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
Best Answer
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.
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 $y_{\alpha}:=\{y_{\beta}: \beta<\alpha \wedge x_{\beta}<x_{\alpha} \ | \ y_{\gamma}: \gamma<\alpha \wedge x_{\gamma}>x_{\alpha}\}$.