Definition of nonstandard models without enumerations

To define nonstandard models of Peano arithmetic or set theory, many articles use enumerations like $$x>0, x>1, \dots$$ where $$x$$ is said to be a nonstandard natural number, ie a number that is finite from the PA model's point of view, but infinite from outside.

I find those enumerations confusing, especially when trying to construct a subtle notion such as a finite infinity. What kind of infinity is hidden in the dots "$$\dots$$" ? Some articles acknowledge the problem, and quickly offer to repair it by the compactness theorem. Then they consider a theory starting with the Peano axioms and adding a new constant symbol $$c$$, with axioms $$c>0, c>1, \dots$$ And that's even more confusing.

So far, the best definition I found of a nonstandard PA model is a structure $$(M,0,S,+,\times,\leq)$$ that satisfies the Peano axioms, and such as the order $$\leq$$ on $$M$$ is not a well-order. It may be more abstract than an enumeration "$$\dots$$", but I think it has the merit of being precise.

Is there a similar definition of a nonstandard model of ZF, not using enumerations or "intuitive" natural numbers ? It's more difficult because it concerns a model of PA inside a model of ZF. I don't manage to access the inner PA order $$\leq$$ from the ambiant logic.

• For a simple definition (for the PA case), couldn't you just say a non-standard model of PA is a model which is not isomorphic to the standard model? That tells you nothing about their structure or why they exist, but it's short 😁
– Ned
Oct 12 '18 at 13:47

The problem is that being standard or non-standard is in relation to your meta-theory.

Suppose that $$M$$ is a non-standard model of $$\sf ZFC$$, say with uncountably many "finite ordinals". Then the natural numbers of $$M$$ is a non-standard model of $$\sf PA$$. But as far as $$M$$ is concerned, it is the standard model of $$\sf PA$$. Or, it might be that $$N$$ is a standard model of $$\sf ZFC$$ inside $$M$$, but it is certainly not going to be a standard model outside of $$M$$ because we know that the natural numbers of $$M$$ and $$N$$ are ill-founded.

This is why you need to resort to that "$$...$$" that you dislike. Because you're appealing to the meta-theory's integers. The simplest way to formalize this is by saying that there exists an $$x$$ such that $$x$$ is not a numeral, where a numeral is a constant term in the language of arithmetic which is obtained by iterative application of the successor function symbol to the $$0$$ symbol.

Now, as for a non-standard model of $$\sf ZFC$$, yes there is such a thing. But first you need to learn that the term "standard model" is different in set theory and in arithmetic. In set theory it simply means that the model's $$\in$$-relation is the meta-theory's one. Usually we require that the model is also transitive, but $$\in$$ is a well-founded and extensional relation, so by the Mostowski collapse lemma a standard model is isomorphic to a transitive model, and often the two are used interchangeably.

A non-standard model, therefore, is one whose $$\in$$ is not the real $$\in$$. Or rather, not isomorphic to a standard model. In other words, it is a model of $$\sf ZFC$$, $$(M,E)$$ such that $$E$$ is not a well-founded relation on $$M$$. The axiom of regularity, however, does tell us that it is internally well-founded. There is no set in $$M$$ which is a counterexample to well-foundedness.

There are two additional points here:

1. It is possible that there are models of $$\sf ZFC$$ which are not standard models, but their $$\omega$$ is in fact standard. These are called $$\omega$$-models.

2. The consistency of "There is a standard model" or even "there is an $$\omega$$-model" is greater than that of "there is a model of $$\sf ZFC$$", specifically because an $$\omega$$-model (and thus a standard model too) agree with the universe on things like consistency statement. So if there is an $$\omega$$-model, it must satisfy $$\operatorname{Con}\sf (ZFC)$$, which is something that $$\sf ZFC$$ by itself cannot prove.