Assume one has. for $V$ and for some transitive class $M$, an elementary embedding

$j$: $V$$\rightarrow$$M$ and that $j$$\neq$$id$, where $id$ is the identity.

If $V$ and $M$ satisfy $ZFC$ then the following Theorem holds

Thm. Let $\alpha$ be an ordinal.

(i) For every $\alpha$, $j$($\alpha$)$\ge$$\alpha$

(ii) $j$ moves some ordinal.

Let $\delta$ be the least ordinal moved by $j$. $\delta$ is called the critical point of $j$.

It can be proven in $ZFC$ that $\delta$ is always a cardinal.

Suppose now that $j$: $V$$\rightarrow$$M$, and $V$ and $M$ both satisfy $ZF$+$\lnot$$AC$. How can the critical point $\delta$ be defined when there exist incomparable cardinals?

I hope that this is not too silly a question. If it turns out to be silly, I will happily delete it.

(Addendum: Regarding Noah's answer to my question ("...even though the cardinalities of $V$ may not be well-ordered, the cardinals and ordinals definitely will be..."), in $ZF$+$\lnot$$AC$, there are cardinals which are not ordinals, and these may not be comparable. Since under a nontrivial elementary embedding no ordinal would be moved, could such cardinals be critical points of such an embedding?)

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    $\begingroup$ $\delta$ is in that case defined as the least ordinal moved by $j$. $\endgroup$ – Andrés E. Caicedo Oct 2 '15 at 2:54
  • $\begingroup$ @AndrésCaicedo: Regardless of whether or not there incomparable cardinals? $\endgroup$ – Thomas Benjamin Oct 2 '15 at 2:56
  • $\begingroup$ @ThomasBenjamin Why does the usual definition seem unsatisfactory if choice fails? $\endgroup$ – Noah Schweber Oct 2 '15 at 2:57
  • $\begingroup$ @NoahSchweber: I am trying to get at the reason(s) (perhaps too indirectly--I really should come out and ask...) why it is so difficult to prove whether $ZF$ admits Reinhardt cardinals or not. I guess I have difficulty picturing what a $V$ satisfying $ZF$+$\lnot$$AC$ 'looks like' (if in fact one can actually picture such a thing...) $\endgroup$ – Thomas Benjamin Oct 2 '15 at 3:13
  • $\begingroup$ @ThomasBenjamin Why not ask that then? I'd certainly be interested in the answer. All I really know is why the standard proof doesn't work without choice (see page 82 of math.bu.edu/people/aki/d.pdf for a beautifully short proof, the most understandable I've seen), and the fact that I can't think of how to fix it. :P $\endgroup$ – Noah Schweber Oct 2 '15 at 3:20

You can define the critical point in exactly the same way.

First, note that even though the cardinalities of $V$ may not be well-ordered, the cardinals and ordinals definitely will be. EDIT: BY "cardinality," I mean cardinality in the most general sense; by "cardinal," I mean cardinality of a well-orderable set, that is, "$\aleph$-number." Equivalently, by "cardinal" I mean "ordinal which is not in bijection with any smaller ordinal."

So we still have the result that $j(\alpha)$ is an ordinal $\ge\alpha$, for every ordinal $\alpha$.

Moreover, by transfinite induction (which doesn't require choice) if $j(\alpha)=\alpha$ for every ordinal $\alpha$ then $j$ is trivial - again, this doesn't use choice.

So the answer is, the critical point can be defined in exactly the same way - as the least ordinal moved by $j$.

  • $\begingroup$ Is this because $V$ and $M$ are well-founded? $\endgroup$ – Thomas Benjamin Oct 2 '15 at 3:01
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    $\begingroup$ It's more because $V$ thinks it is well-founded. By elementarity, $j$ sends ordinals to ordinals. Now, if $j(\alpha)<\alpha$, we must have $j^2(\alpha)<j(\alpha)$, and $j^3(\alpha)<j(\alpha)$, and so forth - and this can be recognized within $V$! So, since $V$ thinks itself well-founded, and $M$ is an inner model of $V$ (so $V$ thinks $M$ is well-founded), we must have $j(\alpha)\ge \alpha$. Similarly, the transfinite induction argument is internal, and doesn't depend on $V$ actually being well-founded. $\endgroup$ – Noah Schweber Oct 2 '15 at 3:05
  • $\begingroup$ I generally object treating cardinals as ordinals in ZF. $\aleph$ numbers, $\aleph$ cardinals, can already be used to designate these. But what about cardinal arithmetic? Do you call it "cardinality arithmetic" now? $\endgroup$ – Asaf Karagila Oct 3 '15 at 7:29
  • $\begingroup$ @AsafKaragila Personally, I've always found it intuitively clearer to keep the word "cardinal" reserved for specifically well-orderable cardinalities, but I think this exchange indicates that you're right pedagogically. :) $\endgroup$ – Noah Schweber Oct 3 '15 at 7:38
  • $\begingroup$ @Noah: Hah! :-P $\endgroup$ – Asaf Karagila Oct 3 '15 at 7:46

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