How can a field be Cauchy complete and non Archimedean The Wikipedia page for the completeness of the Real numbers, says that “ there are non Archimedean fields that are ordered and Cauchy complete.” However, in many other places, I’ve read that non Archimedean fields must be incomplete with the example of $(1/n)$ converging to infinitely many infinitesimal values because NA fields have infinite numbers which have infinitesimal inverses.
This makes me wonder if there Wikipedia page got it wrong, I misread it, or if there really is a way to get a NA ordered field to be Cauchy complete. Which—if any—is it? If there are NA CC ordered fields, then are there any comprehensive resources on them?
 A: In an arbitrary ordered field $\mathbb{F}$, we define Cauchy completeness relative to the field itself. A sequence $\{x_n\}_{n \in \mathbb{N}}$ in $\mathbb{F}$ is “Cauchy” if for all $\epsilon \in \mathbb{F}$ with $\epsilon > 0$, there is $N$ sufficiently large that $|x_n - x_m| < \epsilon$ for all $m,n > N$. Notice that if the field $\mathbb{F}$ has infinitesimals, we allow $\epsilon$ to be infinitesimal! We say that $\mathbb{F}$ is Cauchy Complete if every Cauchy sequence converges.
This is different from Cauchy completeness of a metric space, where we are looking at a real-valued distance function. In this metric space sense, you are correct that a non-Archimedean field, viewed as a pseudo-metric space, is not Cauchy complete as a metric space (really as a pseudo-metric space with the obvious generalization).
An example of a non-Archimedean ordered field which is Cauchy complete is given the field of Laurent series $\mathbb{R}(x)$ where $x$ is interpreted as an infinitesimal. More details are in this post. Example of a complete, non-archimedean ordered field
