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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.

Thank You in Advance.

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)?

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    $\begingroup$ Can you link to / sketch a definition of the omnific integers? $\endgroup$
    – Qiaochu Yuan
    Commented Aug 11, 2011 at 15:37
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    $\begingroup$ @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}. $\endgroup$
    – Doug Chatham
    Commented Aug 12, 2011 at 16:22
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    $\begingroup$ 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? $\endgroup$
    – Joel David Hamkins
    Commented Aug 14, 2011 at 20:54
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    $\begingroup$ Emil, why not post your observations as an answer? $\endgroup$
    – Joel David Hamkins
    Commented Aug 15, 2011 at 16:40
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    $\begingroup$ 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). $\endgroup$
    – Timothy Chow
    Commented Aug 15, 2011 at 17:57

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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).

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    $\begingroup$ 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. $\endgroup$
    – Ali Enayat
    Commented Aug 15, 2011 at 23:27
  • $\begingroup$ Emil, what is a complete extension, and why does this preclude the theory of omnific integers from being recursively axiomatizable? $\endgroup$
    – Keshav Srinivasan
    Commented Oct 10, 2011 at 0:12
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    $\begingroup$ 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. $\endgroup$
    – Emil Jeřábek
    Commented Oct 11, 2011 at 7:58
  • $\begingroup$ Very interesting! Do they have a second-order axiomatization? $\endgroup$
    – Mike Battaglia
    Commented Feb 15 at 21:22

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