I've seen an example of an elementary embedding such that $\omega_1$ is the critical point.

I was wondering what's wrong with the following proof that this cannot be:

Let $\phi(x_1,x_2)$ be the first-order formula: $\phi(x_1,x_2)=x_1\ is\ cardinal\wedge x_2\ is\ cardinal\wedge\nexists\kappa\left(\kappa\ is\ cardinal\wedge\ x_1<\kappa<x_2\right)$.

Let $j:M\to N$ be an elementary embedding with $crit(j)=\omega_1$, $j(\omega_1)=\kappa>\omega_1$

Then $M\models\phi(\omega_0,\omega_1)$, but $N\nvDash\phi(j(\omega_0),j(\omega_1))=\phi(\omega_0,\kappa)$

Contradiction to elementarity.

(By this [false] proof, $crit(j)$ cannot be a successive cardinal).


  • 2
    $\begingroup$ Also see this question. $\endgroup$ Nov 17, 2015 at 21:02
  • 1
    $\begingroup$ Perhaps it is obvious, but the mistake in your argument is in assuming that $N$ does not satisfy the formula. Perhaps you believe that $\omega_1$ witnesses that $\phi(\omega_0,\kappa)$ holds. That this is not the case tells us that what we are calling $\omega_1$ is not a cardinal of $N$. If we assume our models are transitive, then $\omega_1$ is not a cardinal of $V$. This is not a contradiction, but instead tells us that $j$ does not reside in $V$. Most likely the example you have seen is of a generic embedding, living in a forcing extension (where $\omega_1^M$ was collapsed). $\endgroup$ Nov 18, 2015 at 1:48

1 Answer 1


Indeed, you cannot have an elementary embedding $j:V\rightarrow M$ with critical point $\omega_1^V$. However, there are some subtleties to keep in mind:

  • We could have an elementary embedding $j:W\rightarrow M$, where $W$ is an inner model of $V$, with $crit(j)=\omega_1^V$ - for instance, let $W\models$"$\kappa$ is measurable", and let $V$ be a forcing extension of $W$ in which $\kappa$ is collapsed to $\omega_1$. (In fact, we can do even better - see Joel's comment below.)

  • Also, note that, while in ZFC having a $\kappa$-complete ultrafilter on $\kappa$ means that $\kappa$ is the critical point of an elementary embedding, this fails in ZF alone - so, while ZF+AD proves "$\omega_1$ is measurable," this does not mean that in ZF+AD there is an elementary embedding of $V$ into an inner model $M$ with critical point $\omega_1$.

May I ask what argument you have seen?

  • 1
    $\begingroup$ In your measure case, the embedding isn't elementary from $M$ to $M$, but to some $N\subset M$. But meanwhile, you can have elementary $j:M\to M$ with critical point $\omega_1$, for example, under $0^\sharp$, we have $j:L\to L$ with critical point $\omega_1$, since it is a Silver indiscernible. $\endgroup$ Nov 17, 2015 at 20:09
  • $\begingroup$ @JoelDavidHamkins Quite right, I have been sloppy with target models - this is what happens when I try to do math without coffee! $\endgroup$ Nov 17, 2015 at 20:14
  • $\begingroup$ The example I've seen was in "Handbook of Set Theory (Foreman,Kanamori)" Chapter 12, Thm 10.2 $\endgroup$ Nov 18, 2015 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.