Theorem 19.7 (by Gaifman) is the following result:
Let $U$ be a $\kappa$-complete nonprincipal ultrafilter on $\kappa$. Then for every $\alpha$, the $\alpha$th iterated ultrapower $\operatorname{Ult}^{(\alpha)}$ is well-founded.
The first paragraph of the proof states the following:
Clearly if $\operatorname{Ult}^{(\alpha)}$ is well-founded, then $\operatorname{Ult}^{(\alpha+1)}$ is well-founded. Thus if $\gamma$ is the least $\gamma$ such that $\operatorname{Ult}^{(\gamma)}$ is not well-founded, then $\gamma$ is a limit ordinal.The ordinals of the model $\operatorname{Ult}^{(\gamma)}$ are not well-ordered; let $\xi$ be the least ordinal such that the ordinals of $\operatorname{Ult}^{(\gamma)}$ below $i_{0,\gamma}(\xi)$ are not well-founded.
My question is: How do we know that the ordinals of the model $\operatorname{Ult}^{(\gamma)}$ is not well-ordered? The naive approach is to take a decreasing sequence of sets $x_0 \ni x_1 \ni \cdots$, then take rank to yield $\operatorname{rank}(x_0) > \operatorname{rank}(x_1) > \cdots$, but I don't think $\operatorname{rank}$ is defined for non-well-founded models.
Any help is appreciated.
Best Answer
There are two ways in which well-foundedness can fail.
Internally, where the model knows that it fails, i.e. the axiom of foundation is false in the model.
Externally, where the model thinks that the axiom of foundation is true, but we know from outside the model that the relation is not really well-founded.
In the latter case you do have a rank function, since the rank function is internal, and there are no decreasing sequences of ordinals in that model. But externally there is a decreasing sequence nonetheless.
This is akin to non-standard models of $\sf PA$. The model is not well-ordered, and we know that, but the induction axioms are equivalent to stating that any definable set which is not empty has a least element, so this is not something the model knows about.
Note that in the context of ultrapowers, the models are all models of $\sf ZFC$, or a significant fragment thereof which contains the axiom of foundation, and therefore they are all going to fall into the second category.
Consider, if you will, a free ultrafilter $U$ on $\omega$, and consider $V^\omega/U$. This is a definable class, with $E$ denoting its membership relation also being definable. This is an ultrapower of the universe. But it is not well-founded, externally, since $\omega^\omega/U$ is already not well-ordered.
(In the former case you will have a witness that is not well-founded, yes. And the ordinals may or may not be well-founded.)