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Recall the definition of critical point for set theory:

A critical point of an elementary embedding of one transitive class into another transitive class is the smallest ordinal not mapped to itself. (This is from the Wikipedia article "Critical point (set theory)", which claims this definition is from Jech's Set Theory (2002 edition)).

What theorems about critical points of elementary embeddings can be proven in $ZF$ without recourse to the Axiom of Choice? To be specific, are these theorems enough to prove anything useful regarding critical points of nontrivial elementary embeddings $j$: $V$$\rightarrow$$V$?

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  • $\begingroup$ A related question is mathoverflow.net/questions/237662/… . $\endgroup$
    – Rahman. M
    Jun 21, 2016 at 14:02
  • $\begingroup$ Your question is unclear. You ask what we can prove about critical points, and then you specify and ask if the things we can prove are enough to prove anything "useful" about Reinhardt cardinals. You're practically asking the same thing twice, but the repetition is odd, and off, and it makes it very unclear as to what's going on there. $\endgroup$
    – Asaf Karagila ♦
    Jun 21, 2016 at 14:36
  • $\begingroup$ @AsafKaragila: The most important thing to prove about the critical points of nontrivial elementary embeddings $j$:$V$$\rightarrow$$V$ is whether they exist or not. The question I should have asked is whether or not what one can prove in $ZF$ without Choice regarding critical points of elementary embeddings is enough to prove whether Reinhardt cardinals exist or whether one can prove the Kunen inconsistency without recourse to Choice. If this clarifies matters I will change the second question. $\endgroup$
    – Thomas Benjamin
    Jun 21, 2016 at 15:28
  • $\begingroup$ @Rahman.M: I'd like to see the proof, also.... $\endgroup$
    – Thomas Benjamin
    Jun 21, 2016 at 15:29
  • $\begingroup$ Why the downvote (after three years....)? $\endgroup$
    – Thomas Benjamin
    Nov 12, 2019 at 22:56

2 Answers 2

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If $j\colon V\to M$ is a nontrivial elementary embedding with $M\subseteq V$ a transitive class, then $j$ has a critical point:

Suppose there isn't one, then by induction on rank we prove that $j(x)=x$ for all $x\in V$. First note that $\operatorname{rank}(x)=\operatorname{rank}(j(x))$ for all $x$, since the rank function is absolute and if $\operatorname{rank}(x)<\operatorname{rank}(j(x))$, then we have an ordinal which was moved by $j$.

Suppose that $j(y)=y$ for all $y\in x$, then $M\models j(y)\in j(x)$ and therefore $x\subseteq j(x)$; on the other hand, if $z\in j(x)$, then the rank of $z$ is less than the rank of $x$, therefore $j(z)=z$ and therefore $M\models j(z)\in j(x)$, so $V\models z\in x$. Therefore $j(x)\subseteq x$.

In particular, $j\colon V\to V$ without a critical point must be the identity.

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    $\begingroup$ Note that there is a crucial assumption here that $j$ is an internal elementary embedding, that is, that $j$ is definable: otherwise the induction on rank can't go through if $V$ is ill-founded. External elementary embeddings have much more freedom - for example, Cohen's construction of a model of ZF with an automorphism of finite order. There the automorphism does not move any ordinals (exercise), but is still nontrivial. Very little can be said about external elementary embeddings or automorphisms (although not nothing - e.g. the motivation for Cohen's result was that choice must fail!). $\endgroup$
    – Noah Schweber
    Jun 21, 2016 at 17:07
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    $\begingroup$ Well, definable is a big word. I think that "a class in a NBG/MK model" would be enough. :-) But good point nonetheless. $\endgroup$
    – Asaf Karagila ♦
    Jun 21, 2016 at 17:11
  • $\begingroup$ @AsafKaragila: So in other words (assuming $j$ is internal), if $j$ in your theorem has a critical point, then either $rank(x)$$\lt$$rank(j(x))$ or $rank(j(x))$$\lt$$rank(x)$? $\endgroup$
    – Thomas Benjamin
    Jun 22, 2016 at 6:38
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    $\begingroup$ No, this is not true. If it had a critical point, then is some $x$, e.g. the critical point, such that $j(x)$ had a greater rank than $x$. It's not for every $x$. And if there is some $x$ with $j(x)$ having a greater rank, then the ordinal of that rank is moved by $j$. In either case, the rank cannot decrease. $\endgroup$
    – Asaf Karagila ♦
    Jun 22, 2016 at 6:47
  • $\begingroup$ @NoahSchweber: Under what conditions would an external elementary embedding move an ordinal? Also, can one always adjoin(?) (by forcing) an external elementary embedding that moves an ordinal? $\endgroup$
    – Thomas Benjamin
    Jun 22, 2016 at 11:27
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In ZF, many of the usual arguments about critical points still go through.

For example, every critical point $\kappa$ of an elementary embedding $j:V\to M$ is regular, since if $\kappa$ is the supremum of a short sequence $s$ below $\kappa$, then it is easy to see that $j(s)=s$, which would imply $j(\kappa)=\kappa$, contradicting the assumption that $\kappa$ is the critical point.

If $\kappa$ is the critical point of $j:V\to M$, then $\kappa$ is weakly inaccessible. To see this, note that if $\kappa=\delta^+$, then $j(\kappa)=(\delta^+)^M$, which of course is at most $\kappa$, since $P(\delta)^V=P(\delta)^M$, and this contradicts $\kappa<j(\kappa)$. So $\kappa$ is a regular limit cardinal.

If $\kappa$ is the critical point of $j:V\to M$, then $\kappa$ is a measurable cardinal. To see this, let $\mu$ be the set of $X\subset\kappa$ for which $\kappa\in j(X)$, and the usual arguments show that this is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, without using the axiom of choice. So $\kappa$ is measurable. (Note: in ZF measurability is weaker than you might expect, since even $\omega_1$ can be measurable, so it doesn't imply inaccessibility.)

If $\kappa$ is the critical point of $j:V\to V$, then $\kappa$ is $<j(\kappa)$-supercompact, in the sense that there is a normal fine measure on $P_\kappa\lambda$ for every $\lambda$ in the interval $[\kappa,j(\kappa))$, because you can let $\nu$ be the set of $X\subset P_\kappa\lambda$ with $j"\lambda\in j(X)$, and the usual arguments show that this is a normal fine measure. (Note, without AC, the equivalence formulations of supercompactness break down, since one can't always establish that the ultrapowers are elementary.)

Similarly, if $\kappa$ is the critical point of $j:V\to V$, then $\kappa$ is (weakly) extendible, since for every $\lambda>\kappa$ we may look at $j\upharpoonright V_\lambda:V_\lambda\to V_{j(\lambda)}$, which is elementary and verifies weak extendibility (weak = we do not insist on $\lambda<j(\kappa)$, which is equivalent in ZFC using the Kunen inconsistency).

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    $\begingroup$ I'm probably just being dense, but I don't follow the supercompactness proof if $\lambda\geq j(\kappa)$. For $\nu$ to be an ultrafilter on $P_\kappa(\lambda)$, it needs to have $P_\kappa(\lambda)$ as an element, which means $j''\lambda\in j(P_\kappa(\lambda))= P_{j(\kappa)}(j(\lambda))$, so the cardinality $\lambda$ of $j''\lambda$ needs to be $< j(\kappa)$. $\endgroup$
    – Andreas Blass
    Jun 21, 2016 at 16:55
  • $\begingroup$ You are right, I was thinking only of $\lambda<j(\kappa)$. I'll edit. $\endgroup$
    – Joel David Hamkins
    Jun 21, 2016 at 16:57
  • $\begingroup$ But if $\lambda\geq j(\kappa)$ and $\lambda<j(\lambda)$, you can still get a normal fine measure on $P_\lambda\lambda$, which is a large cardinal property in ZFC. $\endgroup$
    – Joel David Hamkins
    Jun 21, 2016 at 17:12
  • $\begingroup$ In ZFC, isn't "normal fine measure on $P_\lambda\lambda$" the same as measurability of $\lambda$? $\endgroup$
    – Andreas Blass
    Jun 21, 2016 at 17:29
  • $\begingroup$ @AndreasBlass On a common meaning of those words, yes, but there are weaker versions, which I had in mind. For example, if $j:V\to M$ has critical point $\kappa$ and $M^{\lambda}\subset M$, where $\lambda=j(\kappa)^+$, a situation that is weaker than 2-huge, then the measure $\mu$ defined by $X\in\mu\iff j"\lambda\in j(X)$ is (i) fine in the sense that $\{\sigma\mid \alpha\in\sigma\}\in\mu$ for all $\alpha<\lambda$ and (ii) normal in the sense that whenever $f(\sigma)\in\sigma$ for $\mu$-a.e. $\sigma$, then $f$ is constant on a set in $\mu$. The first holds because $j(\alpha)\in j"\lambda$,... $\endgroup$
    – Joel David Hamkins
    Jun 21, 2016 at 19:32

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