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).

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    $\begingroup$ In the Grothendieck group (or ring?) of the ordinals, is there an element $x$ satisfying $x+x=\omega\vee x+x+1=\omega$? $\endgroup$ Mar 18 at 5:17
  • $\begingroup$ I guess not! Good point. So the monster model must be larger than the Grothendieck group then, but smaller than the omnific integers. $\endgroup$ Mar 18 at 6:11
  • $\begingroup$ 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 mathoverflow.net/a/72942 . $\endgroup$ Mar 18 at 6:35
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    $\begingroup$ 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. $\endgroup$ Mar 18 at 7:53
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    $\begingroup$ @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. $\endgroup$
    – Alec Rhea
    Mar 18 at 8:36

2 Answers 2


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 $\sigma \bar{a} = \bar{b}$.

  • $\begingroup$ 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. $\endgroup$ Mar 31 at 17:59
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    $\begingroup$ @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. $\endgroup$ Mar 31 at 21:35
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    $\begingroup$ @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. $\endgroup$ Mar 31 at 21:37
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    $\begingroup$ @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). $\endgroup$ Apr 4 at 15:13
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    $\begingroup$ @MikeBattaglia Yes. The fully general construction requires the ultrafilter lemma, but constructions of specific set-saturated models don't always require it. $\endgroup$ Apr 4 at 21:54

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.

  • $\begingroup$ 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 𝐍𝐨 of surreal numbers" and also saying "An easier task would be to try to identify models of Peano Arithmetic (PA) within 𝐍𝐨." 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? $\endgroup$ Apr 4 at 19:40
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    $\begingroup$ @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 $\mathbb{N}$. $\endgroup$
    – nombre
    Apr 4 at 20:44
  • $\begingroup$ 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? $\endgroup$ Apr 4 at 21:09
  • $\begingroup$ @MikeBattaglia Yes, if you say non-standard model of $\mathbb{N}$, then people will understand that you are talking about true arithmetic. $\endgroup$
    – nombre
    Apr 4 at 21:17

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