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?

  • 2
    $\begingroup$ @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/… $\endgroup$ – nombre Nov 29 '18 at 9:05
  • $\begingroup$ 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. $\endgroup$ – 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.

Corollary 3.2.18 in ADH:

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 $x<y$, 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.


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