At the risk of this question having an easy answer, I am asking the following:

Suppose we have a measurable cardinal $\kappa$ and a (possibly regular) cardinal $\lambda>\kappa$. Is there any condition, like the existence of an appropriate ultrafilter, that implies the existence of an elementary embedding $j:V\to M$ with critical point $\kappa$ so that $j(\kappa)=\lambda$?

An example is when $\kappa$ is huge: if there is a normal $\kappa$-complete ultrafilter $U$ on $P(\lambda)$ with the property that $\{X\subseteq P(\lambda)\mid ot(X)=\kappa\}\in U$, then the ultrapower embedding $j:V\to M\simeq Ult(V,U)$ satisfies $j(\kappa)=\lambda$ (and $~{}^{j(\kappa)}M\subseteq M$)..

However, I am interested in the case where $\kappa$ is only measurable, maybe not even strong.

I am assuming Choice and whenever I write $j:V\to M$ I mean that $M\subseteq V$ is a transitive model of ZFC.


1 Answer 1


Here is a necessary and sufficient criterion: $\lambda>2^\kappa$. This is easily seen to be necessary, since if $j:V\to M$ has critical point $\kappa$, then the power set $P(\kappa)$ is contained in $M$, and from this it follows that $j(\kappa)>2^\kappa$ since $j(\kappa)$ is inaccessible in $M$. Conversely, we can hit every cardinal above $2^\kappa$, by the following.

Theorem. If $\kappa$ is a measurable cardinal, then every cardinal $\lambda>2^\kappa$ is the image $\lambda=j(\kappa)$ of some elementary embedding $j:V\to M$ with critical point $\kappa$.

Proof. Let $\mu$ be a normal measure on $\kappa$ and consider the class of images $j_\alpha(\kappa)$, where $j_\alpha$ is the $\alpha$-iterated ultrapower by $\mu$. These are the ordinals that form the critical sequence. This class of ordinals is closed and unbounded. It is clearly unbounded, since by iterating further, we can push $j(\kappa)$ as high as desired. It is closed, since the iterations are defined to take the direct limit at limit stages, and this makes the critical sequence continuous.

Finally, I claim that every cardinal above $2^\kappa$ is on the critical sequence. To see this, it suffices to argue that at each stage, we don't jump over the next cardinal. If $j_\alpha:V\to M_\alpha$ is the $\alpha^{th}$ iterate, then the extender representation shows that every element of $M$ has the form $j(f)(s)$, where $f:\kappa\to V$ and $s$ is a finite sequence from the critical sequence below $\kappa_\alpha$, which are ordinals below $j_\alpha(\kappa)$. If we go one more step, to $j_{\alpha+1}:V\to M_{\alpha+1}$, then we only need to add one more generator or seed, namely $\kappa_\alpha$ itself, and so $|j_{\alpha+1}(\kappa)|^V\leq |\kappa^\kappa|\cdot|(\alpha+1)^{<\omega}|$, which has the same size as $j_\alpha(\kappa)$. So at successor stages, we don't get to the next cardinal, and so we reach all the cardinals at limit stages. So we'll get every cardinal above $2^\kappa$. QED

Lastly, note that although you had asked about cardinals $\lambda$ that are the images of $\kappa$ under an embedding, nevertheless such images are not always cardinals. For example, if $\mu$ is a measure on $\kappa$ with ultrapower $j:V\to M$, then $j(\kappa)$ is never a cardinal, since it is strictly between $2^\kappa$ and $(2^\kappa)^+$. So one might want a criterion for recognizing when an ordinal $\lambda$ is the image of $\kappa$ under an embedding, and I think that is a much subtler question.

  • $\begingroup$ Thanks Joel, very nice answer! I had not realised that when we iterate a normal measure, the critical sequence is a club. Although this is a totally satisfying answer, I am still hoping that the existence of a suitable ultrafilter (on $P(\lambda)$ or on $[\lambda]^\kappa$) could also give the same result, like in the case of hugeness. Also, I agree with your final remark, it would be interesting to have absolute control on where the critical point is mapped. $\endgroup$ Commented Feb 1, 2017 at 15:23
  • $\begingroup$ That is a subtle point, @Stamatis, that the embedding might not be the result of an ultrapower by a normal (or any) measure. $\endgroup$
    – Asaf Karagila
    Commented Feb 1, 2017 at 19:04
  • $\begingroup$ Actually, it's a theorem @StamatisDimopoulos that the existence of a $\kappa$ such that ultrapowers have $j(\kappa)$ arbitrarily large is equiconsistent with the existence of a strong cardinal. Such a $\kappa$ is called a "tall" cardinal. $\endgroup$ Commented Oct 5, 2018 at 0:03
  • $\begingroup$ @KeithMillar Retrospectively, I agree with you, we need at least a strong cardinal. However, isn't there a distinction between what Hamkins calls tall and strongly tall cardinals? In the former case we have merely extender embeddings, while in the latter we have ultrapower embeddings. I had the impression that we still don't know the exact consistency strength of strongly tall cardinals, do you know if that has changed? $\endgroup$ Commented Oct 5, 2018 at 10:16
  • $\begingroup$ @StamatisDimopoulos I believe this is a debate of semantics; Jech's set theory (the textbook I used mainly) considers ultrapowers to include extender embeddings. However, I also think that when I wrote this comment I did in fact mean ultrapowers on measures, and I'm not sure what I was thinking. $\endgroup$ Commented Oct 7, 2018 at 4:16

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