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Let $F$ be an ordered field with the property that for all Dedekind cuts $(A,B)$ and all $n\in \mathbb{N}$ there are $a\in A$ and $b\in B$ such that $b-a< \tfrac{1}{n}$. I would like to conclude that $F$ is Archimedean but after numerous attempts I am not sure if this is true.

If $F$ is any ultrapower of $\mathbb{R}$ then in the cut determined by 0 you will find such elementes so we'd have to employ the assumption that it holds for all cuts, not only some particular ones.

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Suppose not, then there are $0<\alpha<\beta$ in $F$ such that for all $n\in\mathbb{N}$, $n\cdot \alpha<\beta$.

Either $\alpha$ is infinitesimal and $\dfrac{1}{\alpha}>n\cdot 1$ for all $n\in \mathbb{N}$, or $\alpha$ is not infinitesimal, $m\cdot \alpha\geq 1$ and thus $\beta>n\cdot 1$ for all $n\in\mathbb{N}$.

Consider the set $A=\displaystyle{\bigcup_{n\in\mathbb{N}}}(-\infty,n\cdot 1)$. Then $A$ is a closed downward and $B=F\setminus A$ is non empty since either $\frac{1}{\alpha}\in B$ or $\beta\in B$. So, $(A,B)$ is a Dedekind cut.

By hypothesis, there are $a\in A,b\in B$ such that $b-a<1$, but this implies $$b<a+1\leq n\cdot 1 + 1=(n+1)\cdot 1$$ for some $n\in\mathbb{N}$, and we conclude that $b\in A$. obtaining a contradiction.

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  • $\begingroup$ I think so. For the converse let $F$ be archimedean and (A,B) a Dedekind cut. We may assume without loss of generality that $(-infty,0]\subsetneq A$. Pick $\alpha\in A$ with $0<\alpha<1$, and choose $m\in\mathbb{N}$ minimal such that $(m+1)\cdot \left(\frac{\alpha}{n}\right)\in B$ (that can be done because $F$ is archimedean and $B\neq \emptyset$. Then, we have $a=\frac{m}{n}\alpha\in A$, $b=\frac{m+1}{n}\alpha\in B$ and $b-a=\frac{\alpha}{n}<\frac{1}{n}$. $\endgroup$ – Darío G Mar 15 '16 at 13:23

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