Dedekind and Cauchy completeness

Question

I know that the following equivalence holds: $$\text{Cauchy-complete ordered Archimedean field}\Leftrightarrow\text{Dedekind-complete ordered field}$$ I would like to know some concrete examples of a Cauchy-complete ordered field that is not Dedekind-complete.

Answer 1

Generic example: the Cauchy-completion of a non-Archimedean ordered field (see also this answer).

Concrete example: the Laurent series ring $\mathbb{R}((x))$ (see also this answer and this answer). In fact, the proof does not use any special properties of the reals, so we have the following extension:

Proposition 1. Let $R$ be an ordered field. Then $R((x))$, ordered by the positive cone $$ R((x))_+ = \{0\} \cup \left\{\sum_{i=k}^\infty \alpha_i x^i \, : \, \alpha_k > 0\right\}, $$ is a Cauchy-complete ordered field.

The unit $1 \in R((x))$, and therefore the prime field $\mathbb{Q} \subseteq R((x))$, are contained in the subfield $R = \{\alpha_ix^0 \, : \, \alpha_i \in R\} \subseteq R((x))$, so it is clear from the definition that $x^{-1} > q$ ($\, = qx^0$) for all $q\in\mathbb{Q}$. Therefore:

Proposition 2. Let $R$ be an ordered field. Then $R((x))$, ordered as in Proposition 1, is not Archimedean.

Answer 2

The set $$\mathbb{R}((\mathbb{Q})):=\{f:\mathbb{Q}\to \mathbb{R}\ |\ supp(f)\mbox{ is well-ordered}\},$$ where $supp(f):=\{x\in \mathbb{Q}\ | \ f(x)\neq 0\}$, is a field under the addition and multiplication defined as follows: for every $f,g\in \mathbb{R}((\mathbb{Q}))$ and $x\in \mathbb{Q}$,

1. $(f+g)(x):=f(x)+g(x)$,
2. $fg(x):=\displaystyle\sum_{a+b=x}f(a)g(b)$

Consider $\lambda:\mathbb{R}((\mathbb{Q}))\setminus\{0\}\to \mathbb{Q}$, $\lambda(f)=\min\{supp(f)\}$. For $f,g\in \mathbb{R}((\mathbb{Q}))$ we define: $$f<g\Leftrightarrow f\neq g\mbox{ and }(g-f)(\lambda(g-f))>0.$$ Then $(\mathbb{R}((\mathbb{Q})),\leq)$ is a non-Archimedean ordered field called Hann field.

If the map $|\;\;|:\mathbb{R}((\mathbb{Q}))\to \mathbb{Q}$ is defined by $$ |f|:=\begin{cases} e^{-\min\{supp(f)\}}&, f\neq 0 \\ 0 &, f=0 , \end{cases} $$
then $(\mathbb{R}((\mathbb{Q})),|\;\;|)$ is a Cauchy-complete non-Archimedean valued field.

It is important to say that the order topology and the valuation topology coincide.

Another good example is the Levi-Civita field. $$\mathcal{R}:=\{f:\mathbb{Q}\to \mathbb{R} \:\vert\: supp(f)\cap(-\infty,n]\mbox{ is finite for every }n\in\mathbb{Z}\}$$ is a subfield of $\mathbb{R}((\mathbb{Q}))$. When we restrict the valuation of $\mathbb{R}((\mathbb{Q}))$ to $\mathcal{R}$, the latter becomes a Cauchy-complete, non-Archimedean valued field. Similarly, it is also a non-Archimedean ordered field.

Both fields are real-closed field extensions of $\mathbb{R}$ and their complexifications $\mathbb{C}((\mathbb{Q}))$ and $\mathcal{R}+i\mathcal{R}$ are algebraically closed. Also, $\mathbb{R}((\mathbb{Q}))$ is the spherical completion of $\mathcal{R}$. The technical details and more information can be found in Comicheo, A. Barría, and K. Shamseddine. "Summary on non-archimedean valued fields." Advances in Ultrametric Analysis Contemp. Math 704 (2018): 1-36.