Is every order type of a PA model the \omega of some ZFC model?

Question

Let $N$ be a model of first-order Peano arithmetic, and let $\sigma$ be its order-type. Does it follow that there is a (non-transitive, expect when $M$ is the standard model) $ZFC$-model $M$ such that $\omega^{M}=\sigma$? What if $PA$ is strengthened to true arithmetic (i.e. the theory of $(\mathbb{N},+,\cdot)$ or weakened to a subtheory such as $I\Sigma_{1}$, $I\Delta_{0}+EXP$, $I\Delta_{0}$ or $IOpen$? Or when we replace $ZFC$ with something weaker like $KP$?

Answer 1

For countable models, the answer is yes, since all nonstandard such models have the same order type. (For the standard model $N=\mathbb{N}$, assume there is an $\omega$-standard model of ZFC, and in general we must assume Con(ZFC).) Namely, if $N\models\text{PA}$ is nonstandard, then its order type is $\omega+\mathbb{Z}\cdot\mathbb{Q}$, and this is also the order type appearing as the $\omega$ of any $\omega$-nonstandard countable model of ZFC.

It seems to me that this argument would also work for saturated models in larger cardinalities, but I'm unsure of the general case.

A related question would be: which models of PA arise as the full standard structure of arithmetic $\langle\mathbb{N},+,\cdot,0,1,<\rangle^M$ of a model $M\models\text{ZFC}$? In other words, we ask for the full structure, rather than merely the order type as you have. These are known as the ZFC-standard models of PA, and in the countable case they are exactly the standard model of PA (provided there is an $\omega$-model of ZFC), together with the computably saturated models of the arithmetic consequences of ZFC, a theory denoted $\text{Th}(\mathbb{N})^{\text{ZFC}}$. For a discussion of this and applications, see proposition 3 of Satisfaction is not absolute, as well as a 2009 report of Ali Enayat on the topic for the Mittlag-Lefflar institute.

Meanwhile, I was able to get the following, which works even for uncountable models. (Update: I made this an equivalence.)

Theorem. The following are equivalent for any $N\models\text{PA}$.

1. For any finitely many axioms $\varphi_0,\ldots,\varphi_n$ of ZFC, the model $N$ thinks they are consistent.
2. As a model of arithmetic, $N$ is isomorphic to an initial segment of the standard model $\mathbb{N}^M$ of some $M\models\text{ZFC}$.
3. $N$ satisfies all the $\Pi_1^0$ consequences of ZFC.

Proof. ($2\to 3$) If $N$ is an initial segment of $\mathbb{N}^M$ for some $M\models\text{ZFC}$, then $N$ must satisfy all the $\Pi_1^0$ arithmetic statements true in $M$.

($3\to 1$) Those consistency statements are amongst the $\Pi_1^0$ consequences of ZFC, which proves the consistency of any particular of its finite fragments.

($1\to 2$) Suppose that $N$ is a model of PA in which every standard finite fragment of standard ZFC is consistent. (This follows from $N\models\text{Con}(\text{ZFC})$, but it is a strictly weaker hypothesis.) In particular, this hypothesis implies that ZFC is consistent, and so we may assume without loss that $N$ is nonstandard. Consider $N$'s version of the theory ZFC. We may enumerate what it thinks are the axioms, which of course includes many nonstandard axioms. Since we have assumed that $N$ thinks all the standard-finite initial segments of this enumeration are consistent, by overspill there is some nonstandard finite theory $t$ in $N$ which $N$ thinks is consistent, and which includes every standard axiom of ZFC.

Inside $N$, we may now build the usual Henkin completion $T$ of $t$. That is, we add Henkin constants and add every sentence $\sigma$ or its negation $\neg\sigma$ in the new language, in such a way that $N$ thinks $T$ extends $t$ and remains consistent, while having the Henkin property. The theory $T$ is a definable class in $N$, and $N$ is also able to define the corresponding Henkin model $M$.

By meta-theoretic induction, we can show that $M\models\sigma$ just in case $\sigma\in T$, for every standard finite sentence $\sigma$ in the expanded language. In particular, it follows that $M$ is a model of ZFC.

Finally, I claim that $N$ is isomorphic to an initial segment of the $\mathbb{N}^M$, and indeed, $N$ is able to construct this isomorphism. For every $a\in N$, the model $N$ is able to construct the term $\overbrace{1+\cdots+1}^a$, and it has added a Henkin constant $\hat a$ and the assertion $\hat a=\overbrace{1+\cdots+1}^a$ to the theory $T$. Furthermore, $N$ believes that it is able to prove (by induction) that $x<\hat a$ implies $x=\hat b$ for some $b<a$. It also must believe things like $\hat{(a+b)}=\hat a+\hat b$ and $\hat{(a\cdot b)}=\hat a\cdot\hat b$, since these kind of manipulations of terms are provable in PA. Thus, the map $$a\mapsto (\hat a)^M$$ is an isomorphism of $\langle N,+,\cdot,0,1,<\rangle$ to an initial segment of $\mathbb{N}^M$, as desired. QED

Corollary. If ZFC is consistent, then every model of true arithmetic, and indeed every model of $\text{PA}+\text{Con}(\text{ZFC})$, arises as an initial segment of the standard model $\mathbb{N}^M$ of some model $M\models\text{ZFC}$ of set theory.

One can replace ZFC in this argument with other set theories, such as KP or ZFC + large cardinals, and get an equivalence for other theories.

Basically, in the implication ($1\to 2$) of the theorem, the model $N$ thinks it has constructed an $\omega$-nonstandard model of the theory $t$, and it naturally believes that what it thinks is the standard model, itself, is an initial segment of that model.

Answer 2

Joel Hamkins has already answered the question in the positive for countable models of PA; indeed PA can be relaxed to IOpen in his argument.

In the uncountable case, as far as I know, the problem is wide open.

My reason: a closely related question is still (wide) open, namely, is it possible to have distinct consistent completions $T_1$ and $T_2$ of PA such that the collection of order-types of uncountable models of $T_1$ differs from the collection of order-types of uncountable models of $T_2$ ?

This open question was first brought to world attention by Harvey Friedman in his 1975-paper "One hundred and two problems in mathematical logic" (in Journal of Symbolic Logic).

For the state of the art (in relation to order-types of models of PA) see Chapter 11 of the Kossak-Schmerl graduate text on on models of arithmetic. Friedman's problem is mentioned there both in Chapter 11, and also in Chapter 12 (List of 20 open questions).