# Re: Paper and slides on indefiniteness of CH

On Oct 18, 2014, at 10:19 PM, Sy David Friedman wrote:

Just so the mathematics does not drift: I need a proper class of supercompact cardinals. So Super-Reinhardt is OK as is Reinhardt + There is an Extendible.

Do you mean an extendible above the Reinhardt? Aren’t Reinhardt cardinals extendible?

Thanks,
Sy

No just an extendible. For the definition of an extendible cardinal in ZF, one has to be careful. The two possible (equivalent) formulations in ZFC are not equivalent in ZF (unless there are no weak Reinhardt cardinals).

When I say $\kappa$ is an extendible cardinal, I mean:

For all $\alpha > \kappa$, there exists an elementary embedding $j:V_{\alpha} \to V_{j(\alpha)}$ such that $crt(j) = \kappa$ and $j(\kappa) > \alpha$

In ZFC one can drop the clause $j(\kappa) > \alpha$ without altering the notion.

In ZF one can also drop the clause unless $\kappa$ is a weak Reinhardt cardinal where:

Def: $\kappa$ is a weak Reinhardt cardinal if for unboundedly many $\alpha$, there exists $j: V_{\alpha} \to V_{\alpha}$ with $\text{crt}(j) = \kappa$.

Hugh