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

 A: 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.


*

*$\mathcal{L}_1$: ordered sets; $T_1$: theory of linearly ordered sets; $T_1'$: theory of dense linear orders without endpoints.

*$\mathcal{L}_2$: ordered groups; $T_2$: theory of (linearly) ordered (abelian) groups; $T_2'$: theory of divisible ordered groups.

*$\mathcal{L}_3$: 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 $y_{\alpha}:=\{y_{\beta}: \beta<\alpha \wedge x_{\beta}<x_{\alpha} \ | \ y_{\gamma}: \gamma<\alpha \wedge x_{\gamma}>x_{\alpha}\}$.
A: 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.
