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I have read several times that assuming Con(ZFC), and using compactness it can be proved the existence of a model of ZFC with an ill-founded $\omega$. How is that? Any reference will be welcome.

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When you say "ill-founded $\omega$" what exactly do you mean? – Asaf Karagila Oct 2 '10 at 9:53
Not well-founded, i. e. with infinite descending epsylon-chains. In case it is not clear enough, I am asking for some (sketch of a) proof for that result, not for an explanation of why such models exist or for what they look like. – Marc Alcobé García Oct 2 '10 at 10:24
up vote 10 down vote accepted

This is a standard application of the Compactness Theorem, and works basically the same in producing nonstandard models of ZFC as it does for producing nonstandard models of PA or real-closed fields.

Consider the theory $T$, in the language of set theory augmented with an additional constant symbol $c$, consisting of all the ZFC axioms, plus the assertions that $c$ is a natural number, but not equal to $0$, not equal to $1$, and so on, including for each natural number $n$ the statement $\varphi_n$ that $c$ is not equal to $n$. (Note that all such $n$ are definable in set theory, and so we use the definition of $n$ in $\varphi_n$ when asserting that $c$ is not $n$.)

If there is a model $M$ of ZFC, then every finite subtheory of $T$ is consistent, since any finite subtheory of $T$ makes only finitely many assertions about $c$, and we may therefore interpret $c$ as any natural number of $M$ not mentioned in the subtheory.

Thus, by compactness, $T$ has a model. Any such model of $T$ will be $\omega$-nonstandard, since the interpretation of $c$ in the model will be a nonstandard natural number.

There are numerous other ways to produce $\omega$-nonstandard models of ZFC from existing models, the most common being ultrapowers.

The conclusion is that if there is any model of ZFC, then there are nonstandard models of ZFC. The converse of this is not true, for if there are any transitive models of ZFC, then there is an $\in$-minimal such model $M$, and being standard, $M$ will have the same arithmetic as the ambient universe, thus thinking Con(ZFC), but being minimal, will have no transitive models of ZFC inside it.

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This is perfectly right. Another compactness construction, that doesn't go through $\omega$, is to simply add an infinite set of new constant symbols $c_1, c_2, \ldots$ and for each $i$ an axiom $c_{i+1} \in c_i$. – Carl Mummert Oct 2 '10 at 12:46
Carl, yes, that will produce a nonstandard model, but not necessarily an $\omega$-nonstandard model, which the OP requested. But you could just add the assertions that $c_i$ was a natural number. – Joel David Hamkins Oct 2 '10 at 12:57
Agreed; the original question is about $\omega$ and your answer addresses that. In a comment below the question, the issue of infinite descending $\in$ chains is mentioned. The construction I mentioned only gives that, but it avoids the complication of $\omega$. – Carl Mummert Oct 2 '10 at 13:08

I'm not sure, but maybe the question is why a non-ω-model of ZFC must have infinite decreasing $\in$ chains. This follows from two facts:

  • Viewed externally, any nonstandard model of Peano arithmetic is non-well-founded. Given a nonstandard number $n$, consider the sequence $n-1$, $n-2$, ...

  • The standard construction of a model of Peano arithmetic within ZFC uses $\in$ for the order relation on the natural numbers. So if the natural numbers within a model of set theory are not well founded, then the model's set inclusion relation is also not well founded. These are, formally, different meanings of "well founded".

It's interesting to ask how the model can think that its natural numbers are well founded, given that the descending sequence above was completely concrete. The answer is that if that sequence is defined within the model, as $a_k = n-k$, then $k$ can be any number in the model, even a nonstandard one, and the model verifies that this sequence reaches $0$ after $n$ steps. Only from an external viewpoint can we limit $k$ to an external set of "standard" natural numbers.

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Thanks to both for your answers. – Marc Alcobé García Oct 2 '10 at 16:09
Shameless plug for a similar question: – Adam Dec 7 '10 at 2:12

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