# Can we axiomatize Omnific Integers without the Surreal Number system?

Omnific integers are the counterpart in the Surreal numbers of the integers. The surreal numbers are usually defined using set theory, and then the omnific integers are defined as a particular subset (or rather subclass) of them. My question is, does it have to be this way? Is it possible to give a first-order axiomatization of the Omnific integers and their arithmetic, without having to define the surreal numbers themselves? I know they form a proper class, so there is a risk that they may be "too big" to describe. But Tarksi gave a first-order axiomatization for the ordinal numbers, which also form a proper class, so at least we have some hope.

The reason I'm interested is because of this question I asked a while back, about finding a nonstandard model of (Robinson) arithmetic whose field of fractions forms a real closed field. The Omnific integers form such a nonstandard model, so I want to find out whether we can axiomatize them.

Any help would be greatly appreciated.

EDIT: To be clear, I don't want an axiomatization of the Omnific Integers that's based on something else, like the real numbers, the surreal numbers, or set theory. I want a theory along the lines of Peano Arithmetic.

EDIT 2: As Emil said, it seems that a recursive axiomatization of the Omnific integers is impossible. So might we define them in some other way, without reference to the surreal numbers (or the real numbers)?

• Can you link to / sketch a definition of the omnific integers? – Qiaochu Yuan Aug 11 '11 at 15:37
• @Qiaochu: According to p. 45 of Conway's On Numbers and Games (2nd edition, ISBN 1-56881-127-6), a surreal number x is an omnific integer if x = {x-1|x+1}. – Doug Chatham Aug 12 '11 at 16:22
• Perhaps it would help focus the question to inquire what fragment of the first order theory PA holds of Om. For example, presumably Robinson's Q holds in Om. Does any nontrivial induction principle hold? e.g. induction for quantifier-free assertions? – Joel David Hamkins Aug 14 '11 at 20:54
• Emil, why not post your observations as an answer? – Joel David Hamkins Aug 15 '11 at 16:40
• Keshav, just to verify, are you sure you're looking for first-order axioms for the omnific integers, or are you just looking for a direct axiomatic definition of the omnific integers that does not refer to surreals, etc.? In case the distinction I'm making is not clear, first-order Peano Arithmetic does not define the natural numbers (there exist nonstandard models). – Timothy Chow Aug 15 '11 at 17:57

Here’s a couple of observations from my comment above.

First, the theory of omnific integers is a complete extension of Robinson’s arithmetic, hence it is not recursively axiomatizable. This makes it rather unlikely that we can describe its full axiomatization in any reasonable way.

Surreal numbers No form a real-closed field, and omnific integers Oz are its subring, hence they make an ordered ring. In fact, it is known that Oz is an integer part of No (i.e., for any surreal number $r$ there exists a unique omnific integer $n$ such that $n\le r< n+1$), which—by a well-known result of Shepherdson—means that Oz is a model of IOpen (the theory of discretely ordered rings + induction for open formulas in the language of ordered rings). Moreover, the fraction field of Oz (namely, No) is real-closed; this can be expressed by a first-order axiom schema (let’s call it A), with one axiom for each degree. (This set of axioms can be simplified: in the presence of A, IOpen is equivalent to the theory of discretely ordered rings + division with remainder.)

On the other hand, Oz does not satisfy induction for larger classes such as $E_1$ (bounded existential formulas), nor does it satisfy algebraic axioms such as normality or gcd. The reason is that such axioms contradict A (or even its corollary that $\sqrt2$ is in the fraction field of Oz).

• This is a footnote to Emil's answer. There are MANY integer sets in No whose quotient fields are real closed . This follows from coupling the following facts: (1) No is a saturated model, (2) Oz is an integer set of No whose fraction field is real-closed, and (3) the Chang-Makkai Theorem. Here MANY means "same size as the class of ordinals". Moreover, one can show that any saturated real closed field of size $\kappa$ has $2^{\kappa}$ integer sets whose fraction fields are real closed. – Ali Enayat Aug 15 '11 at 23:27
• Emil, what is a complete extension, and why does this preclude the theory of omnific integers from being recursively axiomatizable? – Keshav Srinivasan Oct 10 '11 at 0:12
• A complete extension of a theory is an extension that is a consistent complete theory (includes every sentence or its negation, but not both). No recursively axiomatized extension of Robinson’s Q is complete, due to Gödel’s incompleteness theorem. – Emil Jeřábek Oct 11 '11 at 7:58

You can construct the Omnific Integers by starting with the Ordinal numbers and then expanding the set to define subtraction everywhere. This is similar to how you can expand the Natural Numbers to the Integers by defining subtraction everywhere: e.g., 2−3 is not defined within the Natural Numbers; but defining −1 such that it equals 2−3, you expand the Natural Numbers to encompass all of the Integers. In a similar way, ω is an Ordinal and so is ω+1; but ω−1 is not an Ordinal. But by defining it as “that number which, when you add 1 to it, gives you ω”, you can expand the Ordinals to include subtracting 1 from ω. Extend this to include answers for every possible difference of two Ordinals, and you have the Omnific Integers.

Put another way, the Ordinals are a subset of the Omnific Integers that can be constructed strictly through addition (or rather, “next”) and never subtraction (or “prior”); and you get the rest of the Omnific Integers by introducing a “prior” for everything.

And once you have the Omnific Integers, you can use them to define the Surreal Numbers: On Numbers and Games includes a proof that shows that every Surreal Number can be expressed as the ratio of two Omnific Integers. And yes, this includes Irrational Numbers like π, ℯ, and √2. I don’t happen to know what those particular ratios are, and it wouldn’t surprise me if writing them out was at least as involved as writing out the decimal representations of these numbers; but they exist. It does, however, mean that “Irrational” only makes sense in the context of (finite) Integers; once you allow the full range of Omnific Integers, nothing is Irrational.

• I once made this same mistake, but the omnific integers are not equivalent to the ordinals plus negation. To see this very simply, observe that $\sqrt{\omega}$ is an omnific integer but trivially not a finite linear combination of the negations of any ordinals. – Alec Rhea Mar 27 '19 at 4:25
• For a more in depth reference, see mathoverflow.net/questions/103057/… – Alec Rhea Mar 27 '19 at 4:31