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Compare these two results:

Theorem (Scott): $ZFC+V=L\vdash \nexists~\text{Measurable cardinal}$

Theorem (Kunen): $ZFC+AC\vdash \nexists~\text{Reinhardt cardinal}$

Now compare these two chains:

$V=L\rightarrow AC\rightarrow \cdots$

$Con(ZF+ \exists~\text{Measurable cardinal})\leftarrow Con(ZF+\exists~\text{Reinhardt cardinal})\leftarrow \cdots$

By an imaginary correspondence it seems we may assume:

Large cardinals are contradictory with choice-like axioms. i.e. Larger large cardinals are contradictory with weaker versions of choice. (See the imaginary diagram below).

In the other words, based on the relation $V=L\rightarrow GCH\rightarrow AC\rightarrow AC_{\omega}\rightarrow \cdots$ we may expect the followings:

(a) There is a large cardinal axiom above the Reinhardt cardinals (e.g. Berkeley cardinals) which is provably inconsistent with the axiom of countable choice, $AC_{\omega}$.

(b) There is a large cardinal axiom between measurable and Reinhardt cardinals which is inconsistent with $GCH$.

In some sense, the statement (b) is true if we assume Shelah's Proper Forcing Axiom (PFA) as a large cardinal axiom, however if we remove PFA from our large cardinal hierarchy, it is not known yet.

Let's focus on the statement (a) which is a interpretation-free statement. We may ask two different questions here:

Question 1: What is the weakest version of the axiom of choice which is necessary to prove inconsistency of Berkeley cardinals?

Question 2: Within $ZF$, what is the weakest large cardinal axiom (above Reinhardt) which is provably inconsistent with $AC_{\omega}$?

Remark: For more on equivalent versions and consequences of Berkeley cardinals see here.

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    $\begingroup$ What would be interesting would be to see if the existence of Reinhardt or Berkeley cardinals could be forced to hold in the Gitic model, where choice fails miserably. Or in any 'choiceless' model of ZF, for that matter.... $\endgroup$ – Thomas Benjamin Dec 23 '14 at 11:55
  • $\begingroup$ In your statement of Kunen's result, should "ZFC + AC" be "ZF + AC" (or simply ZFC)? $\endgroup$ – John Bentin Nov 4 '17 at 11:59

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