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Let $E$ be an ordered field and let $F$ be a real closed subfield. We say that $E$ is $F$-archimedean if for each $e\in E$ there is $x\in F$ such that $-x\le e\le x$.

Is it true that if $E$ is $F$-archimedean then every interval in $E$ contains an element in $F$? That is, is it true that for every $e<e'$ in $E$ there is an element $x\in F$ such that $e<x<e'$?


This is known if $F$ is the field of real algebraic numbers (in which case $E$ is an ordered subfield of $\mathbb{R}$), and it seems to me that it should have an easy proof in the general case. However I cannot find neither an easy proof nor a counterexample.

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    $\begingroup$ Have you tried $E = \mathbb{Q}((\infty)), F = \mathbb{Q}((\infty^2))$? $\endgroup$ – user44191 Dec 17 '18 at 19:49
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    $\begingroup$ What is $\mathbb{Q}((\infty))$? Do you mean $\mathbb{Q}(t)$ with the order making $t$ an infinite positive element (I'm not sure this order can be extended to the Laurent series)? $\endgroup$ – Denis Nardin Dec 17 '18 at 19:50
  • $\begingroup$ Yes, sorry, my idiocy. $\endgroup$ – user44191 Dec 17 '18 at 19:51
  • $\begingroup$ No need to be self-deprecating it's a good counterexample... Uhm there might in fact be no number between $t$ and $t+1$. Can you do a counterexample where $F$ is real closed? Sorry for changing the goal posts, I should have put it from the beginning since I was really thinking about that case. $\endgroup$ – Denis Nardin Dec 17 '18 at 19:52
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    $\begingroup$ Try $E$ the real closure of $\mathbb{Q}(x, y)$ with $x > \mathbb{Q}, y > \mathbb{Q}(x)$, and $F$ the real closure of $\mathbb{Q}(y)$? $\endgroup$ – user44191 Dec 17 '18 at 20:00
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Let $F$ be any real-closed field of uncountable cofinality. That is, every countable subset of $F$ is bounded. One can make such a field in a process of $\omega_1$-many field extensions; alternatively, the ultrapower of $\mathbb{R}$ by nonprincipal ultrafilter on $\omega$ also has uncountable cofinality.

Let $E=F^\omega/\mu$ be an ultrapower of $F$ by a nonprincipal ultrafilter $\mu$ on $\omega$. We may identify $F$ with its canonical copy in $E$ using equivalence classes of constant functions $x\mapsto [c_x]_\mu$.

Since every function from $\omega$ to $F$ is bounded by a constant function, it follows that $E$ is $F$-archimedean. But $F$ is not dense in $E$, since there are no constant functions between $[\text{id}]_\mu$ and $[\text{id}+1]_\mu$, where $\text{id}:n\mapsto n$, viewing $\omega\subset F$.

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    $\begingroup$ Oh well... I guess that was too good to be true. Thanks for the answer, I'll wait till tomorrow to see if other interesting answers pop up and then accept it. $\endgroup$ – Denis Nardin Dec 17 '18 at 20:08
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    $\begingroup$ Sure thing, no problem. I liked your question. The argument I give shows that no first-order property of the ordered field (like being real-closed) can have the consequence you want, since we can always find such fields of uncountable cofinality and then take an ultrapower. $\endgroup$ – Joel David Hamkins Dec 17 '18 at 20:22
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Let $E$ be the real closure of $\mathbb{Q}(x, y) = (\mathbb{Q}(x))(y)$, with order given by $x > \mathbb{Q}$ and$y > \mathbb{Q}(x)$. In other words, positivity on $\mathbb{Q}(x, y)$ is determined first by degree in $y$, and then by degree in $x$. Then let $F$ be the real closure of $\mathbb{Q}(y)$.

First, we prove that $E$ is $F$-archimedean. Let $e \in E$. There is some $e' \in \mathbb{Q}(x, y)$ with $e' > e > -e'$. Then $e'$ has degree $n$ in $y$ for some $y$; let $f = y^{n + 1}$. By the order on $\mathbb{Q}(x, y)$, we have $f > e' > e > -e' > -f$. Therefore $E$ is $F$-archimedean.

On the other hand, clearly, there is no element of $F$ between $x$ and $x + 1$.

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    $\begingroup$ Thank you for your answer! Now I'm going to have a hard time deciding which one to accept :) $\endgroup$ – Denis Nardin Dec 17 '18 at 20:35
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    $\begingroup$ I ended up accepting JDH's answer because it provides me with a "machine" to generate counterexamples to similar statements, but your simple answer was very appreciated (and possibly shows that I hadn't thought through the situation as much as I thought...) $\endgroup$ – Denis Nardin Dec 18 '18 at 22:37
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This is similar to user44191's answer but I want to put the emphasis on the fact that there is no reason that the cofinality of $F$ in $E$ (which is what ou call [$E$ is $F$-archimedean]) imply the density of $F$ in $E$.

Indeed, if $F$ is any non-archimedean ordered field, then the field $F(t)$ can be equipped with an order where $t$ is positive infinite but smaller than any positive infinite element of $F$. Thus $F$ is cofinal in $F(t)$ but not dense in it since no element of $F$ is close to $t$. To do so, say that a fraction $\frac{P(t)}{Q(t)}$ is positive if $P(t)$ and $Q(t)$ have the same sign, where the sign of a polynomial $R(t)$ is positive if $R(s)$ is positive for sufficiently large finite element $s$ of $F$.

This is a special case of filling a cut in an ordered field using a simple extension.

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