Let $\pi: V \rightarrow M$ be a non-trivial elementary embedding with critical point $\kappa$, where $M$ is a transitive class. I don't seem to understand a given proof of the following basic observation.

Let $\alpha$ be an ordinal with $cf(\alpha) < \kappa$. Then $\pi(\alpha) = \sup (\pi[\alpha])$. (Here $\pi[\alpha] = \{ \pi(\beta) \mid \beta < \alpha\}$)

The proof goes as follows:

Let $f: cf(\alpha) \rightarrow \alpha$ be cofinal. Then $\pi(f): cf(\alpha) \rightarrow \pi(\alpha)$ is cofinal by elementarity. But $ran(\pi(f)) \subseteq \pi[\alpha]$, so that $\pi(\alpha) = \sup (\pi[\alpha])$. q.e.d.

First of all, we get that $M \models "\pi(f): cf(\alpha) \rightarrow \pi(\alpha) \text{ is cofinal}"$, so we need to see that $V \models "\pi(f): cf(\alpha) \rightarrow \pi(\alpha) \text{ is cofinal}"$. Why is that?

Also, $ran(\pi(f)) \subseteq \pi[\alpha]$ doesn't seem to rely on the cofinality of $\alpha$ - it appears to me as a consequence of the elementarity of $\pi$. Am I mistaken?

How and where does $cf(\alpha)$ come into play? The argument has to use it somewhere, because otherwise it would prove that $\pi(\kappa) = \sup(\pi[\kappa]) = \sup(\kappa) = \kappa$, which is nonsense. Edit: I know that I would have to replace $cf(\alpha)$ with $\pi(cf(\alpha))$ if I don't know that $cf(\alpha)$ is fixed under $\pi$. But I fail to see why this would destroy the given argument.


1 Answer 1


The key point is that $cf(\alpha)<crit(\pi)$ so that $\pi(cf(\alpha))=cf(\alpha)$ This means that the cofinalities of $\alpha$ as viewed in $V$ and $\pi(\alpha)$ as viewed in $M$ are the same.

Now elementarity says that if $f$ witnesses this cofinality in $V$ then $\pi(f)$ witnesses it in $M$. Further $ran(\pi(f))=\pi(ran(f))$ is again a consequence of the fact that the domain of $f$ is fixed.

  • $\begingroup$ Wait, isn't the key point that $ran(\pi(f)) = \pi(ran(f))$? Because if I know this, I always get a $\pi(cf(\alpha))$ long sequence of ordinals below $\pi(\alpha)$ which is cofinal in $\pi(\alpha)$. It might be longer than $cf(\alpha)$, but why would I care? If that's true, then I think I see my problem. Please note: I only try to make sure to get your point by asking this. $\endgroup$ Commented Aug 11, 2014 at 13:13
  • $\begingroup$ Well it amounts to the same thing, if $cf(\alpha)$ were enlarged under $\pi$ the so too would be $im(\pi(f))$. $\endgroup$ Commented Aug 11, 2014 at 13:19
  • $\begingroup$ Yeah, I am aware of that. I just wanted to make sure :) Thank you. Your help is much appreciated! $\endgroup$ Commented Aug 11, 2014 at 14:17

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