4
$\begingroup$

Kunen showed that if $j:V \rightarrow M$ is a nontrivial elementary embedding from the von Neumann universe $V$ into a transitive class $M$, then $M \neq V$, or equivalently that there are no Reinhardt cardinals, assuming the axiom of choice.

Hamkins, Kirmayer, and Perlmutter (2012) cover a wide breadth of generalizations of Kunen's inconsistency result and cover the metamathematical results surrounding its proof ("a fuller power" of the theorem being revealed when understood in MK or NGBC set theory due to those theories' ability to talk about classes in addition to sets), including the proof's reliance on the axiom of choice. At the end it is stated that whether or not Kunen's full result and thus the nonexistence of a Reinhardt cardinal in ZF alone is an open problem:

Perhaps the principal open question is whether one can prove the Kunen inconsistency without using the axiom of choice.

My question is: What is the current progress on proving or refuting the existence of a Reinhardt cardinal in ZF? Are there any known conjectures that once resolved would imply the answer? Or, in lack of concrete results, what is the most current literature on this topic or any current literature that gives an overview as to the technical way that this problem should be approached?

$\endgroup$

1 Answer 1

4
$\begingroup$

The problem seems to be open.

I would like to mention the following result of Woodin:

Theorem (ZF). Assume that ZFC proves the HOD Conjecture. Suppose $δ$ is an extendible cardinal. Then for all $λ > δ$ there is no non-trivial elementary embedding $j : V_{λ+2} → V_{λ+2}.$

From this, Woodin concludes the following: Thus (assuming that ZFC proves the HOD Conjecture) one nearly has a proof of Kunen’s Theorem without using the Axiom of Choice.

See Theorem 30 and the remarks after it of the following: The HOD dichotomy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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