TITLE: Dedekind completion of ordered fields
QUESTION [8 upvotes]: Let $\mathbb S$ be an ordered field of cardinality larger than $\mathbb R$.  Let $\mathbb S^*$ be the completion of $\mathbb S$ via Dedekind cuts.  Now it is well-known that $\mathbb R$ is the unique complete ordered field.  So in what was does $\mathbb S^*$ fail to be a complete ordered field, and why?

REPLY [5 votes]: You may be interested in knowing that there is a kind of completion of $\mathbb{S}$ akin to Dedekind completion:
Instead of considering every cut $(A,B)$, take only good cuts: cuts $(A,B)$ where for any $0<\varepsilon \in \mathbb{S}$, there is $(a,b) \in A \times B$ such that $b-a < \varepsilon$.
Then you really get a dense ordered field extension $\widetilde{\mathbb{S}}$ which is complete in the sense that it has no proper dense ordered field extension. This is the same as completing $\mathbb{S}$ using Cauchy $\lambda$-sequences where $\lambda$ is the least cardinal of a subset of $\mathbb{S}$ without bounds. This is also the same as completing $\mathbb{S}$ seen as an ordered field with its canonical uniform structure.