# Does every total order embed into the surreal numbers?

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

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 $$y_{\alpha}:=\{y_{\beta}: \beta<\alpha \wedge x_{\beta}x_{\alpha}\}$$.

• @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. Commented Apr 28, 2020 at 11:10
• @bof I made it explicit. Commented Apr 28, 2020 at 11:50
• That is really neat! So every ordered group and ordered ring also embeds into the surreal numbers then? Commented Apr 28, 2020 at 13:10
• @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. Commented Apr 28, 2020 at 13:25
• @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. Commented Apr 29, 2020 at 11:57

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.