12
$\begingroup$

A well-known result of Kunen says that there is no non-trivial elementary embedding $j: V \rightarrow V.$ There are several proofs of this theorem (see Kanamori, The higher infinite). I wonder to know if the following argument works:

Assume on the contrary that such an embedding exists. Let $\Phi(\alpha)$ be the following statement:

'' $\alpha=crit(j)$ for some non-trivial $j: V \rightarrow V.$''

Let $\kappa$ be the least such that $\Phi(\kappa)$ and let $j: V \rightarrow V$ witness this. Using $j$ we can conclude that $j(\kappa)$ is also the least such that $\Phi(j(\kappa))$ and this is impossible.

$\endgroup$
1
  • $\begingroup$ No such elementary embedding for $V$ the universe of ZFC sets. $\endgroup$
    – David Roberts
    Commented Feb 27, 2013 at 4:44

2 Answers 2

17
$\begingroup$

Your $\Phi(\kappa)$ uses an existential quantifier over classes (since $j$ is a class), whereas $j$ is only assumed to preserve first order formulae, so there is no reason to conclude that $j(\kappa)$ is also the least such that $\Phi(j(\kappa))$,

$\endgroup$
13
$\begingroup$

Although your argument doesn't prove the full Kunen inconsistency, for the reasons already explained by Ali Enayat, it does prove the following form of the Kunen inconsistency, ruling out definable nontrivial elementary embeddings. What is more, it is often only this form of the result that one finds expressed by set theorists. For example, the version of the Kunen inconsistency stated in Kanamori's text is only this weaker form (see footnote 1 in the paper linked below).

Theorem. There is no definable class $j$ that is a nontrivial elementary embedding $j:V\to V$.

Proof: This is a theorem scheme of ZFC, with one theorem for each possible defining formula of $j$. Suppose first that $j:V\to V$ is a nontrivial elementary embedding and that $j(x)=y$ is defined by a formula $\psi(x,y)$. We may then define the critical point of $j$, just as in your argument, and deduce by elementarity that $j(\kappa)$ would also have to be the critical point, which is a contradiction. Thus, there is no definable nontrivial elementary embedding $j:V\to V$.

One can adapt the argument to show that there is no nontrivial elementary embedding $j:V\to V$ that is definable from parameters. To do this, if $j(x)=y$ is defined by $\psi(x,y,z)$, with parameter $z$, then you may assume $z$ is chosen so that this embedding gives rise to the smallest possible critical point among any parameter for which the definition does define an elementary embedding (and this is itself definable, by asserting that the relation is $\Sigma_1$-elementary and cofinal). This defines $\kappa$ without parameters, and so again one attains a contradiction. QED

In particular, when one chooses to formalize the Kunen inconsistency as a result in ZFC, as opposed to the second order theories GBC or KM, then it admits of this very soft proof. Furthermore, this argument does not use the axiom of choice, and so actually it is provable in ZF. This argument was published by Suzuki in 1998. You will also find an account of it and further generalizations in my paper Generalizations of the Kunen Inconsistency, joint with Norman Perlmutter and Greg Kirmayer.

Our view, which we explain in the paper, is that this ZFC formulation of the Kunen inconsistency does not capture the full power of his argument. Kunen himself proved his theorem in Kelly-Morse KM set theory, although his argument and the other combinatorial proofs can be given in Gödel-Bernays GBC set theory, which gives a stronger result. Our paper contains several further generalizations of this argument, to treat nontrivial embeddings $j:V[G]\to V$ or $j:V\to V[G]$ that might be definable in a forcing extension $V[G]$ and many other cases.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .