# Surreal number: trying to construct complete ordered fields

Let $R$ be a subring of $\mathbf{No}$, the set of surreal number. We try to construct $\tilde{R}$, the Cauchy completion of $R$, just like the ordinary Cauchy completion for metric space.

In the following we only consider the sequences in $R$ indexed by (i.e. with length equal) $\mathrm{Cf}(R)$, the cofinality of $R$. For any Cauchy sequences $(x_{\alpha})$ and $(y_{\alpha})$, we define an equivalent relation: $$x\sim y\;\;\mathrm{ iff }\;\; |x_{\alpha}-y_{\alpha}|\rightarrow0.$$ Let $\tilde{R}$ be the set of all equivalent classes of Cauchy sequences. On $\tilde{R}$ we define addition $[x]+[y]=[x+y]$ and multiplication $[x][y]=[xy]$ of classes. It is standard to check that these operations are well defined, $\tilde{R}$ becomes a ring and is Cauchy complete, and that $R$ is dense in $\tilde R$.

For each ordinal number $\alpha$, denote $O_{\alpha}$ the set of surreal numbers with birthday $<\alpha$. It is known that if $\alpha=\omega^{\omega^{\beta}}$ for some ordinal $\beta$ then $O_{\alpha}$ is a ring, and if $\alpha=\epsilon_{\beta}$ for some ordinal $\beta$ then $O_{\alpha}$ is a field. It is easy to check that in the latter case, $\tilde{O_{\alpha}}$ is not only ring but also a field. The question is, in the case $\alpha=\omega^{\omega^{\beta}}$, is $\tilde{O_{\alpha}}$ actually a field?

It is worth pointing out that if $\beta=0$ then $O_{\alpha}$ is the set of dyadic fraction and hence $\tilde{O_{\alpha}}=\mathbf R$, the set of reals, and is certainly a field. Apparently the difficult part is about the existence of multiplicative inverse.

In Fields of surreal numbers and exponentiation (Fund. Math. 167 (2001), pp. 173-188, doi:10.4064/fm167-2-3), Lou van den Dries and I show that $$O_\alpha$$ is an ordered field if and only if $$\alpha$$ is an epsilon number (see Corollary 4.9). Moreover, for epsilon $$\alpha$$, $$O_\alpha$$ is never Cauchy Complete in the familiar generalized sense you have in mind. On the other hand, for epsilon $$\alpha$$, $$O_\alpha$$ has a Cauchy completion consisting of $$O_\alpha$$ together with all the surreal numbers of tree rank $$\alpha$$ that fill the Dedekind gaps in $$O_\alpha$$ having breadth $$0$$ (where the breadth of a Dedekind cut $$(X,Y)$$ of an ordered abelian group $$G$$ is the largest convex subgroup $$G'$$ of $$G$$ for which $$x+|g'|\in X$$ for all $$x\in X$$ and all $$g' \in G'$$).
Edit: Suppose $$\alpha > \omega$$. Then the Cauchy Completion of $$O_\alpha$$ is an ordered field if and only if $$\alpha$$ is an epsilon number. In particular, for the case you have in mind, consider the following.
By Lemma 4.8 of the aforementioned paper, we have: If $$\beta >1$$ is not an epsilon number, then the tree rank of $$\omega^{-\beta} < \omega^{\beta}$$. Accordingly, if $$\beta >1$$ is not an epsilon number, then $$\omega^{-\beta}\in O_{\omega^{\beta}}$$ but $$\omega^{\beta}$$ is not in $$O_{\omega^{\beta}}$$, so $$O_{\omega^{\beta}}$$ is not an ordered field. Moreover, since $$\omega^{\beta}$$ does not fill any gap in $$O_{\omega^{\beta}}$$ of breadth 0, $$\omega^{\beta}$$, which is the multiplicative inverse of $$\omega^{-\beta}$$, is not in the Cauchy completion of $$O_{\omega^{\beta}}$$, and hence the Cauchy completion of $$O_{\omega^{\beta}}$$ is not an ordered field.
• But in the example I gave, the Cauchy completion of $O_{\omega}$ is a field, even though ${\omega}$ is not an epsilon number. – JSCB Jul 28 '17 at 5:08