TITLE: Are archimedean subextensions of ordered fields dense?
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: 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.