I would like to suggest another argument for Kunen's inconsistency result, and I wonder to know if the argument is correct. I am also interested to see, if the proof is correct, which part of the argument uses AC.

Theorem. There is no non-trivial elementary embedding $j: V \rightarrow V.$

Proof. Assume on the contrary that there is a non-trivial elementary embedding $j: V \rightarrow V$ with $\kappa=crit(j).$ Iterate it to get $\langle \langle M_n: n\leq \omega \rangle, \langle j_{n,m}: n\leq m\leq \omega \rangle \rangle$ where $\langle M_{\omega}, \langle j_{n, \omega}: n\leq \omega \rangle \rangle =dirlim \langle \langle M_n: n<\omega \rangle, \langle j_{n,m}: n\leq m< \omega \rangle \rangle.$ Note that $j=j_{0,1}$ and that for all $n<\omega, M_n=V.$ Let $\kappa_n = j_{0, n}(\kappa), n\leq \omega.$ Then $\kappa_{\omega}=sup_{n<\omega}\kappa_n.$ Also we have $M_{\omega}=V$ and we get a contradiction, since $\kappa_{\omega}$ must be regular in $M_{\omega}=V.$

  • $\begingroup$ Doesn't seem to be correct because you assume that $j$ itself is continuous, in the sense that $j_\omega$ is really just $j$ itself applied $\omega$ times, whereas it is more likely that $M_\omega$ has some "bump". Much like how for cardinal exponentiation it is false that $\sup_{n\in\omega}2^n=2^\omega$. $\endgroup$
    – Asaf Karagila
    Mar 29, 2013 at 12:52

1 Answer 1


This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this is a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

Note that the limit of the iteration is well-founded, by essentially the usual Kunen argument. Namely, if it were ill-founded, then we may consider the least $\gamma$ with $j_{0,\omega}(\gamma)$ in the ill-founded part of the limit. By pushing $\gamma$ into $M_n$, we may conclude on the one hand that $j_{0,n}(\gamma)$ must be least sent into the ill-founded part, but on the other hand we may also drop down to smaller ordinal born at that stage still in the ill-founded part, making a contradiction. (To formalize this precisely, one may avoid the complications caused by $j$ being a proper class simply by chopping off $j$ to the set $j\upharpoonright V_\alpha$, such that the iteration whose iteration is ill-founded.) This argument relies fundamentally on the fact that the iteration of $j$ in $M_0$ produces the same limit as the iteration of $j(j)$ in $M_1$.

Finally, if on the other hand you had formed your iteration by using the same embedding $j_{n,n+1}=j$ at each stage, then this is clearly ill-founded, since the threads starting with $\kappa$ at each stage form a descending sequence of ordinals in the limit.

  • $\begingroup$ Nice answer Joel. By the way, the well-foundedness of the direct limit should also be addressed. $\endgroup$
    – Ali Enayat
    Mar 29, 2013 at 18:20
  • $\begingroup$ It seems that the direct limit is well-founded by the usual Kunen argument. I'll edit to explain. $\endgroup$ Mar 29, 2013 at 19:16
  • 1
    $\begingroup$ I added an explanation that the limit of the iteration is well-founded. $\endgroup$ Mar 29, 2013 at 19:37

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