Upon learning about nonstandard models of arithmetic, one may respond that but yeah you can define ℕ in set theory so #whocare. This is this kicking the can down the road—there are nonstandard models of ZF whose ℕ is some ill-founded thing. 1/

One could then make a (quasi-)philosophical argument that N can be uniquely identified. This thread isn't about that. This thread is about the nonstandard models of arithmetic who are the ℕ of some model of ZF. Let's call these guys ZF-standard. Can we characterize them?

To warmup, let's look for necessary conditions. An easy one is the theory of the model. ZF proves all kinds of number theoretic theorems, e.g. consistency statements, so any ZF-standard model of arithmetic must have those ZF theorems in its theory.

Thinking of another necessary conditions is probably harder. I'll mention one next tweet. (The reader is encouraged to stop reading this thread and think for a moment before continuing on.)

We back?

Tarskian satisfaction classes. ZF proves that you can take any first-order structure and attach to it a class saying what formulae are true of what elements. Formulae are just numbers (by Gödel coding), so we can think of the satisfaction class as a subset of the model.

More, this satisfaction class has to decide the truth of all formulae, even the nonstandard ones. In jargon: it is a full satisfaction class. This is because the recursive definition inside the nonstandard model of ZF goes through all finite stages, even the nonstandard ones.

Punchline: these two conditions are sufficient, for countable models.

Theorem (Ali Enayat): A countable model of arithmetic is ZF-standard iff it satisfies all ZF-theorems of arithmetic and it admits a full satisfaction class. https://www.researchgate.net/publication/281936805_STANDARD_MODELS_OF_ARITHMETIC

I promise a sketch of the proof. But first let's see why this second condition is nontrivial. This is best seen via an equivalent formulation in terms of saturation. Indeed we have two equivalent formulations like this.

(1) Recursive saturation. A structure is recursively saturated if: given any computable set p(x) of formulae φ(x), if "∃x s.t. φ(x) for all φ(x) ∈ p(x)" is consistent with the theory of the structure, then you can find such an x already in the structure.

(2) Resplendency. A structure is resplendent if: given any formula Φ(X) with a new predicate symbol X, if "∃X Φ(X)" is consistent with the theory of the structure, then you can expand it by adding such an X as a predicate.

In the uncountable case, resplendency -> admitting a satisfaction class -> recursive saturation, and all can be separated. But they turn out to be equivalent for countable nonstandard models. Showing such is far beyond a twitter thread :P

To see nontriviality: Consider the computable set of formulae of the form "if φ(y) defines a unique y, then x > y". By compactness this must be consistent, so any rec. sat. model of arithmetic must contain undefinable elements, ruling out e.g. Skolem hulls from being rec. sat.

Now that we know the two conditions are nontrivial, let's prove Enayat's theorem.

*ahem*

By resplendency.

Quod erat demostratum.

In more detail, we want to show our nonstandard model of arithmetic M is ZF-standard. So we need to somehow produce the model of ZF whose natural numbers are exactly M. We will produce this model of ZF as a subset of M.

Namely we want E ⊆ M^2 so that (M,E) satisfies ZF and an isomorphism f: ℕ^(M,E) --> M. We can think of E and f as coded by a single subset X of M, and write down a formula Φ(X) which describes the desired X.

There's work to check that Φ(X) is consistent with the theory of M. But once we've done that, the resplendency of M—because M is countable + admits a satisfaction class—gives us the X witnessing that M is ZF-standard.

What about the uncountable case? Good question, let's chat later.