Kunen showed that Reinhardt cardinals are inconsistent in ZFC. But his proof is a bit technical for a non-set-theorist to follow. Berkeley cardinals are stronger than Reinhardt cardinals. You can refute them in ZFC by observing that every Berkeley cardinal is Reinhardt, and then appealing to Kunen's theorem.

Question 1: Is there a more direct refutation of Berkeley cardinals in ZFC, perhaps one which might be more digestible by a non-set-theorist?

Recall that a definably-Berkeley cardinal is a cardinal $\kappa$ such that whenever $M$ is a transitive set with $\kappa \in M$, there are elementary embeddings $M \to M$ with arbitrarily large critical point $\alpha < \kappa$. A Reinhardt cardinal is when this holds for $M = V_\kappa$ (EDIT: In fact the "Reinhardt cardinal is defined to be the critical point of the embedding). So perhaps there is some other $M$ one can cook up which obviously fails the Berkeley property? Maybe $M = \kappa + 1$, for example?

(A Berkeley cardinal is a cardinal $\kappa$ such that whenver $M$ is a transitive set with $\kappa \in M$ and $A \subseteq M$, there are elementary embeddings $j : (M,A) \to (M,A)$ with arbitrarily large critical point below $\kappa$. Although it seems sometimes one restricts the definition to apply only when $M = V_\alpha$? I'm not sure if that's equivalent or weaker.)

Question 2: Let $\kappa$ be a cardinal (probably uncountable, regular, limit, measurable,etc.). Can it be the case in ZFC that there exists a nontrivial elementary embedding $\kappa + 1 \to \kappa + 1$?

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    $\begingroup$ Actually, I just noticed that since $\kappa \not \in V_\kappa$, it's maybe not so clear why Berkeley cardinals are Reinhardt... $\endgroup$
    – Tim Campion
    Nov 20 at 17:23
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    $\begingroup$ Is your question 2 really just asking about elementary embeddings between ordinals? If so, the answer is yes, $\mathsf{ZFC}$-provably there are lots of nontrivial elementary embeddings from $\kappa+1$ to itself whenever $\kappa$ is an uncountable cardinal. Basically, ordinals by themselves have very little expressive power (although things get more interesting if we move up from FOL, see this old question of mine). $\endgroup$ Nov 20 at 17:53
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    $\begingroup$ Berkeley cardinals are not Reinhardt cardinals as $\mathsf{ZF}$ proves the least Berkeley cardinal is singular but every Reinhardt cardinal must be regular. However, if $\delta$ is a Berkeley cardinal, then there is $\lambda<\delta$ such that $(V_\lambda, V_{\lambda+1})$ is a model of $\mathsf{NGB}$ with a Reinhardt cardinal, which becomes a model of choice if we work over $\mathsf{ZFC}$. $\endgroup$
    – Hanul Jeon
    Nov 20 at 18:02
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    $\begingroup$ Another question in this vein one could ask is whether you can write down a large cardinal notion not known to be inconsistent with ZF with an 'immediate' proof of inconsistency with ZFC. (Obviously this is a somewhat vague question.) $\endgroup$ Nov 20 at 18:46
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    $\begingroup$ Tim, towards the question addressed to @Hanul, if $\kappa\in M$, then this is the definition of Berkeley. If not, then $M\{\kappa\}$ is transitive again, so by the definition of Berkeley we get embeddings. All we need to note that in this case it must be that $j(\kappa)=\kappa$ when $j:M\to M$ is an elementary embedding, since $\kappa$ is the largest ordinal in $M\cup\{\kappa\}$ in that case. $\endgroup$
    – Asaf Karagila
    Nov 20 at 20:05

1 Answer 1


Yes, it is easier to refute Berkeleys than Reinhardts. There is a very simple refutation of Berkeleys in ZFC that is due to Woodin. It is part of the motivation for his contention that Berkeley cardinals should be inconsistent with ZF.

Towards a contradiction, assume ZFC plus $\delta$ is the least Berkeley. For each $\alpha< \delta$, choose a transitive set $M_\alpha$ containing $\delta$ such that there is no $j:M_\alpha \to M_\alpha$ with critical point less than $\alpha$. Let $\gamma$ be the supremum of the ranks of the $M_\alpha$. Since $\delta$ is Berkeley there is an elementary $i : V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $i(\vec M) = \vec M$. Notice that $i$ has no fixed points $\alpha$ between its critical point and $\delta$. Otherwise $i(M_\alpha) = M_\alpha$, but then $i\restriction M_\alpha$ is an elementary embedding from $M_\alpha$ to $M_\alpha$ with critical point less than $\alpha$.

Therefore $\delta$ has countable cofinality: in fact, $\delta = \sup_{n<\omega} \kappa_n$ where $\kappa_n = i^n(\kappa)$. But applying Berkeliness again, we can get another embedding $k: V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $k(\vec N) = \vec N$ where $N_n = M_{\kappa_n}$. Clearly $k$ fixes each $N_n$ (since $k(n) = n$). But taking $n$ large enough that $\kappa_n > \text{crit}(k)$, $k\restriction N_n$ is an elementary embedding that contradicts the definition of $M_{\kappa_n}$.

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    $\begingroup$ Thanks, this is great! Does $\vec M$ denote (the graph of) the function $\delta \to V$, $\alpha \mapsto M_\alpha$? Also, does $i(M_\alpha)$ really mean $i(M_\alpha)$, or does it mean the image of $i$ restricted to $M_\alpha$? $\endgroup$
    – Tim Campion
    Nov 21 at 15:05
  • $\begingroup$ Yes, $\vec M$ is what you guessed. But $i(M_\alpha)$ really means $i(M_\alpha)$!!! Of course since $i(\vec M) = \vec M$, we have $i(M_\alpha)= M_{i(\alpha)}$. $\endgroup$ Nov 21 at 15:52
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    $\begingroup$ @TimCampion Choice is being used in the step where we choose the sequence of $M_\alpha$'s uniformly. $\endgroup$ Nov 21 at 18:22
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    $\begingroup$ (2) Well, you can make $\gamma$ as big as you want. Depending on how you code the sequence of $M_\alpha$'s, the $\gamma$ I chose will work. But you could go up $\omega$ ranks and never worry about a pairing function again. (3) I'm not sure what you mean about definably Berkeley, $M_\alpha$ isn't definable. (4) If $i(M_\alpha) = M_\alpha$ then $i\restriction M_\alpha$ is elementary from $M_\alpha$ to $M_\alpha$ since the $M_\alpha\vDash \varphi(a)$ if and only if $V_{\gamma+1}$ satisfies the relativization $\varphi^{M_\alpha}(a)$ iff it satisfies $\varphi^{M_\alpha}(i(a))$. $\endgroup$ Nov 21 at 18:53
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    $\begingroup$ (6) Well it may not be definable, but $\vec \kappa$ is in $V_{\gamma+1}$ and $i(\vec \kappa) = \vec \kappa\restriction \mathbb N^+$ (roughly), so $i(\sup \vec \kappa) = \sup (\vec \kappa\restriction \mathbb N^+) = \sup \vec \kappa$. $\endgroup$ Nov 21 at 18:54

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