I have been trying to write up some notes on completion of ordered fields, ideally in the general case (i.e., not just completing $\mathbb{Q}$ to get $\mathbb{R}$ but considering the completion via Cauchy sequences of any ordered field). I have found the technical details of this to be surprisingly thorny, especially compared to the relatively clean construction of the completion of a metric space (which takes $\mathbb{R}$ as given).

Finally I found a clean, concise treatment of the general case in $\S$ 8.7 of P.M. Cohn's text *Basic Algebra: Groups, Rings and Fields*. He does the general case and also includes the universal property of completion. Following his terminology, let's say an ordered field is **complete** if every Cauchy sequence in that field is convergent (I would say "sequentially complete" to differentiate both from Dedekind completeness and also the possibility of considering more general Cauchy nets).

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda: K \rightarrow \tilde{K}$ such that to each order-embedding $f: K \hookrightarrow L$ into a complete ordered field $L$ there is a unique order-embedding $f': \tilde{K} \rightarrow L$ such that $f = f' \circ \lambda$.

I am happy with the existence of $\lambda$. However, Cohn's proof of the universal property says:

"Finally, let $f: K \hookrightarrow L$ be an order-embedding in a complete field $L$. Any element $\alpha$ of $\tilde{K}$ is obtained as the limit of a Cauchy sequence $\{a_n\}$ in $K$; it is easily seen that $\{f(a_n)\}$ is a Cauchy sequence in $L$ and so has a limit..."

This sounds reasonable at first glance...but it's false, I think. Suppose $K = \mathbb{Q}$ and $L = \mathbb{R}((t))$, the latter given the unique ordering in which $t$ is positive and smaller than every positive real number. It is well-known that $L$ is complete (all I'm using about $L$ is that it is complete and non-Archimedean, and such fields certainly exist, so if you prefer just take any such $L$). Let $a_n = \frac{1}{n}$, so of course $f(a_n) = \frac{1}{n}$. But we have a problem: the sequence $\frac{1}{n}$ is not in fact Cauchy (or equivalently, convergent) in $L$ *because* $L$ is non-Archimedean: the open interval $(-t,t)$ about $0$ contains no terms of the sequence!

[In fact one can make the problem even less subtle. It is known that ultraproducts and such construct ordered fields which have uncountable cofinality (equivalently, are not first countable in the order topology). In such a field, the only Cauchy sequences are the eventually constant ones. In particular such fields are automatically complete. However, such a field contains subfields like $\mathbb{Q}$ which have plenty of not eventually constant convergent sequences. Clearly an order embedding cannot make a not-eventually-constant sequence eventually constant!]

Thus I claim the distressing fact that a homomorphism of ordered fields need not be even sequentially continuous with respect to the given order topologies: *a fortiori* it need not be continuous.

So the question is: am I making some silly mistake here? (Cohn is a gold-standard algebraist and his book is lovely and authoritative, even compared to many of the more familiar standard texts.) If not, is the result even true? I think it isn't: instead of $\mathbb{R}((t))$ we could take $\mathbb{Q}((t))$ which I believe is still complete -- a sequence is Cauchy iff for each $n$ the sequence of $n$th Laurent series coefficients is eventually constant; hence Cauchy sequences are convergent -- and then the embedding from $\mathbb{Q}$ will not extend to $\mathbb{R}$. If the result is false, I wonder:

What is the appropriate universal property of the completion of an ordered field? (Note that I asked a similar question about metric spaces here a while back.) We could require for instance the maps to be "sequentially Cauchy" (i.e., to preserve Cauchy sequences) and then the proof goes through. But that seems a little weak...

**Added**: One fix is to impose the condition that $f$ too be a dense embedding (i.e., such that in between any two distinct elements of $L$ lies an element of the image of $f$). This condition is sufficient for Cauchy sequences to map to Cauchy sequences (I think it is not necessary: imagine embedding a field of uncountable cofinality into another field of bigger cofinality). If we put this in then essentially we get a rather wordy way of saying that the completion is unique, but that is indeed the main application of this result anyway, so far as I know.

**Further Added**: The following recent note seems to face some issues on the topology of ordered fields more head-on than what I've seen otherwise in the literature. Here is the MathSciNet review:

Tanaka, Yoshio

Topology on ordered fields. Comment. Math. Univ. Carolin. 53 (2012), no. 1, 139–147. An ordered field is a field $(K,+,∗)$ equipped with a linear order $<$ and the open interval topology $λ(<)$ where the algebra and order are related as follows: if $a,b,c \in K$ and $a<b$, then $a+c<b+c$; and if $c>0$ then $a∗c<b∗c$. Consequently, any ordered field contains an algebraic copy of $\mathbb{Q}$, the usual field of rational numbers. A key issue in this paper is that for a subset $A \subset K$, the topology $λ(<)|A$ that $A$ inherits as a subspace of $(K,λ(<))$ might, or might not, be the same as the open interval topology defined on $A$ by the restricted order $<|A$. The author discusses the Archimedean property of an ordered field, showing, for example, that K is Archimedean if and only if Q is dense in K, and that for any ordered field K, either Q is dense in K or is closed discrete in K. Other results concern Dedekind completeness of K and another result shows that if K is an Archimedean ordered field and if f:K→K′ is a surjective algebraic homomorphism, then the following are equivalent: f is continuous; f is a homeomorphism; and f is order preserving. The paper ends with examples of various ordered fields and homomorphisms.

I haven't been able to get my hands on a copy, but if someone can and it seems relevant, I'd take it as a favor if they'd let me know.