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 $-x\le e\le x$.



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> 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 $e<x<e'$?
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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.



Are archimedean subextensions of ordered fields dense?
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 $\omega\subset F$.


