# Nonstandard Models of Arithmetic 27

MW: Enayat’s second major result is:

Theorem 7: Every countable recursively saturated model of PA+ΦT is a T-standard model of PA.

Recall the definition of ΦT: the set of formulas of the form

{φ → Con(Tn) : φ∈L(PA), n∈ℕ}

where Tn is the first n statements of T, and φ is φ∈L(PA) translated into a formula of L(ZF).

Notice the countability condition, which was not needed for the reverse result, Proposition 6. Enayat mentions that the quickest way to prove the theorem is via a replendence argument. Someday I may do a post about this, but the proof he gives is worth our attention. It falls into four parts. Let N be a countable recursively saturated model of PA+ΦT. Here’s a quick preview—details to come, so don’t worry if not all of it makes sense yet.

1. Using the Arithmetized Completeness Theorem, plus recursive saturation, he obtains a model U of T whose ω is both an end-extension of N and elementarily equivalent to it: ωUeN and Th(ωU)=Th(N). (Recall that Th(N) is the set of all sentences of L(PA) that hold in N, and likewise for Th(ωU).)
2. Next he observes that Proposition 6 applies to ωU, so it’s recursively saturated.
3. Since ωUeN, their standard systems are the same: SSy(ωU)=SSy(N). The standard system of a nonstandard model of PA is the family of subsets of ℕ that can be coded by elements of the model.
4. Finally, he notes that any two countable recursively saturated models of PA are isomorphic if they have the same first-order theory and the same standard system. That is, if Th(M)=Th(N) and SSy(M)=SSy(N) and M and N are countable and recursively saturated, then MN. So N is isomorphic to the ω of a model of T. But that’s the definition of T-standard. QED

The rest of this post expands on (1), with (2)–(4) relegated to later posts. Let’s start with the Arithmetized Completeness Theorem (ACT). I wrote a Topics post about this, but it’s pretty long so I’ll just cherry-pick the info we need.

The Completeness Theorem says that a consistent theory T has a model U. The ACT says that a consistent arithmetic (aka representable) theory has an arithmetic model. Arithmetic here means definable via formulas of L(PA). Thus, the set of Gödel numbers of the axioms of T is definable by a formula of L(PA): φ∈T iff θ(⌜φ⌝) holds, for some θ(x)∈L(PA). Likewise, the domain, relations, constants, and functions of U are all specified by formulas in L(PA).

Moreover, the ACT says this can all be proved inside PA. The proof of the ACT shows how, starting with a formula defining a consistent T, we can whip up formulas for the all the semantic aspects of U. Foremost among these is the truth predicate SatU: SatU:(⌜φ⌝) for a sentence φ expresses the fact that U satisfies φ.

It’s convenient to gloss over the distinction between a formula and its Gödel number, so I’ll mostly omit the brackets ⌜⌝ below. Imagine that formulas are numbers.

Because all this is happening inside PA, we can apply it to any model N. For nonstandard N, matters take a curious turn. First off, we’ll have nonstandard formulas, including some of nonstandard length. (If Formula(x) is the formalization of “x is a formula”, then any nonstandard d satisfying Formula(d) represents a nonstandard formula.)

The truth predicate SatU defines satisfaction for all sentences in the language of T, including nonstandard ones. The proof of the ACT shows that SatU extends ordinary standard satisfaction: if φ is standard, then SatU(φ) iff U⊧φ. To avoid confusion, I will use the symbol ⊧ only for satisfaction in the standard sense.

The notion of consistency also demands a second look. Con(T), a sentence in the language of arithmetic, says that there is no proof of a contradiction, not just no standard proof. An instructive example: by Gödel’s Second Incompleteness Theorem, the theory PA+¬Con(PA) is consistent. Suppose N is a model of it. Then N⊧¬Con(PA). So in N, there is a proof of a contradiction, but no standard proof.

The ACT demands that N⊧Con(T) before it goes to work. Showing that T is consistent in the standard sense doesn’t cut it.

OK, let’s look at Enayat’s step (1). We need to construct a model of T, starting with a model of PA. Not only that, but ΦT  is knocking on the door of Con(T). This suggests using the ACT. Larry Manevitz observed (in the 70s) something similar. Let N be a model of PA. Then

If N⊧Con(ZF), then there is a model of ZF whose ω is an end extension of N.

If N⊧Con(ZF), the ACT immediately hands us a model U of ZF. Manevitz’s result falls out from looking at the recursion that imbeds ℕ as an initial segment of any model (say M) of PA. You recursively define a map nnM with Successor(n)↦SuccessorM(nM). A straightforward induction shows that the image is an initial segment. This can all be formalized in PA, so it works for any model N of PA. In particular, N an be imbedded as an initial segment of ωU. That’s Manevitz’s result. It works without change for any arithmetic extension T of ZF: if N⊧Con(T), then there is a model of T whose ω is an end extension of N.

We just need to deal with two issues to get Enayat’s step (1):

• ΦT gives us Con(Tn) for any n∈ℕ. (Just let φ ≡ 1=1.) This is not quite the same as Con(T).
• We also need Th(ωU)=Th(N).

For the first bullet point, we might think we’re in the clear since Con(Tn) for all n∈ℕ implies that T is consistent. But we saw above that N⊧Con(T) is stronger. Overspill comes to the rescue. Since N⊧Con(Tn) for all n∈ℕ, we have N⊧Con(Td) for some nonstandard d. We can apply the ACT to Td to get a model U of it. But Td includes all the standard formulas of T, so UT with T interpreted in the standard sense.

For the second bullet point, we appeal to recursive saturation. The trick here: code Th(N) via a nonstandard element of N. Any element v of N can be regarded as a bitstring; let’s write vi for the i-th bit of v. Also write φi for the sentence with Gödel number i. If v is to code Th(N), we need vi=1 when N⊧φi and vi=0 when N⊧¬φi. In other words,

N⊧(φivi=1).

Now, the set {φivi=1 : i∈ℕ} is a recursive set of formulas with the free variable v. It is obviously finitely satisfiable, so it’s a recursive type. Recursive saturation says there is an element aN realizing this type:

N⊧(φiai=1), for all i∈ℕ.

Next step is to whip up a theory combining T and Th(N). Once again we do this first for finite fragments of the theory. Let

Γ(n) = Tn+{φi : i<n and ai=1}

N⊧ΦT tells us that N⊧Con(Γ(n)) for all n∈ℕ. By overspill, N⊧Con(Γ(d)) for some nonstandard d. Apply the ACT to Γ(d). It hands us a U satisfying Γ(d). A fortiori U is a model of T (standardly speaking) and of exactly those φi that hold in N. For the record, note that N “believes” there are nonstandard sentences φe, but if e>d then Γ(d) says nothing about them. Fortunately, we care about φi only for standard i. Summing up, Th(ωU)=Th(N), and by the Manevitz argument, ωUeN. (1) is done!

Finally, a point to ponder. It looks as though the ACT-provided truth predicate SatU offers a way around Tarski’s Undefinability Theorem. Namely, N⊧φ iff SatU). What gives?

The answer lies buried in the details of Topics 4 (the Addendum) and Topics 10 (a remark at the end of the Proof Sketch). I’ll explain next time.