This is, alas, in large part a series of questions on unpublished work of Hugh Woodin; it's also quite frivolous if Reinhardt cardinals turn out inconsistent.


Call $\kappa$ an $I-1(\kappa,\delta)$ cardinal if it is the least ordinal lifted by a nontrivial elementary embedding $j:V_{\delta+2} \to V_{\delta+2}$ where $\delta$ is an uncountable strong limit cardinal of countable cofinality and the supremum of the critical sequence $\kappa_n$.

Call $\kappa$ a weak Reinhardt cardinal if it is $I-1(\kappa,\delta)$, there is an ordinal $\gamma \gt \delta$, and elementary substructures $V_{\kappa} \prec V_{\delta} \prec V_{\gamma}$.

Call $\kappa$ a Reinhardt cardinal if it is the critical point of $j:V \to V$.

Call $ZFR$ the base theory $ZF$ in the language $(\in,j)$ augmented with a scheme asserting $\{\exists\kappa\mid \kappa\text{ is Reinhardt}\}$

Question 1: What does $ZFR$ imply?

Woodin folklore mentions a proof that $ZFR$ implies the existence of models of all large cardinal axioms not known to refute Choice. What is the proof $ZFR \implies$ "There exists a transitive model of $ZFC + I1(\kappa,\delta)$", and why does it not immediately extend to a proof of "There exists a transitive model of $ZF + DC_{\delta} + I-1(\kappa,\delta)$"?

Reinhardt cardinals (trivially) satisfy any large cardinal property derived from $j:V \to M$ and so are themselves at least $I2$ in some sense, but I don't know how to proceed further, and Woodin frequently mentions that there is no known proof of the expected implication Reinhardt $\implies$ weak Reinhardt. And while David Asperò has recently published in A short note on very large large cardinals a nice proof that Reinhardt cardinals imply many $j:V_{\mu} \to V_{\mu}$ with the target of the embedding moved arbitrarily high, these $\mu$ are poorly constrained and $V_{\mu}$ in general satisfies neither Replacement nor Choice. They may, therefore, not even be rank-into-rank in the usual sense.

Question 2: What's "next" after $ZFR$?

The rare tome On Woodin's Investigations into "Here there be Dragons" contains a proof that $j:HOD \to HOD$ follows from a strong extension of $ZFR$. What is this extension?

The strongest such axiom I recall encountering in the literature is a scheme asserting that there is an elementary substructure $V_{\kappa} \prec V$. However perilously strong, this axiom is not super $n$-helpful for determining what sort of cardinal structure the Reflecting Reinhardt cardinal is duplicating. Thus I'm hoping the answer might be something else, more useful for establishing answers to Question 1.

Question 3: What is the point?

Do these very large large cardinals hint at whether the Axiom of Choice constrains the consistency strength of $ZFC$, or not?

Here we arrive at the soft and murky motivation behind these questions. In light of Kunen's Inconsistency Theorem, it's possible to ask if the Axiom of Choice is close kin to the Axiom of Constructibility, something that places a sharp bound on the amount of consistency strength $ZF$ can support. Under this interpretation, if $I-1(\kappa,\delta)$ is consistent with $ZF$ then there is no equiconsistent large cardinal axiom consistent with $ZFC$. And although there are various shenanigans we can perform to sneak the strength back in (e.g. work in $ZFC +$ Con($ZFR$)), these don't tell us much about what $V$ is like. Further, they raise the question of why we oughtn't just work in $ZFC + V=L$ and get our measurables from some $j:M \to N$, since the Axiom of Constructibility is doing the same job as Choice in this context, only better.

Contrariwise, it's also possible to ask if it's the definability restriction of $V=L$ that renders it so inhospitable to large cardinals, and to assume that if a prospective axiom is too strong for the notoriously non-constructive Axiom of Choice, then it was outright inconsistent to begin with and Choice has merely facilitated the proof. Woodin's $HOD$ Conjecture is work in this vein (it refutes $I-1(\kappa,\delta)$), and even if the $HOD$ Conjecture is false, $I-1(\kappa,\delta)$ might be inconsistent for some currently unknown reason.

The sort of relevant evidence I'm hunting for is a large cardinal property or forcing axiom that:
1. is very strong in ZFC
2. is weaker but not trivial in ZF
3. indicates that some property a Reinhardt cardinal should have has leaked away through the absent wellordering of some $P(\aleph_{\alpha})$

The first two criteria are satisfied by a scant few things I can think of, including for example some generalizations of Chang's Conjecture beyond $(\omega_{3},\omega_{2}) \to (\omega_{2},\omega_{1})$; these generalizations, however, either have a cardinal structure that doesn't illuminate the third requirement or they have a cardinal structure that I find baffling and unwieldy. (And yes, that third requirement is so vague that the question may not admit an answer even if an example exists. If I knew how to make it precisely state some such property I would, but in general we don't know what those properties are!)

First post to this site; no snub intended if I don't respond in comments =)


Hello Joel, and thank you for the warm welcome. For $ZFR$ I wanted
1. All the Axioms of $ZF$ in the language $\{\in\}$
2. Replacement for formulae in the language $\{j,\in\}$
3. An Axiom scheme asserting $j:\langle V,\in \rangle \to \langle V,\in \rangle$
...and not that there are many critical points. This is, however, probably a mistake (or at least overly specific) and I'll gladly substitute any formalization of "$ZF$ + There is a Reinhardt cardinal" that doesn't force the embedding, by elementarity, to be the identity, even if that formalization is implicit and be understood in a fully two-sorted metatheory. Particularly if that's what Woodin had in mind for his proofs.

And yes, the first two requirements should easy enough to formalize (though I also would like $ZF + \psi \vDash$ Con($ZF$) $\wedge$ $ZFC + \varphi \vDash$ Con($ZF + \psi$) at the very least). It's the third requirement that I don't know how to formalize without an applicable, rigorous notion of "large cardinal property". Woodin's won't suffice in full generality when I don't know how to express a given property as some $j:V \to M$ with appropriate constraints on $M$. And in the case of Reinhardt cardinals, I'm not even sure which properties to expect.

Corazza's Wholeness Axiom is almost exactly what I'm looking for, thank you for the reminder. The failure of Replacement for $j$-classes acts as a pressure valve, disallowing the critical sequence from forming a set, and thereby preventing the absurdity of cof$(Ord) = \omega$ and reducing the consistency strength of Reinhardt cardinals to something weaker than $I3(\kappa,\delta)$ and stronger than a cardinal super $n$-huge for all $n \in \omega$, which is quite manageable. Unfortunately, it rules out the same hugeness that Kunen's theorem does, and doesn't supply new information about how a fully Reinhardt cardinal must appear.

I think? It would be nice to be wrong about that! And you've studied it more than most.

  • 1
    $\begingroup$ Welcome to MO! Could you clarify your formalization of ZFR? You have $j$ in the language---do you express "$j$ is elementary" just with $\Sigma_1$-elementary (and then appeal to meta-theoretic induction), or do you express it as a scheme in the manner of the Wholeness axiom? Also, I don't understand what you mean by "a scheme asserting $\{\exists\kappa\mid \kappa\text{ is Reinhardt}\}$." I would think you just want to assert that this particular $j$ has a critical point, not that there might be a Reinhardt cardinal for some other embedding (and that is not first-order expressible anyway). $\endgroup$ Feb 1, 2012 at 13:53
  • $\begingroup$ Isn't it easy to satisfy your properties (1) and (2) at the end in a purely formal way? Let $\varphi$ and $\psi$ be any two assertions, and consider the "axiom" asserting: $(\text{AC}\to\varphi)\wedge(\neg\text{AC}\to\psi)$, which in ZFC asserts $\varphi$, but in ZF+$\neg$AC asserts $\psi$. $\endgroup$ Feb 1, 2012 at 14:17
  • $\begingroup$ Hi Ekki, Would it be possible for you to email me a copy of "On Woodin's ..."? Thanks! $\endgroup$ Apr 8, 2012 at 3:48
  • $\begingroup$ Hi Andres! I've cc'd the document to your University of Tlon.edu address; it should arrive by Tuesday. Definitely no later than Tuesday. $\endgroup$
    – Ekki
    Apr 12, 2012 at 1:39
  • $\begingroup$ Would it be inappropriate to request a copy of "On Woodin's..." from either Ekki or Andres? $\endgroup$ Aug 9, 2012 at 1:48

5 Answers 5


I prove several new consequences of Reinhardt cardinals in my paper "Measurable cardinals and choiceless axioms," some of which are not known to be consistent relative to any weaker theory. TLDR:

Theorem (ZFR). There is a proper class of regular cardinals.

Proposition (ZFR). For all cardinals $\nu$, for all sufficiently large regular cardinals $\delta$, the club filter on $\delta$ is $\nu$-complete.

If $F$ is a filter, a set $S$ is an atom of $F$ if $\{A\cap S : A\in F\}$ is an ultrafilter on $S$. A filter $F$ is atomic if every $F$-positive set includes an atom of $F$.

Theorem (ZFR). The club filter on any sufficiently large regular cardinal is atomic.

Corollary (ZFR). For a club class of cardinals $\lambda$, either $\lambda$ or $\lambda^+$ is measurable.

In particular, there is a proper class of measurable successors of singular cardinals. Actually, much more is true:

Theorem (ZFR). For a club class of cardinals $\lambda$, every $\lambda$-complete filter on an ordinal extends to a $\lambda$-complete ultrafilter.

  • 4
    $\begingroup$ For a moment I wondered what set theory "TLD" is. $\endgroup$ Aug 1, 2021 at 6:27

The following is due to Woodin:

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.

  • $\begingroup$ Can you give a reference for this theorem (or the ideas of the proof)? $\endgroup$
    – Yair Hayut
    Jul 14, 2014 at 8:11
  • 1
    $\begingroup$ The proof follows from the results of Woodin in his paper "Suitable extender models I" $\endgroup$ Jul 14, 2014 at 10:27

First, let's show that I-1 implies I0:

Suppose $I-1(\kappa,\delta, j)$; we may assume, by forcing if necessary, that $V_\kappa$ (and hence $V_\delta$) satisfies the Axiom of Choice. Now if j is a rank-into-rank embedding $V_\delta \prec V_\delta$, it has a unique extension to a $j' : V_{\delta+1} \prec_{\Delta^1_0} V_{\delta+1}$ (ie it only preserves properties with quantifiers in $V_\delta$) given by $j'(S) = \cup_i j(S \cap \kappa_i)$; if $I1(\kappa,\delta,j)$ then the embedding will be fully ("second-order") elementary.

So letting $j'$ be the embedding $j|V_{\delta+1}$, we must have that $j' \in HOD(V_{\delta+1})$, and it can there be extended to some $k : HOD(V_{\delta+1}) \to M \subseteq HOD(V_{\delta+1})$, for some transitive class $M$. By the Condensation Lemma, $k$ restricts to an embedding $L(V_{\delta+1})$ to itself, and we have proven that $I0(\kappa, \delta)$ holds. In addition, $HOD(V_{\delta+1})$ must satisfy Choice, because any surjection in $V_{\delta+1}$ will have inverse in $M \cap V_{\delta+2}$ by the elementarity of $k$.

Now if $\kappa$ is Reinhardt, then $I-1(\kappa, \delta)$ holds, since $\delta$ is a fixed point of $j$ (by construction, it is the least such fixed point beyond $\kappa$).

Of course, we don't know that $k = j$, so we can't conclude from this argument that $k : HOD \prec HOD$. Likewise, we don't get $DC(V_\delta)$, since we only know that Choice holds in $V_\delta$.

  • $\begingroup$ Could you explain how you recover AC without loss? $\endgroup$ Feb 9, 2012 at 1:43
  • $\begingroup$ Hmmm. It may not be so easy after all. I was figuring that any generic well-ordering of $V_\delta$ would do; but it also needs, at least, a property ensuring consistency with $j$. In fact, the failure of Choice on $V_{\delta+2}$ should transfer to $V_{\kappa+2}$, and hence to a stationary sequence of smaller sets as well. So a better approach would be to use forcing over HOD to get a restriction of $j$ into the extended model. $\endgroup$ Feb 10, 2012 at 0:46
  • $\begingroup$ Hi Ben, and thanks. For naturality's sake, I'm interested in the "Reinhardt cardinals are any $j\colon V_{\delta+\eta} \rightarrow V_{\delta+\eta}$ they can be" approach. Is it still effective when Choice fails badly? Assume the critical sequence $\kappa_{n}$ has no supremum, there is a minimal upper bound $\delta$ with the usual properties, but also at least one cardinal $\xi \subset V_{\delta+1}$ that is a minimal upper bound but not an ordinal. Can we still conclude that a Reinhardt cardinal fully satisfies $I1$ by construction? $\endgroup$
    – Ekki
    Feb 10, 2012 at 20:09
  • $\begingroup$ Yeah, that still works, since $j(rank(S)) = rank(j(S))$ for any set $S$. $\endgroup$ Feb 11, 2012 at 18:35
  • $\begingroup$ Oh, but we don't necessarily know that $\delta$ is strong limit; if $\lambda < \kappa$, then $j(2^\lambda) = 2^j(\lambda) = 2^\lambda$, but this doesn't preclude the possibility that $2^\lambda > \delta$, nor that $2^\lambda$ is an alternate least fixed point of $j$ beyond the $\kappa_n$. So I guess we only get a partial version of the rank-into-rank axioms. But I think Aspero's result implies that there is some $\mu$ which has, for each $\eta$, $i_\eta:V_{\nu+\eta} \to V_{\nu+\eta}$ (\nu the LFP for i_\eta), and $i_\eta(\bar{\kappa})$ can be chosen greater than the Hartog cardinal for $\mu$. $\endgroup$ Feb 11, 2012 at 19:04

This answer is confined to Question 3. I suggest that the Axiom of Choice (AC) be regarded as placing a sharp bound on reflection. In particular, for an elementary embedding j with critical point κ and multiple iterations of j(κ), and where reflection is from the nth iterate of j(κ) to the mth iterate, for m less than n, AC places the bound "n less than ω" on such reflection, as Kunen's theorem shows; i.e., AC requires that n here be an integer.

I take this to be an analogue, within the language of ZF, to the following informal limit on reflection: if we say that the universe V is such that all true statements S about V are reflected, we need to restrict the range of S to (e.g.) sentences of the language of ZF, since otherwise we might be led to view the sentence "All true statements S about V are reflected" as being itself in the range of S and hence as getting reflected, which leads to an obvious conflict with Regularity. In other words, we need our reflection principles to be free of self-reference.

Kunen's theorem too deals with a situation involving a kind of self-reference, or circularity. Namely, for λ the ωth iterate of j(κ), we have j(λ)=λ, so that j(λ) is, in effect, present within the expression "j(λ)" itself. And the reflection of λ is, at the same time, the reflection of j(λ), which is what ultimately leads to the contradiction that Kunen obtains. So, using the above analogy, AC can be seen as performing the desirable role, within the language of ZF, of avoiding self-reference in reflection; and as such, AC is doing something different than the Axiom of Constructibility. Corazza's dropping of Replacement is another way to avoid such self-reference, but within a modified version of the language of ZF.

What I'm proposing here is that there are limits on reflection, and that within the language of ZF, AC is valuable (in part) because it clearly indicates a major limit of this kind. In my opinion, dropping AC does not remove or free us from the limit itself; it only prevents us from seeing or grasping the limit. Hence, I'm skeptical about the tenability of Reinhardt cardinals.

  • 2
    $\begingroup$ It seems like you are claiming that Kunen's theorem says $\omega$-huge cardinals are inconsistent with AC, but this is not the case. Kunen's theorem says there cannot be a nontrivial elementary embedding from $V_{\lambda+2} \to V_{\lambda+2}$. As far as we know there can be a nontrivial elementary embedding from $V_{\lambda +1} \to V_{\lambda +1}$. (By Kunen's theorem $\lambda$ must be the $\omega$-th iterate of the critical point.) $\endgroup$ Jul 21, 2012 at 16:48
  • 1
    $\begingroup$ Kunen's theorem entails that λ does not witness the 𝜔-hugeness of the critical-point cardinal κ; as such, this theorem does indeed rule out the existence of 𝜔-huge cardinals. However, this failure of λ to witness 𝜔-hugeness can only be shown or derived at Vλ+2, and not at Vλ+1, since the function 𝜔-Jonsson for λ that yields Kunen's contradiction is in Vλ+2, but not in Vλ+1 (see A. Kanamori, The Higher Infinite, pp. 321-2). That's why there can be a nontrivial elementary embedding j: Vλ+1 → Vλ+1. The latter embedding does not negate the fact that λ fails to witness an 𝜔-huge cardinal. $\endgroup$
    – wmitt
    Feb 24, 2020 at 23:24
  • $\begingroup$ I must have been thinking of a different definition of "$\omega$-huge cardinal" (see e.g. en.wikipedia.org/wiki/Huge_cardinal#%CF%89-huge_cardinals). Although now that I think about it, that seems like a bad definition. So I agree with your recent comment. $\endgroup$ Feb 25, 2020 at 17:25

While link-wandering for something completely unrelated I found the following answer to Question 2 on Wikipedia, heh. If I'm reading it correctly we have:

$J1$: There exists $\kappa$ such that for each ordinal $\alpha$, there is a corresponding $j:V \to V$ with $\kappa$ as its critical point and $j(\kappa) \gt \alpha$

We then proceed by "adding a generic wellordering of V" (thereby destroying the $J1$ cardinal and probably much more) and are left with $V$ satisfying $ZFC + \exists j:HOD \to HOD$. Presumably Global Choice is added by forcing (but which forcing, and what does it preserve?) and the language is Kelly Morse - Choice rather than $ZF$. No references, sadly, nor any expected beyond "private communication".

This does at least suggest that investigating the structure of $V$ below the first Reinhardt cardinal should illuminate which extensions it can admit.

Implausibilia: $\kappa$ the critical point of $j: V_{\delta+n+1} \to V_{\delta+n+1}$ is preceded by stationarily many $\bar{\kappa}$ that are critical points of embeddings $j: V_{\bar{\delta}+n} \to V_{\bar{\delta}+n}$. It's true for $n=0$.

Woodin has somewhere proved that $J1$ implies the consistency of $J2$: $\kappa$ is Reinhardt, $\lambda$ is as usual the first fixed point of the embedding, and Dependent Choice holds for sequences of length $\lambda$. If Reinhardt cardinals definitionally satisfy $j: V_{\delta+3} \to V_{\delta+3}$, Implausibilia implies many $j: V_{\delta+2} \to V_{\delta+2}$ below the least Reinhardt cardinal. Then $DC_{\lambda}$ could not hold, and $J2$ would be inconsistent. So one way to produce abductive evidence for these looming hypotheses would be to exhibit a model in which the least $I1(\kappa,\delta)$ cardinal is $I-1(\kappa,\delta)$, or a model with no $I-1(\kappa,\delta)$ cardinals below the least Reinhardt cardinal (there must be stationarily many $I3(\kappa,\delta)$ cardinals at least).

Too difficult for me right now! I'll hold off on accepting an answer just a bit longer in hopes of exasperating someone into correcting all this.


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