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The surreal numbers are the largest ordered field, and have the unique property that every ordered field is isomorphic to a subfield of the surreal numbers.

Do they also have the property that every possible total order embeds into the surreal numbers?

My thinking: I am basically wondering if they have a property similar to that of the rational numbers, which is that every countable total order embeds into the rationals. It sure seems like a similar thing would be true for the surreal numbers, but is it? Does the answer differ based on what set theory you are using?

Why I think this is neat, if true:

  • An order is a total order iff it can be injected into the surreal numbers in an order-preserving way (not just if, but iff)
  • A total order can easily be built, for any set $S$, by giving each element a unique surreal-valued "ranking" - aka an injection from $S \to \mathbf{No}$
  • In general, a total order on some set $S$ can be viewed as a particular equivalence of injections from $S$ to the surreals
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2 Answers 2

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In this answer, I consider a first order language $\mathcal{L}_i$, a theory $T_i$ in $\mathcal{L}_i$ and its model companion $T_i'$ which is complete. Moreover the natural interpretation of $\mathcal{L}_i$ in the class $\mathbf{No}$ of surreal numbers yields a saturated model of $T_i'$.

Every set-sized model of $T_i$ embeds into a model $T_i'$, which by saturation embeds in ZFC into the $\mathcal{L}_i$-structure $\mathbf{No}$. Thus $\mathbf{No}$ contains every model of $T_i$, although probably not in a canonical way. In NBG with global choice, the set-size restriction can be discarded. It is hard to believe that this would work without choice however.

This works for the three examples below.

  1. $\mathcal{L}_1$: ordered sets; $T_1$: theory of linearly ordered sets; $T_1'$: theory of dense linear orders without endpoints.
  2. $\mathcal{L}_2$: ordered groups; $T_2$: theory of (linearly) ordered (abelian) groups; $T_2'$: theory of divisible ordered groups.
  3. $\mathcal{L}_3$: ordered rings; $T_3$: theory of ordered domains; $T_3'$: theory of real-closed fields.

I don't know if something of the sort can be said of $\mathbf{No}$ as an ordered exponential field without using additional structure. Same question for valued differential rings. I suppose this would take a lot of work to prove, but $\mathbf{No}$ may also be saturated as a model of the theory of transseries. edit: it is not since the field of constants is a set ($\mathbb{R}$).

To prove your result more succintly, pick an enumeration $(x_{\alpha})_{\alpha<\kappa}$ of your linear order, and send each $x_{\alpha}$ inductively onto $y_{\alpha}:=\{y_{\beta}: \beta<\alpha \wedge x_{\beta}<x_{\alpha} \ | \ y_{\gamma}: \gamma<\alpha \wedge x_{\gamma}>x_{\alpha}\}$.

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    $\begingroup$ @bof $\mathbf{No}$ is standard for the surreal numbers, yes. Also the part at the very bottom is a pretty direct argument answering the question. $\endgroup$
    – Mark S.
    Commented Apr 28, 2020 at 11:10
  • $\begingroup$ @bof I made it explicit. $\endgroup$
    – nombre
    Commented Apr 28, 2020 at 11:50
  • $\begingroup$ That is really neat! So every ordered group and ordered ring also embeds into the surreal numbers then? $\endgroup$
    – Mike Battaglia
    Commented Apr 28, 2020 at 13:10
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    $\begingroup$ @MikeBattaglia Yes, and there's a similar model-theoretically light proof: Start with an enumeration of the divisible hull of said group / the real-closure of said field. Define $y_{\alpha}$ similarly except you choose $\alpha$ as the least ordinal such that all $x_{\beta},\beta<\alpha$ lie in the divisible hull / real closure of the subset of elements that have already be embedded. Such $x_{\beta}$, may be embedded into $\mathbf{No}$ in a unique way, so this defines a unique $y_{\beta}$. This unicity is specific to this case, it does not follow from the conditions in the top of my answer. $\endgroup$
    – nombre
    Commented Apr 28, 2020 at 13:25
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    $\begingroup$ @MikeBattaglia My bad, what I am calling "ordered ring" is what I should perhaps call "ordered domain", i.e. we should have $(a>0$ and $b>0) \Rightarrow a b >0$. I will edit my answer to reflect this. $\endgroup$
    – nombre
    Commented Apr 29, 2020 at 11:57
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Yes; we can define a natural set $S_d$ of surreals for every ordinal $d$ so that any total order $X$ can embed into some $S_d$.

If we accept the general continuum hypothesis, then we can embed $X$ in $S_d$ where $|S_d|=|X|$.

To define $S_d$, consider each surreal $s$ as an ordinal $d$ together with a function $s:d\to\{+,-\}$. Write $D(s)$ to denote the ordinal $d$ of $s$ (the domain of $s$ as a function). Then $S_d := \{s: D(s) < \aleph_d\}.$

Details: Surreal Canonical Linear Orders.

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