I am reading the paper Ramsey-like cardinals II by Victoria Gitman and Philip Welch (Journal of Symbolic Logic, vol. 76, no. 2. pp. 541-560, 2011) and maybe I am missing something.

According to the deffinitions provided there:

An $M_0$-ultrafilter $U_0$ is $0$-good if its ultrapower $M_1$ is well-founded.

$U_0$ is $1$-good if it is weakly amenable and $0$-good.

$U_0$ is $2$-good if $U_1=j_{01}(U_0)$ determines a well-founded ultrapower $M_2$ of $M_1$.

Hence, if $U_0$ is $2$-good, we have two well-founded ultrapowers, $M_1$ and $M_2$ and, in general, if $U_0$ is $n$-good we have $n$ well-founded ultrapowers, $M_1,\ldots, M_n$.

However, $U_0$ is $\omega$-good if it provides a sequence of well-founded ultrapowers $M_1,M_2,\ldots$, but the inductive limit $M_\omega$ is not necessarily well-founded. When this happens, $U_0$ is said to be $\omega+1$-good, but the ultrapower $M_{\omega+1}$ must be ill-founded. $U_0$ is $\omega+2$-good if $M_{\omega+1}$ is well-founded, and so on.

So, if I have understood the definition, for $n<\omega$, when $U_0$ is $n+1$-good the ultrapower $M_{n+1}$ is well-founded, but for $\beta\geq \omega$, when $U_0$ is $\beta+1$-good the ultrapower $M_\beta$ is well-founded, but $M_{\beta+1}$ may be ill-founded.

I insist on this because, for instance, in Theorem 4.1 we have an $\alpha$-good ultrafilter, with $\alpha=\beta+1$, and so, if I am right, the ultrapower $M_\beta$ is well-founded, but we cannot say that $M_\alpha$ is also well-founded. However, the proof requires to make a complex construction within $M_\alpha$, and it seems strange to me that nowhere in the text is said something like "notice that $M_\alpha$ maybe ill-founded". In fact, I am not sure about how to handle this situation, since until now I have worked with well-founded ultrapowers only, but before trying to adapt the proof of Theorem 3.11 as indicated, I have preferred to ask here, just in case $M_\alpha$ is well-founded and I am missing why it is so.

  • $\begingroup$ Carlos, thanks for your careful reading. I think that after writing down the definition I completely ignored the subtlety of the limit stages, which you observed. $\endgroup$ Oct 4, 2014 at 18:32

1 Answer 1


The objection is correct. The argument of Theorem 4.1 uses that $M_\alpha$ is well-founded, which may not be the case if $\beta$ is a limit, according to the definition of $\alpha$-iterable cardinals given in the paper. There are two ways to address this. It does make sense to redefine the limit stages of the hierarchy to say that the direct limit is well-founded. I believe that in this case, Theorem 4.1 is correct. The new definition would say that a cardinal is $\alpha$-iterable if for every $A\subseteq\kappa$, there is a weak $\kappa$-model $M$ with $A\in M$ and a weakly amenable $M$-ultrafilter on $M$ whose $\alpha^{\text{th}}$-iterate model is well-founded. I am not sure how satisfied I am with changing the definition. I am still hoping that the increasing strength of the hierarchy can be proved with the original definition. The question is this: suppose that $\beta$ is a limit cardinal and $M_\beta$ is the well-founded direct limit of an iteration of some $M$ by an $M$-ultrafilter $U$. Does it hold in $M_\beta$, that every $a\subseteq\kappa$ is contained in a weak $\kappa$-model $m$ for which there is a weakly amenable $m$-ultrafilter $u$ producing an iteration of length $\beta$ (with a possibly ill-founded direct limit)? Currently, I don't know how to do it, although I have some ideas that I will write about later if they work out. This is related to another erroneous claim made later in the paper that we can assume in the definition of $\alpha$-iterable cardinals for $\alpha\geq 2$ that $M\models{\rm ZFC}$ (as opposed to ${\rm ZFC}$ without powerset). It is my conjecture that this assumption actually increases the consistency strength of the large cardinal notion. It seems that the argument of Theorem 4.9, which shows the statement in the case of $1$-iterable cardinals should generalize to all $\alpha$-iterable cardinals particularly with the revised definition.

Update: I now think that Theorem 4.1 is correct for the definition of $\alpha$-iterable cardinals given in the paper. To fill the gap in the original argument, we need to show the following. Suppose that $M$ is a weak $\kappa$-model with an $M$-ultrafilter $U$ and $\beta$ is a limit ordinal such that $U$ can be iterated $\beta+1$-many times, meaning that the direct limit $M_\beta$ of the first $\beta$-many steps of the iteration is well-founded. We must show that in $M_\beta$, every subset of $\kappa$ can be put into a weak $\kappa$-model $M$ for which there is a $M$-ultrafilter producing $\beta$-many well-founded iterates with a possibly ill-founded direct limit. For the tree construction of the proof of Theorem 4.1 to go through, it suffices to show that all embeddings $j_{\xi\gamma}^i:M^i_\xi\to M^i_\gamma$ are elements of some set in $M_\beta$ (so that we can construct the tree on a set, whose branch will gives the desired directed system). Let $\{\kappa_\xi\mid\xi\leq\beta\}$ be the critical sequence of the iteration and assume that $V_\kappa\in M$. I want to argue that all $j_{\xi\gamma}$ are in $V_{\kappa_\beta}^{M_\beta}$. For clarity, I will give the argument for $\beta=\omega$.

First, consider $f=j_{01}^i:M^i_0\to M^i_1$. We know $f\in M_1$ and has size $\kappa_1$ there (because $M^i_0$ has size $\kappa$ and $M^i_1=j_{01}(M^i_0)$ has size $\kappa_1$). Thus, $f\in V_{\kappa_2}^{M_2}$. The map $j_{2\omega}:M_2\to M_\omega$ has critical point $\kappa_2$, from which it follows that $j_{2\omega}(f)=f\in j_{2\omega}(V_{\kappa_2})=V_{\kappa_\omega}$. Next, consider the general case $f=j_{mn}^i:M^i_m\to M^i_n$. We know that $f\in M_n$ and has size $\kappa_n$ there. Thus, $f\in V_{\kappa_{n+2}}^{M_{n+2}}$, and rest of the argument is the same.

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    $\begingroup$ Thank you for your answer, Victoria. I feel bad for having found that gap in your beautiful paper. I was convinced that the problem was mine. I also feel that the result is valid for the original definition, but unfortunately I am just an amateur, and there is little I can do. But I hope you will find the right proof soon. Of course, I will be pleased to read it when available. $\endgroup$
    – Carlos
    Oct 4, 2014 at 20:44
  • $\begingroup$ I am very glad that you are reading the paper carefully :). I am losing hope that the result is valid for the original definition and actually starting to think that maybe the more robust definition is the second one. The new definition seems to have the property that it is easy to see that assuming ZFC models increases consistency strength, which is a nice property. $\endgroup$ Oct 4, 2014 at 21:16
  • $\begingroup$ Congratulations!!! I was confident that you would find the right proof. I am a bit bussy now, but I will continue with your paper soon. $\endgroup$
    – Carlos
    Oct 7, 2014 at 22:53
  • $\begingroup$ @Carlos are you still interested in that detail in Kanamori's proof? I looked at it, but can't figure it out yet. I want to think about it a bit more. $\endgroup$ Oct 7, 2014 at 23:33
  • $\begingroup$ Oh, you are very kind. I would not want to waste your time. I have found a simpler proof in Boos' paper, so understanding Kanamori's proof is not too important for me anymore, but, of course, it is a pity having understood a long proof but a single step, so I am still curious about it. But I am afraid that my curiosity is not worth wasting your time. $\endgroup$
    – Carlos
    Oct 8, 2014 at 0:02

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