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Let $k$ be an ordered field of cofinality $cf(k)$ whose Cauchy $cf(k)$-sequences are convergent.$^{(1)}$

Let $\mathcal{R}(k)$ be its real closure. As an algebraic extension of $k$, it has the same cofinality.$^{(2)}$

I wonder if $\mathcal{R}(k)$ has the same Cauchy completeness property as $k$, and if so, if this is true of any algebraic (ordered field) extension of $k$?

$^{(1)}$A map $u: cf(k) \rightarrow k$ is Cauchy if $\forall \varepsilon >_k 0(\exists \alpha \in cf(k)(\forall \beta,\gamma > \alpha(|u(\beta) - u(\gamma)| <_k \varepsilon)))$. It is convergent to $l \in k$ if $\forall \varepsilon >_k 0(\exists \alpha \in cf(k)(\forall \beta > \alpha(|u(\beta) -l| <_k \varepsilon)))$.

$^{(2)}$ If $(K,\rho)$ is an extension of $k$ then an induction on $d$ yields: $\forall d \in \mathbb{N}^*(\forall x \in K(\exists P \in \rho(k)_d[X](P(x) \leq 1 \longrightarrow [x;+\infty[ \cap \rho(k) \neq \varnothing)))$ so if $K / k$ is algebraic, $\rho(k)$ is cofinal. Therefore, $cf(K) \leq cf(k)$.

Now, given $E,E'$ cofinal in $k,K$ of order type $cf(k),cf(K)$, the range of the map $f:\alpha \in cf(K) \mapsto \min(\{\rho(x) \in \rho(E) \ | \ \rho(x) > E'^{\alpha}\})$ is cofinal in $\rho(E)$ so $|f(cf(K))| = cf(k)$, and $cf(K) \geq cf(k)$.

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Note that for a nonarchimedean ordered field, this notion of completeness is equivalent to its completeness as a valued field under the natural valuation induced by the order (that is, with valuation ring $\{a\in k:\exists n\in\mathbb N\,(-n\le a\le n)\}$).

Finite algebraic extensions of complete discretely valued fields are complete, but infinite extensions typically are not. The canonical example is the complete field of $p$-adic numbers $\mathbb Q_p$, whose algebraic closure is not complete. This is not an ordered field, but it’s easy to cook up ordered counterexamples as well.

For example, let $k$ be the field of formal Laurent series $\mathbb Q(\!(x)\!)$, ordered so that $x$ is positive infinitesimal, and $K$ be its real closure. Let $R\subseteq\mathbb R$ be the real closure of $\mathbb Q$. Since the field of Puiseux series $R\langle\!\langle x\rangle\!\rangle$ is real-closed, $K\subseteq R\langle\!\langle x\rangle\!\rangle$. In fact, since the residue field of a finite extension of $k$ is a finite extension of $\mathbb Q$, $K$ is even smaller: $$K\subseteq\bigcup\{F\langle\!\langle x\rangle\!\rangle:F\supseteq\mathbb Q\text{ finite ordered}\}\subsetneq R\langle\!\langle x\rangle\!\rangle.$$ On the other hand, since Puiseux series can be approximated by Puiseux polynomials over $R$, which are algebraic over $k$, the completion of $K$ includes all of $R\langle\!\langle x\rangle\!\rangle$, and by the same argument also the larger field of those Hahn series $f\in R[\![x^{\mathbb Q}]\!]$ whose support is either finite or tends to $+\infty$. So, $K$ is incomplete.

It might be worth mentioning that all becomes well-behaved if you swap the roles of algebra and topology: that is, when $k$ is a real-closed field, the completion of $k$ is also real-closed.

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  • $\begingroup$ (I had been trying to prove that $\mathcal{R}(\widetilde{k}) = \widetilde{\mathcal{R}(k)}$ (where $\widetilde{.}$ denotes the completion or an ordered field) but I could only get dense embeddings $\mathcal{R}(k) \rightarrow \mathcal{R}(\widetilde{k})$ or $\mathcal{R}(\widetilde{k}) \rightarrow \widetilde{\mathcal{R}(k)}$, that's why I was interested in $\mathcal{R}(\widetilde{k})$ being complete.) $\endgroup$ – nombre Jun 28 '15 at 23:37
  • $\begingroup$ Right. Using these two operators, we can construct from any ordered field $k$ the four fields $\mathcal R(k)$, $\tilde k$, $\mathcal R(\tilde k)$, and $\widetilde{\mathcal R(k)}=\widetilde{\mathcal R(\tilde k)}=\mathcal R(\widetilde{\mathcal R(k)})$. There are natural embeddings $k\to\tilde k\to\mathcal R(\tilde k)$, $k\to\mathcal R(k)\to\mathcal R(\tilde k)$, and $\mathcal R(\tilde k)\to\widetilde{\mathcal R(k)}$, but in general all these may be strict extensions, and there are no other embeddings. $\endgroup$ – Emil Jeřábek 3.0 Jun 29 '15 at 9:05

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