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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).

Thanks.

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    $\begingroup$ Also see this question. $\endgroup$ – Miha Habič Nov 17 '15 at 21:02
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    $\begingroup$ 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). $\endgroup$ – Andrés E. Caicedo Nov 18 '15 at 1:48
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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?

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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Nov 17 '15 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! $\endgroup$ – Noah Schweber Nov 17 '15 at 20:14
  • $\begingroup$ The example I've seen was in "Handbook of Set Theory (Foreman,Kanamori)" Chapter 12, Thm 10.2 $\endgroup$ – Nadav Sherman Nov 18 '15 at 7:10

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