# 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.

• Very cool -- does having breadth zero for a Dedekind cut correspond to the notion of being 'bridgeable', in the sense mentioned here: math.stackexchange.com/questions/389253/… ? Jul 28, 2017 at 2:16
• Yes, these are the gaps Joel calls "bridgeable." They (along with the corresponding completeness condition) go back to G. Veronese's late 19th-century pioneering work on non-Archimedean geometry (1889, 1891) and they have been rediscovered and carefully studied many times since. Following non-standard analysts, Joel calls the completion "Scott completeness". However, Dana Scott's rediscovery and beautiful treatment was quite late in the day. For many references on "Veronese Completeness", see my paper "Dedekind cuts of Archimedean complete ordered abelian groups," Jul 28, 2017 at 2:57
• Algebra Universalis 37 (1997), pp. 223-234. Also, for the early history of these cuts, see my "The Rise of non-Archimedean Mathematics and the Roots of a Misconception I: The Emergence of non-Archimedean Systems of Magnitudes," Arch. Hist. Exact Sci. 60 (2006) 1–121. Jul 28, 2017 at 2:57
• 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, 2017 at 5:08
• Jason: I have clarified. I meant to say that there is an ordinal not in the structure whose multiplicative inverse is in the structure. I hope this helps. Jul 28, 2017 at 16:29