The following question was posted to MathStackExchange (original here). As there were no comments/answers on the original, I have ported it unedited.

I am interested in determining the cardinality of $j(\kappa)$ when $j\colon V\to M$ is an elementary embedding arising from an ultrapower embedding obtained through $\kappa$, with $\kappa$ an $\aleph_1$-strongly compact cardinal.

We say that $\kappa$ is $\aleph_1$-strongly compact if for all $\lambda\geq\kappa$ there is a fine $\sigma$-complete ultrafilter $\mathcal{U}$ on $\mathscr{P}_\kappa(\lambda)=\{x\subseteq\lambda\mid|x|<\kappa\}$. Here, $\mathcal{U}$ is fine if for all $\alpha<\lambda$, $\{x\in\mathscr{P}_\kappa(\lambda)\mid\alpha\in x\}\in\mathcal{U}$. Here I am taking $\lambda$ to have cofinality at least $\kappa$ and such that $|\mathscr{P}_\kappa(\lambda)|=\lambda$ (though I am very interested in being able to violate either of these rules).

When taking the ultrapower embedding, that is $j\colon V\to M$ obtained from $\mathcal{U}$ (where $M$ is the transitive collapse of the ultrapower), we have that $j``\lambda\subseteq[\operatorname{id}]\in M$, and $M\vDash|[\operatorname{id}]|<j(\kappa)$, so certainly $\lambda\leq j(\kappa)$. Furthermore, since $j(\kappa)=\{[f]\mid f\colon\mathscr{P}_\kappa(\lambda)\to\kappa\}$, we have $|j(\kappa)|<|\kappa^{\mathscr{P}_\kappa(\lambda)}|^+=(2^\kappa)^+$.

This gives us that $\lambda\leq j(\kappa)<(2^\lambda)^+$, so my question is: Can we control this further? Perhaps by imposing more restrictions on $\mathcal{U}$ before implementing the ultrapower.

My hope is that either for any such $\lambda$ we can guarantee that $2^\lambda\leq j(\kappa)$; or that for any such $\lambda$ we can guarantee that $|j(\kappa)|=\lambda$.

$\aleph_1$-strong compactness is not a very strong hypothesis, so if we are unable to control $j(\kappa)$ in any meaningful way, would we be able to do so with a stronger hypothesis? I know that, e.g. strong compactness would be sufficient, so can we do better than that?

  • $\begingroup$ Your variable names for $\lambda$ and $\kappa$ are backwards from the usual convention, for which one considers fine measures on $P_\kappa\lambda$ rather than $P_\lambda\kappa$. $\endgroup$ Oct 31, 2023 at 13:10
  • $\begingroup$ @JoelDavidHamkins This is an unfortunate consequence of me following some old work of Kunen in which he uses $\lambda$ for the strongly compact cardinal. I probably should have swapped them in the question before posting it. $\endgroup$ Oct 31, 2023 at 13:56
  • 3
    $\begingroup$ I think it might be worthwhile still to swap them, since this convention is quite established now. I think many MO readers will appreciate it. $\endgroup$ Oct 31, 2023 at 14:02
  • $\begingroup$ Are you interested in examples where a tighter bound is possible? I can do that. $\endgroup$ Oct 31, 2023 at 14:42
  • $\begingroup$ @JoelDavidHamkins That would be great! $\endgroup$ Oct 31, 2023 at 14:53

2 Answers 2


If $2^\gamma \leq \kappa$ for all $\gamma < \kappa$ (e.g., if GCH holds and $\kappa$ is the least $\aleph_1$-strongly compact), then $|j(\kappa)| \geq 2^\lambda$. To see this, let $\sigma = [\text{id}]$. By elementarity, $M\vDash j(\kappa) \geq |P^M(\sigma)|$. But there is an injection $i : P(\lambda)\to P^M(\sigma)$ given by $i(A) = j(A)\cap \sigma$, so $|P^M(\sigma)| \geq 2^\lambda$.

  • $\begingroup$ GCH seems quite a strong additional assumption on top of the existence of an $\aleph_1$-strongly compact cardinal; do you know if it is forcable from merely 'there exists an $\aleph_1$-strongly compact cardinal'? $\endgroup$ Nov 1, 2023 at 10:11
  • $\begingroup$ In fact, it is an open question whether given a model containing a strongly compact cardinal, it is possible to force GCH and preserve the strongly compact (see Apter-Dimopoulos-Usuba's "Strongly compact cardinals and the continuum function," Question 5.1). I think it is an interesting question whether one can start with an $\omega_1$-strongly compact and force to make the first $\omega_1$-strongly compact a strong limit cardinal; maybe this is possible just by "pushing up" the first $\omega_1$-strongly compact. More information in Gitik's "On $\sigma$-complete uniform ultrafilters." $\endgroup$ Nov 1, 2023 at 14:58

I want merely to point out an unusual feature of your large cardinal conception. Namely, since you have not required the measure to be $\kappa$-complete, it seems that according to your definition the $\aleph_1$-strongly compact cardinals are closed upward. That is, if $\kappa_0$ is $\aleph_1$-strongly compact, then any larger $\kappa\geq\kappa_0$ is also $\aleph_1$-strongly compact, since $P_{\kappa_0}(\lambda)\subseteq P_{\kappa}(\lambda)$, and we can just use the measure arising from $\kappa_0$, but as a witness for $\kappa$. It will still be fine and $\sigma$-complete.

The embedding $j$ could have critical point $\kappa_0$, or less, rather than $\kappa$.

This kind of situation can be used to make interesting examples that relate to the weakenings of your assumptions that you mention. For example, perhaps although $\lambda^{<\kappa_0}=\lambda$, we might have $\lambda^{<\kappa}$ much bigger, but all the calculations with the measure will arise from $\kappa_0$.

  • 2
    $\begingroup$ I guess we have Bagaria-Magidor to thank for the unusual definition, not OP. $\endgroup$ Oct 31, 2023 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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