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One characteristic of the surreal numbers is that they are a monster model of the first-order theory of real numbers, according to Joel David Hamkins in this post. Thus they are real-closed, and every other real-closed field embeds into them with very nice properties. So we may ask if one can similarly create a monster model of the first-order theory of natural numbers as a kind of "surnaturals."

One natural question to ask is if the non-negative elements in Conway's ring of "omnific integers" would fit the bill, given that they are often promoted as the "surreal version of the integers" or something like that. However, it is rather easy to see that this is not true: the omnific integers have the interesting property that their field of fractions is the entire surreal number field, thus there are two omnific integers whose quotient is the square root of 2, whereas no such integers exist in any model of the naturals.

Thus in some sense, the omnific integers are not quite the most direct correspondant of the natural numbers within the surreal numbers. So one question is if the monster model can be built constructively, similarly to the surreals. I would suspect that such a construction exists, given that Joel David Hamkins was able to explicitly construct a monster model of all groups (!) in the above post, which would seem to suggest a monster model of the integers also exists, and thus the naturals as those non-negative integers.

How can one build such a model?

EDIT: I initially asked if the Grothendieck ring of the ordinals with commutative addition/multiplication could be a monster model for the integers, but it is apparently too small, since for $\omega$ we should have that $\omega$ is either even or odd. This means there should be some element $x$ such that $x + x = \omega$, or $x + x = \omega + 1$, so either $\omega/2$ or $(\omega + 1)/2$ should be in our set. Also, there will need to be some element in the monster model which is divisible by every standard finite number, and if that were $\omega$, we'd thus need to have $\omega \cdot q$ for every rational q. Thanks to Noah Schweber and Emil Jeřábek for pointing this out, and also to Emil for clearing up some confusion I had about whether the non-unique factorization of omnific integers necessarily implies they are not a model of the naturals (apparently it does not, but there are other reasons why, such as there are two omnifics whose ratio is sqrt(2), which is provably not true in any model of PA).

Mike Battaglia
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    In the Grothendieck group (or ring?) of the ordinals, is there an element $x$ satisfying $x+x=\omega\vee x+x+1=\omega$? – Noah Schweber Mar 18 '22 at 05:17
  • I guess not! Good point. So the monster model must be larger than the Grothendieck group then, but smaller than the omnific integers. – Mike Battaglia Mar 18 '22 at 06:11
  • Unique factorization is not a first-order property. But yes, there are other basic properties of standard natural numbers that fail in omnific integers, see https://mathoverflow.net/a/72942 . – Emil Jeřábek Mar 18 '22 at 06:35
  • Ordinal addition is not even commutative, so it is not really appropriate to speak of the Grothendieck group, but if we understand it to be the universal abelian group that the ordinals map into, well, since $1 + \omega = \omega$ then the image of $1$ must be $0$, so by induction the finite ordinals are all annihilated. – Zhen Lin Mar 18 '22 at 06:42
  • @EmilJeřábek I thought this answer from Carl Mummert showed how to formalize it in first order logic? https://math.stackexchange.com/questions/532350/showing-unique-prime-factorization-in-first-order-logic – Mike Battaglia Mar 18 '22 at 07:19
  • @ZhenLin that is why I referred to commutative addition and multiplication above (also called Hessenberg addition and multiplication). – Mike Battaglia Mar 18 '22 at 07:20
  • You can formulate something to that effect using sequence encoding, but this will not have the usual algebraic meaning of “unique factorization”. No nonstandard model of (sufficiently strong) arithmetic is a UFD: for example, a nonstandard power of $2$, or a product of all primes below a nonstandard number (as usually formalized using sequence encoding), cannot be written as a product of finitely many primes. – Emil Jeřábek Mar 18 '22 at 07:39
  • @EmilJeřábek yes, it would be "hyperfinitely" many primes, I guess. Well, your example is better, which is that there are two omnific integers whose quotient is sqrt(2), so I will put that instead. – Mike Battaglia Mar 18 '22 at 07:47
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    The example does not work whether you say “finitely” or “hyperfinitely” many. Let me put it differently: the fact that $\omega$ is divisible by all standard positive integers does not in any way contradict its being an element of a monster model of arithmetic. In fact, a monster model (or any nonstandard model, for that matter) of arithmetic must contain such an element. – Emil Jeřábek Mar 18 '22 at 07:53
  • @EmilJeřábek that is true. So basically, you are saying that there must be some element which is divisible by all standard positive integers, and if that element is $\omega$, then that means we would at least have to have $\omega \cdot q$ for all positive rationals $q$. I guess you could also easily build such an element which is a perfect square, a perfect cube, and a perfect n-th power for all standard positive n, such as $({1!}^{1!}, 2!^{2!}, 3!^{3!}, 4!^{4!}, ...)$. Then if $\omega$ is hat number, that would mean we would also have to have $\omega^q_1 \cdot q_2$ for rationals $q_1, q_2$. – Mike Battaglia Mar 18 '22 at 08:14
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    @ZhenLin The ordinals only get fun algebraically when considered together with their natural operations — roughly speaking, expand an ordinal in Cantor normal form then treat it like a polynomial in the indeterminate $\omega$. The ordinals together with natural addition/multiplication are isomorphic to the polynomial ring over $\mathbb{N}$ in a proper class of variables. – Alec Rhea Mar 18 '22 at 08:36

2 Answers2

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I asked (and also answered) a more general version of this question a while ago. To summarize the answer, some results of Kanovei and Shelah have the following corollary:

Fact. In $\mathsf{ZFC}$ there is a uniform procedure for building 'set-saturated,' class-sized elementary extensions of arbitrary structures. That is to say there are formulas $S(M,L,x)$ and $F(M,L,f,x)$ in the language of set theory such that in any model $V \models \mathsf{ZFC}$ if $L \in V$ is a language and $M \in V$ is an $L$-structure, then the following hold (where $M^\ast = \{x \in V : V \models S(M,L,x)\}$):

  • $M \subseteq M^\ast$,
  • if $\varphi \in V$ is an $L$-formula with free variables $x_0,\dots,x_n$ and $\bar{a} \in M^\ast$ is an $n$-tuple, then $V \models F(M,L,\exists x_n\varphi,\bar{a})$ if and only if $V \models (\exists x \in M^\ast) F(M,L,\varphi,\bar{a}x)$ (where we are using some fixed coding of tuples in $\mathsf{ZFC}$),
  • furthermore, if $\bar{c} \in M$ is an $(n+1)$-tuple, then $V \models F(M,L,\varphi,\bar{c})$ if and only if $V \models “M \models \varphi(\bar{c})”$ (in particular, if $\varphi$ is a sentence, then $V \models F(M,L,\varphi,\varnothing)$ if and only if $V \models “M \models \varphi”$),
  • $F$ is compatible with Boolean combinations (i.e., $V\models F(M,L,\varphi\wedge \psi,\bar{a})$ if and only if $V\models F(M,L,\varphi,\bar{a})\wedge F(M,L,\psi,\bar{a})$, etc.), and
  • if $A \subseteq M^\ast$ is a set and $p(x)$ is a finitely satisfiable set of $L_A$-formulas with free variable $x$, then there is $b \in M^\ast$ such that for any $\varphi(x,\bar{a}) \in p(x)$, $V \models F(M,L,\varphi,b\bar{a})$.

So to state it informally, $S(M,L,x)$ defines the universe of a class-sized elementary extension of $M$ and $F(M,L,f,x)$ is its truth predicate.

Applying this to the naturals tells us that there is a formula that defines a proper class monster model of $\mathrm{Th}(\mathbb{N})$ in any model of $\mathsf{ZFC}$.

One thing to note, though, is that without global choice (which makes my original question trivial), it's unclear whether there's always a definable isomorphism between different set-saturated class-sized models of a given theory. I believe this is related to an unanswered MathOverflow question of Hamkins. That said, if $M$ and $N$ are $L$-structures and $M \equiv N$, then there will be an isomorphism between $M^\ast$ and $N^\ast$ that is definable with certain parameters.

Another thing to note is that some constructions that model theorists commonly use with the monster model are unclear in the context of these class monster models. There isn't necessarily a good way to talk about arbitrary global types, for instance. You do, however, get a good homogeneity property: There is a subgroup $G$ of $\mathrm{Aut}(M^\ast)$ that can be represented as a class in a definable way which has the property that if $\bar{a}$ and $\bar{b}$ are set-sized tuples that realize the same type, then there is a $\sigma \in G$ such that $\sigma \bar{a} = \bar{b}$.

James Hanson
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  • Thanks @JamesHanson. When you say that a formula exists, do you mean something more tangible than, for instance, a proper class sized ultrapower of the naturals? I am kind of curious what that would look like, or if it could be related to the "birthday" structure on the surreal numbers. – Mike Battaglia Mar 31 '22 at 17:59
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    @MikeBattaglia First of all, the existence of $\kappa$-saturated elementary extensions of arbitrary structures entails the ultrafilter lemma, which suggests that any construction like this will necessarily involve ultrafilters. – James Hanson Mar 31 '22 at 21:35
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    @MikeBattaglia The actual construction is a sort of iterated ultrapower (in the more general sense that set theorists use it). The key lemma is this: There is a uniform procedure that given a structure $M$ and a cardinal $\kappa$ produces an elementary extension $M' \succeq M$ such that for any ultrafilter $\mathcal{U}$ on $\kappa$, $M^{\mathcal{U}}$ embeds into $M'$ in a way that fixes $M$. So in other words, it's possible to 'do all ultrapowers of a given size at the same time.' Ultimately you just iterate this. – James Hanson Mar 31 '22 at 21:37
  • thanks, that makes sense. I guess the thing I don't get is, you're saying that these kinds of structures entail the ultrafilter lemma, but the surreals exist even in ZF. Is perhaps the idea that, even if the surreals exist, the ultrafilter lemma is still required to prove that they have nice properties like being the monster model of the reals, or that kind of thing? – Mike Battaglia Apr 01 '22 at 01:17
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    @MikeBattaglia I don't believe you need any choice to show that the surreals are a set-saturated model of $\mathsf{RCF}$. (This should follow from o-minimality, which is really an arithmetic fact about $\mathsf{RCF}$.) You may need some choice to show that they are a set-universal model of $\mathsf{RCF}$ (i.e., every set-sized model of $\mathsf{RCF}$ elementarily embeds into the surreals). – James Hanson Apr 04 '22 at 15:13
  • I'm not quite following - I thought you were saying above that you need the ultrafilter lemma. Is it just that you need the ultrafilter lemma for arbitrary structures, but not for the first-order theory of the reals (for which the surreals suffice even in ZF)? – Mike Battaglia Apr 04 '22 at 19:37
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    @MikeBattaglia Yes. The fully general construction requires the ultrafilter lemma, but constructions of specific set-saturated models don't always require it. – James Hanson Apr 04 '22 at 21:54
  • After reading your comment in the other post I think I realize how the wording in my question was unclear. The question I was meaning to ask is if it's possible to build either a model of PA, or a constructible model of the entire first-order theory of the naturals (e.g. true arithmetic), and hopefully to do so in some kind of "constructive" manner in the same way that the surreals are constructive. (1/2) – Mike Battaglia Apr 05 '22 at 04:24
  • I am not sure what I was hoping for - it would be very neat if there were something clever involving "birthdays", and ordinals, and "left and right sets" or something, similarly to the surreals, but which leads to a model of (some reasonable theory of) the naturals rather than the reals. But I am now seeing that these subtleties lead to some very different results regarding constructiveness, and even existence with respect to the ambient set theory. So I would just ask: is there any way to explicitly construct some nonstandard model of at least, let's say PA, using any of these methods? – Mike Battaglia Apr 05 '22 at 04:27
  • @MikeBattaglia The existence of a non-standard model of $\mathsf{PA}$ can be established in fairly weak theories, but you do need something that would typically be considered mildly non-constructive. You need to be able to find a completion of a given consistent theory. – James Hanson Apr 05 '22 at 21:35
  • @MikeBattaglia These are very classically minded notions of constructivity, but evidence of my claim is Tennenbaum's theorem: There are no computable non-standard models of $\mathsf{PA}$. In terms of reverse math, this means that $\mathsf{RCA}_0 + \mathrm{Con}(\mathsf{PA})$ is not enough to establish the existence of a non-standard model of $\mathsf{PA}$. On the other hand, $\mathsf{WKL}_0 + \mathrm{Con}(\mathsf{PA})$ is enough to establish it. – James Hanson Apr 05 '22 at 21:38
  • I didn't mean constructivity in such a strict sense - I don't think the ordinals wouldn't even be "constructive" in the sense I think you're talking about, let alone the surreals. But I see what you are saying; I guess I am just wondering if there is any way to do this that's different from the usual ultrapower-based method, I guess. – Mike Battaglia Apr 06 '22 at 01:03
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I think it should be said that the first order theory of the semi-ring $\mathbb{N}$ of non-negative integers is much more difficult to work with than that of the real ordered field.

The existence of certain proper elementary extensions of $\mathbb{N}$ usually relies on non-constructive methods, so it is unlikely that one could give a nice description of such an object within the class $\mathbf{No}$ of surreal numbers.

An easier task would be to try to identify models of Peano Arithmetic (PA) within $\mathbf{No}$ (the class of non-negative omnific integers only satisfies the fragment of PA called Open Induction, which only contains weak instances of the induction axioms). This seems already difficult enough, but for this one at least there is some literature. Specifically about integer parts of real-closed fields which are models of (the full integers version of) PA. I don't know this literature, but I know Salma Kuhlmann and Paola D'Aquino have look into related questions.

nombre
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  • Thanks @nombre, the references to integer parts of real-closed fields looks very useful and I will read that. I am not quite following the first part though, where you are saying "it is unlikely that one could give a nice description of such an object within the class of surreal numbers" and also saying "An easier task would be to try to identify models of Peano Arithmetic (PA) within ." Aren't these the same thing? Or are you somehow drawing a distinction between the first-order theory of $\Bbb N$ as a semi-ring vs the first-order theory of PA? – Mike Battaglia Apr 04 '22 at 19:40
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    @MikeBattaglia Yes, PA usually refers to the recursively enumerated theory consisting in Roinson arithmetic + the induction scheme, whereas "true arithmetic" is sometimes used to refer to the first order theory of $\mathbb{N}$. – nombre Apr 04 '22 at 20:44
  • I think I've been confused about this terminology all this time! When we take an ultrapower of the naturals, for instance, to form the hypernaturals, and we say this is a nonstandard model of the naturals, are we then saying that it's a nonstandard model of "true arithmetic" i.e. stronger than just a nonstandard model of PA? – Mike Battaglia Apr 04 '22 at 21:09
  • @MikeBattaglia Yes, if you say non-standard model of $\mathbb{N}$, then people will understand that you are talking about true arithmetic. – nombre Apr 04 '22 at 21:17