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I recently claimed somewhere that any two complete ordered fields are isomorphic. Someone pointed out that here we need to distinguish between Cuachy completeness (every Cauchy sequence converges) and Dedekind completeness (any set bounded above has a least upper bound). In fact what's true is that any two Dedekind-complete ordered fields are isomorphic.

My first reaction was that the two notions of completeness are equivalent. Before saying so I realized that no, this equivalence depends on the Archimedean property. Of course a Dedekind-complete ordered field is Archimedean, hence Cauchy-complete. This raises the question of what ordered field is Cauchy complete but not Dedekind complete.

The simplest non-Archimedean ordered field I can think of is $\Bbb R(x)$, where we say that $r>0$ if there exists $A\in\Bbb R$ such that $r(x)>0$ for every $x>A$.

This is certainly not Archimedean: $1/x>0$ but $1/x<1/n$ for every $n\in\Bbb N$. (And hence it's not Dedekind complete. To be explicit, let $S=\{n/x:n\in\Bbb N\}$. Then $S$ is bounded above by $1$, but if $b$ is an upper bound then $b-1/x$ is a smaller upper bound, so $S$ has no least upper bound.)

But thinking about it a bit I haven't been able to decide whether $\Bbb R(x)$ is Cauchy complete. True, false, trivial, hard?

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  • $\begingroup$ Theorem 6 here gives 4 equivalent characterisations of Cauchy completeness: hindawi.com/journals/aaa/2016/6023273 $\endgroup$ – Arnaud Mortier Feb 19 '18 at 15:20
  • $\begingroup$ Also, in his answer here, Twiffy mentions that the field of Laurent series is the simplest example of a non-Archimedean Cauchy complete field: math.stackexchange.com/questions/17687/… $\endgroup$ – Arnaud Mortier Feb 19 '18 at 15:29
  • $\begingroup$ @ArnaudMortier Thanks. At first it seemed that what you say about Laurent series contradicts what I thought I'd proved below, giving a divergent Cauchy sequence of Laurent series. But no, the definition of the order is different... $\endgroup$ – David C. Ullrich Feb 19 '18 at 15:35
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    $\begingroup$ @ArnaudMortier No, that's not it. The order I defined on $\Bbb R(x)$ is the same as the restriction to $\Bbb R(x)$ of that order on Laurent series. There's no contradiction because my $(r_n)$ is convergent in the field of formal Laurent series. (That field seems like it must be the Cauchy completion of $\Bbb R(x)$ (???)) In any case thanks, that does give the example I wanted $\endgroup$ – David C. Ullrich Feb 19 '18 at 15:47
  • $\begingroup$ You're welcome. I buy this interpretation more easily than the previous one :) $\endgroup$ – Arnaud Mortier Feb 19 '18 at 15:48
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Sometimes one sees an answer immediately after asking the question. No, it's not Cauchy complete.

Let $$r_n(x)=\sum_{k=0}^n2^{-k}x^{-k}.$$

I think it's clear that $(r_n)$ cannot be convergent in $\Bbb R(x)$, since the pointwise limit is not rational - I'll leave a formal justification of that to anyone who wants to do it. But $(r_n)$ is Cauchy:

Say $\epsilon\in\Bbb R(x)$ and $\epsilon>0$. Since $\epsilon(x)$ is a rational function which is positive for large $x$ there exist $c,A\in \Bbb R$ with $c>0$ and $k\in \Bbb N$ such that $$\epsilon(x)\ge cx^{-k}\quad(x>A).$$

But if $n>m$ then $$r_n(x)-r_m(x)\le2^{-m}x^{-m}\quad(x\ge1);$$hence $|r_n-r_m|=r_n-r_m<\epsilon$ if $m$ is large enough.

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  • $\begingroup$ You only asked the question $20$ minutes ago :) $\endgroup$ – Mr Pie Feb 19 '18 at 15:28
  • $\begingroup$ @user477343 Not sure what your point is - I said I saw the answer as soon as I'd asked the question... $\endgroup$ – David C. Ullrich Feb 19 '18 at 15:48
  • $\begingroup$ Exactly. Ahh nvm... it’s $3$ a.m. in the morning here. $\endgroup$ – Mr Pie Feb 19 '18 at 15:52

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