (Of course, it makes no difference to the consistency strength, which is what you asked about originally.) $\endgroup$ – Andreas Blass Jun 13 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.