Proof that all ordered fields are in the Surreals It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field?
 A: A sketch answer only:
The critical model-theoretic fact is that the theory of real closed fields has quantifier elimination.
That means that if you have a real closed field $K$ which you are trying to extend (for an inductive proof) with a new element $b \notin K$ (or indeed if we happen to be looking at some $b \in K$), everything that $K$ can tell you about $b$ is determined by:


*

*formulas $P(x, a_1, \ldots, a_n) = 0$ for all the polynomials which $b$ satisfies, with coefficients $a_i$ in $K$;

*formulas $\neg Q(x, a_1, \ldots, a_n) = 0$ for all the polynomials which $b$ does not satisfy;

*all the order inequality formulas $a < x$ and $x < a$ which $b$ satisfies.


This collection of formulae is known as the type of $b$; the polynomial equalities and non-equalities together form the algebraic type and the order formulas on their own determine the cut.
These ideas should look fairly familiar from the construction of $\mathbb{R}$ from $\mathbb{Q}$. But we can also have elements that are greater than everything in $k$, for a non-Archimidean extension.
Moreover if $b \notin K$ we know that $b$ is actually transcendental over $K$, because $K$ is by hypothesis real-closed: meaning that it already contains all the roots of polynomials over $K$ that can possibly exist in an ordered field. So the type of $b$ over $K$ is exactly its cut, plus all the statements $\neg P(x, a_1, \ldots, a_n)=0$ for all polynomials and choices of parameters $a_i$ from $K$.
The construction of the surreals is a transfinite recursion that keeps defining new elements to go into every cut. So the embedding of any real closed field $k$ into the surreals is easy by transfinite induction:


*

*well-order $k = \{b_\alpha: \alpha < N\}$ (using Choice, of course, where the infinite ordinal $N$ is the cardinality of $k$);

*in your sequence of partial mappings, for each new $b_\alpha$ you find its image in the cut you want, and also the images of all members of $k$ that are algebraic over $\{b_\beta : \beta \leq \alpha\}$;

*take unions to taste.


That all looks straightforward. You embed $k$ into an initial segment of the surreal tree of the same cardinality $N$ as $k$, so the fact that the surreals as a whole are a proper class doesn't cause problems.
