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My understanding (please correct me if I'm wrong) is that if you have some transitive set M which is an $\epsilon$-model of ZFC, and you take an ultrapower of it using an approprate ultrafilter, you wind up with a new model whose membership relation is not the $\epsilon$ relation of the ambient set theory, but still satisfies ZFC. Furthermore, if the membership relation of the ultrapower is well-founded, one can always use the Mostowski collapse theorem to produce an isomorphic $\epsilon$-model.

My question is this: how could one possibly end up with a model of ZFC which satisfies the axiom of regularity ("every set is disjoint from one of its members"), yet whose membership relation isn't well-founded?

I'm struggling to imagine this; the most I can come up with is that for no set in the infinite chain is its transitive closure also a set (of the model). But I'm skeptical about whether or not that can be the case, because it seems like you ought to be able to construct the transitive closure using definition by transfinite recursion, letting $f(0)$ be any set in the chain and $f(n+1)=\bigcup f(n)$ (axiom of union). Then $f(\omega)$ (axiom of infinity) ought to contain all the sets needed to build a contradiction to regularity.

Sorry if this question sounds like I'm arguing with myself. This has been bothering me for a few days now.

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For an ultrapower of $V$ by an ultrafilter $\mathcal{U}$ there is an exact characterization of when the ultrapower will be well-founded: precisely when $\mathcal{U}$ is closed under countable intersections.

As for how a model $M$ could possibly satisfy regularity but not be wellfounded, the problem is that there may be infinite descending chains that $M$ cannot 'see': each individual object may belong to $M$ but the chain itself may not. I can be a little more precise. Let's say $R$ is what $M$ understands to be the $\in$-relation. There may well be $x_0,\ldots x_n,\ldots $ belonging to $M$ so that $x_{n+1}Rx_n$ for all $n\in\omega$; as long as the sequence $\langle x_n:n\in\omega\rangle$ does not belong to $M$ the axiom of regularity from $M$'s point of view need not be violated.

You don't need ultrapowers to construct such models. Assuming CON(ZFC) you can build one using the compactness theorem for first order logic.

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    $\begingroup$ Thanks Justin! But, as I mentioned in my question, if all the sets of the chain are sets of the model, and the model has an infinite ordinal (whose image under the membership relation of the model will necessarily be infinite, since [in]finiteness is absolute), then the sequence must also belong to the model by the axiom of replacement. $\endgroup$
    – Adam
    Commented Feb 8, 2010 at 9:12
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    $\begingroup$ Not true; just because $x_0,x_1...x_n...$ all belong to a model there is no reason to believe the set containing them all belongs to the model. (Otherwise a model should contain all its countable subsets, and then how could we have countable models of ZFC?) Look at what you are saying more formally and try to see where it breaks down. What set and formula are you attempting to apply replacement to? $\endgroup$
    – Justin Palumbo
    Commented Feb 8, 2010 at 9:21
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    $\begingroup$ Transitive closure is definable, and if say x is the first element of the chain then indeed tc(x) (or more precisely what M believes to be the transitive closure) will belong to M. But even with the entire chain a subset of tc(x), that still won't violate regularity inside M, because the chain itself need not belong to M. $\endgroup$
    – Justin Palumbo
    Commented Feb 8, 2010 at 9:39
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    $\begingroup$ The ill-founded model has a fake omega, a nonstandard omega with nonstandard natural numbers, that look infinite from the outside, but which the model believes to be finite. It can produce what it thinks is a finite descending sequence properly extending the actual descending epsilon sequence, but this still doesn't violate Foundation because it would need an infinite sequence using its own omega for a counterexample. $\endgroup$
    – Joel David Hamkins
    Commented Feb 8, 2010 at 14:05
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    $\begingroup$ For ultrapowers of the universe, you can prove that they are ill-founded if and only if they have an ill-founded omega. This is because both of these are equivalent to the countable incompleteness of the ultrafilter. The ill-foundedness, if it exists at all, must exist already in the natural numbers. But for other nonstandard models of set theory, which are not obtained by ultrapowers, the ill-foundedness can begin much higher up. $\endgroup$
    – Joel David Hamkins
    Commented Feb 8, 2010 at 18:40

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