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It's well known that Reinhardt cardinals are inconsistent, provided that we have access to axiom of choice, but, as far as I know, we are clueless about this when we don't assume choice. For me, the fact that Reihardt cardinals are inconsistent with choice already makes them, in a way, "wrong", that is, I'd believe personally that Reinhardt cardinals cannot exist in $V$ anyways. However, I have never seen any claims like this. So my question is:

Is it believed that Reinhardt cardinals can be consistent without choice? Are there any solid arguments for/against them? (other than "we haven't found them inconsistent, so they should be consistent")

Thanks in advance.

Note: I added soft-question tag because this question is more about philosophical views. Feel free to remove it if you think otherwise.

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    $\begingroup$ Woodin's current inner model theory project is said to provide evidence for their inconsistency with ZF, provided that certain parts of the project work out as expected. I'll let someone more knowledgable flesh this out as an answer. $\endgroup$
    – Monroe Eskew
    Commented Dec 7, 2014 at 11:10
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    $\begingroup$ I think the question is completely open. $\endgroup$
    – Joel David Hamkins
    Commented Dec 7, 2014 at 12:19
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    $\begingroup$ @JoelDavidHamkins I wouldn't be surprised if it were the case. My question is asking about personal opinions and views, and possibly some arguments to support them. $\endgroup$
    – Wojowu
    Commented Dec 7, 2014 at 12:56
  • $\begingroup$ This is just speculative, but even if some large cardinals are found to be inconsistent with ZF, there could be weaker set theories where appropriate versions are consistent. I'm thinking more precisely of Aczel's constructive set theory CZF, based on intuitionistic logic, over which appropriate notions of regular, inaccessible and even Mahlo cardinals are defined (see "Inaccessibility in constructive set theories and type theories" by Rathjen-Griffor-Palmgren). In this case, the addition of AC turns the theory in just ZFC plus the usual notions...(cont.) $\endgroup$
    – godelian
    Commented Dec 7, 2014 at 18:03
  • $\begingroup$ ...But it is conceivable that a large cardinal hierarchy could be mirrored in an intuitionistic setting where more notions of large cardinals can be defined without inconsistency (they will have to be non-classical though). $\endgroup$
    – godelian
    Commented Dec 7, 2014 at 18:04

1 Answer 1

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The following theorems of Woodin may be related:

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

Thus (assuming that $ZFC$ proves the $HOD$ Conjecture) one nearly has a proof of Kunen's inconsistency theorem without using the Axiom of Choice.

Theorem. Assume $ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals is consistent. Then there exists a genric extension of the universe which satisfies $ZF$ + the axiom of choice + there exists a proper class of supercompact cardinals, and such that in it Woodin's $HOD$ conjecture fails.

As far as I know, Woodin believes the $HOD$ conjecture is true (at least the current methods can not be used to solve the problem), so by the above theorem we may expect to show that at least "$ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals" is not consistent.

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  • $\begingroup$ Does the $HOD$ conjecture fail in the Gitik model? $\endgroup$
    – Thomas Benjamin
    Commented Dec 24, 2014 at 18:46
  • $\begingroup$ @ThomasBenjamin In Gitik's model, there are no extendible cardinals, and in fact Gitik's forcing kills them all, even if we assume their existence, so we can not talk of HOD conjecture in it. $\endgroup$
    – Mohammad Golshani
    Commented Jan 2, 2015 at 10:46
  • $\begingroup$ Thanks. Does this mean that in the Gitik model, there might a way to force the existence of Reinhardt cardinals, or does the fact that there are no extendible cardinals in that model imply that Reinhardt cardinals cannot exist there either? $\endgroup$
    – Thomas Benjamin
    Commented Jan 2, 2015 at 13:42
  • $\begingroup$ I don't see it, the model is constructed by Prikry type forcing construction, and a basic fact is that such forcings usually kill very large cardinals below them, say strongly compact or supercompact cardinals. Also we can't extend Gitik's model to a ZFC model by forcing without collapsing all cardinals. $\endgroup$
    – Mohammad Golshani
    Commented Jan 4, 2015 at 12:22
  • $\begingroup$ Are there then models of $ZF-Foundation +BAFA$, for example, in which we might reasonably expect to find Reinhardt cardinals? Also regarding the Gitik model, what would be the purpose of collapsing all cardinals to form a new $ZFC$ model? Would that become a new ground model? $\endgroup$
    – Thomas Benjamin
    Commented Jan 4, 2015 at 14:42

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