# Completion of surreal subfields

Let $\kappa$ be a regular uncountable ordinal. Let $No(\kappa)$ denote the field of surreal numbers of birthdate $< \kappa$.

In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(\kappa)$ is isomorphic to $\mathbb{R}((x))^{No(\kappa)}_{<\kappa}$ which is the subfield of Hahn series of length $< \kappa$ over $\mathbb{R}$ with value group $No(\kappa)$.

Let $S$ be the subset of $\mathbb{R}((x))^{No(\kappa)} = \mathbb{R}((x))^{No(\kappa)}_{<{\kappa}^+}$ of Hahn series of either length $<\kappa$ or of length $\kappa$ whose $\kappa$-sequence of exponents is cofinal in $No(\kappa)$.

It is not too difficult to see that $S$ is stable under $+$ and $-$.

I wonder if it is a subfield of $\mathbb{R}((x))^{No(\kappa)}$, in which case it would be an example of completion of $No(\kappa)$. Does anyone know how to prove/disprove this?

• @meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/… – nombre Nov 29 '18 at 9:05
• Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful. – meowzz Dec 3 '18 at 3:10

So as said in the comments, this follows from the following general result in valuation theory:

Let $$F$$ be a field, let $$(G,+)$$ be a non trivial ordered group, and let $$F[[\varepsilon^G]]$$ be the field of Hahn series with value group $$G$$ and coefficients in $$F$$. This is equipped with the uniform structure and topology induced by the standard valuation $$v$$ whose valuation ring is the set of series whose support is a subset of $$G^{\geq 0}$$.

For $$f \in F[[\varepsilon^G]]$$ and $$g \in G$$, we let $$f|_g$$ denote the series with support $$\operatorname{supp} f \cap G^{> g}$$ which coincides with $$f$$ on this set. This is a truncation of $$f$$ as a series.

If $$T$$ is a subfield of $$F[[\varepsilon^G]]$$ which is stable under truncation and contains $$\varepsilon^G$$, then the set $$\overline{T}:=\{f \in F[[\varepsilon^G]]:\forall g \in G, f|_g \in T\}$$ is a completion of $$T$$, that is, a maximal dense valued field extension of $$T$$.
In the case when $$F$$ is ordered, the same definition gives a maximal dense ordered field extension of $$T$$, that is, the Cauchy-completion of $$T$$ as an ordered field.
Let's apply this to the case $$F=\mathbb{R}$$, $$G=\mathbf{No}(\kappa)$$ and $$T=\mathbf{No}(\kappa)$$. We have $$\mathbf{No}(\kappa)=\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{<\kappa}$$ which is stable under truncation in $$\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]=\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{<\kappa^+}$$ and contains $$\varepsilon^{\mathbf{No}(\kappa)}$$.
Let $$f \in \overline{\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{<\kappa}}$$. If $$\operatorname{supp} f$$ is cofinal in $$\mathbf{No}(\kappa)$$, then let $$\alpha$$ denote its order type and assume $$\kappa< \alpha$$. The ordinal $$\alpha$$ must be limit so $$\kappa+1<\alpha$$. By the definition of the completion, we have $$f|_{x_{\kappa+1}} \in \mathbf{No}(\kappa)$$ where $$x_{\kappa+1}$$ is the $$(\kappa+1)$$-th element of $$\operatorname{supp} f$$. But the order type of $$\operatorname{supp} f|_{x_{\kappa+1}}$$ is $$\kappa$$, which contradicts the identity $$\mathbf{No}(\kappa)=\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{<\kappa}$$. So $$\alpha\leq\kappa$$, and thus $$f \in S$$. Else $$\operatorname{supp} f$$ is has an upper bound $$x$$ in $$\mathbf{No}(\kappa)$$ and then $$f|_{x+1}=f \in \mathbf{No}(\kappa)$$ so $$f \in S$$. This proves the inclusion $$\overline{\mathbf{No}(\kappa)} \subseteq S$$.
Conversely we have $$\mathbf{No}(\kappa) \subset \overline{\mathbf{No}(\kappa)}$$. For $$f \in \mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{\leq \kappa}$$ whose support is cofinal in $$\mathbf{No}$$, the order type $$\alpha$$ of $$\operatorname{supp} f$$ satisfies $$\kappa=\operatorname{cof}(\kappa)\leq\alpha\leq \kappa$$ so $$\alpha=\kappa$$. For $$x \in \mathbf{No}(\kappa)$$, there is $$y \in \operatorname{supp} f$$ with $$x, so $$\operatorname{supp} f|_x<\operatorname{supp} f|_y\leq \kappa$$. Thus $$f|_x$$ lies in $$\mathbf{No}(\kappa)$$. We deduce that $$f \in \overline{\mathbf{No}(\kappa)}$$.
Hence $$S=\overline{\mathbb{R}[[\varepsilon^{\mathbf{No}(\kappa)}]]_{<\kappa}}=\overline{\mathbf{No}(\kappa)}$$ as desired.