Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In $\textit{Set Theory}$ by Jech 1978 edition, in the proof of Lemma 32.5 which you can hopefully see at the Google book link.

In the course of the proof using the tree property, he produces from any weakly compact cardinal $\kappa$ a non principal $L_\alpha$-ultrafilter $U$ on $\kappa$, which is $L_\alpha$-$\kappa$ complete, and moreover the intersection of $\kappa$ many elements of $U$ (taken in $V$) is nonempty.

The latter implies that countable intersection (taken in $V$) is nonempty. This fact implies iterability for example by 19.13 of Kanamori. So $L_\alpha$ with this ultrafilter can be iterated of length $\text{Ord}$

However, $0^\sharp$ follows from the existence of a mouse (iterable premouse). But $0^\sharp$ has stronger consistency strength than a weakly compact.

Is this $(L_\alpha, U)$ not a pre-mouse? I have found varying definition of pre-mouse. One definition has the addition condition that $\kappa$ should be the largest cardinal of $L_\alpha$. This seems like it may not hold since at the beginning of the proof, the proof chooses an $\alpha$ such that $L_\alpha \models ZF^-$.

I would like to have the $\text{Ord}$ length iteration for what I am trying to do; however, I am troubled by whether or not this would give an iterable premouse and hence imply sharps.

Thanks for any clarification you can provide.

share|improve this question
It should be pointed, I suppose, that if $L$ miscalculates the successor of a weakly compact cardinal then $0^\#$ exists. –  Asaf Karagila Feb 18 at 22:09

2 Answers 2

up vote 7 down vote accepted

This is a very nice question that reveals a subtle point about iterating ultrapowers. Namely, the issue is that in order to iterate the ultrapower construction, one needs the ultrafilter $U$ to be weakly amenable to the structure, which means that the structure has $\{\alpha\lt\kappa\mid X_\alpha\in U\}$, whenever it has $\langle X_\alpha\mid\alpha\lt\kappa\rangle$. In other words, the structure knows about how the ultrafilter works on its size $\kappa$ families of sets. The need for weak amenability arises naturally when you want to define the second-step of the ultrapower, specifically, when you want to define the image of $U$ on the ultrapower structure. One can show that weak amenability amounts to the assertion that the power set of $\kappa$ does not enlarge under the ultrapower. And this is clearly stronger than weak compactness, since it supports reflection arguments, like this: if $j:M\to N$ has critical point $\kappa$ and $P(\kappa)^M=P(\kappa)^N$, then $\kappa$ must have the tree property in $N$, since every $\kappa$-tree has a $\kappa$-branch in $N$, and the trees and branches are the same in the two models. Thus, there must be many weakly compact cardinals below $\kappa$. The iterable cardinals, introduced by Victoria Gitman, are concerned precisely with this issue.

Getting back to your question, the situation is that in Jech's context, with a weakly compact cardinal, while you do get the $\kappa$-complete ultrafilter, nevertheless you do not in general get an ultrafilter that is weakly amenable to the structure $L_\alpha$ you mention. In particular, the power set of $\kappa$ in $L_\alpha$ and the target structure $L_\beta$ may not be the same. And without weak amenability, there is no natural way to define the ultrafilter on the ultrapower and no way to continue the iteration, even for two steps.

Meanwhile, in Kanamori's context, the weak amenability hypothesis is a part of the framework in chapter 19, and so there is no contradiction here.

share|improve this answer
Sorry for the redundancies, Joel! I didn't see that you posted more details before I posted my reply. –  Victoria Gitman Feb 18 at 21:33
No problem! It is great to have fuller information, and I'm glad that you posted... –  Joel David Hamkins Feb 18 at 21:35
Very nice answers (both of you Victoria and Joel). The issue of amenability is indeed important. For potential premice having extenders, we want the extenders to be amenably coded in the structure. It allows us to check Los Theorem for $\Sigma_0$ formulas and he satisfaction predicate for $\Sigma_1$ formulas is $\Sigma_1$. –  Carlo Von Schnitzel Feb 19 at 2:34

I would like to elaborate a little on Joel's answer. To carry out the iterated ultrapower construction with an ultrafilter that is external to the model, as it is in the case of $L_\alpha$ and $U$, requires making extra assumptions. In the case of an ultrafilter that is an element of your model, the successor stages of the construction simply use the image of the ultrafilter from the previous stage. But we cannot apply $j$ to $U$ because $U$ is not an element of $L_\alpha$. To define the successor stage ultrafilter we need that $U$ is "partially internal" to $M$. This translates into the technical requirement that $U$ is weakly amenable: for every $X\in L_\alpha$ such that $|X|^{L_\alpha}=\kappa$, we have $X\cap U\in L_\alpha$. Weak amenability is equivalent to $L_\alpha$ and its ultrapower having the same subsets of $\kappa$, as Joel mentioned.

It is easy to see that assuming that every $A\subseteq\kappa$ is contained in a transitive model $M$ of size $\kappa$ for which there is a weakly amenable $M$-ultrafilter on $\kappa$ (measuring sets in $M$) with a well-founded ultrapower is a much stronger large cardinal assumption than a weakly compact cardinal. I call such cardinals 1-iterable. Assuming that such weakly amenable $M$-ultrafilter is additionally countably complete characterizes Ramsey cardinals, which imply $0^\sharp$. Meanwhile, it is possible to have $M$-ultrafilters with varying degrees of iterability. An $\alpha$-iterable cardinal is characterized by the existence of weakly amenable $M$-ultrafilter with $\alpha$-many well-founded iterated ultrapowers and $\alpha$-iterable cardinals form a hierarchy of strength below Ramsey cardinals. We introduced and studied these cardinals together with Philip Welch.

share|improve this answer
Call $U$ $\Sigma_1$-amenable with respect to $M$ if $U$ is $\boldsymbol \Sigma_1(M)$. Is it possible to characterize a large cardinal using $\Sigma_1$-amenability and countable completeness? –  Carlo Von Schnitzel Feb 19 at 4:04
Carlo, can you clarify what you mean by $\Sigma_1$-amenable? –  Victoria Gitman Feb 19 at 13:18
I want to say something like: the ultrafilter $U$ is boldface $\Sigma_1$ definable from parameter in $M$, or we can find a $\Sigma_1$ formula $\phi$ such that $X_{\alpha} \in U$ iff $\phi(\alpha, x)$, $x$ some parameters in $M$. When $U$ is weakly amenable then $L_{\alpha}$ and its ultrapower have the same subsets of $\kappa$, and when $U$ is $\Sigma_1$-amenable then $L_{\alpha}$ and its ultrapower should have the same subsets of $\kappa$ which are $\Sigma_1$ definable. One property can hold without the other (IIRC). I was curious to see that even if one property can hold without the other... –  Carlo Von Schnitzel Feb 19 at 22:40
...the "$\Sigma_1$ amenability" would still characterize a (possibly) weaker large cardinal than Ramsey. I am trying to make this idea precise. –  Carlo Von Schnitzel Feb 19 at 22:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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