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.



Surreal number: trying to construct complete ordered fields
In *Fields of surreal numbers and exponentiation* (Fund. Math. 167 (2001), pp. 173-188, doi:[10.4064/fm167-2-3](https://doi.org/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. 


