For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$

Let $F$ be a complete ordered field (Complete ordered field (least-upper-bound property)) and let $\epsilon \in F$ with $\epsilon >0$. Then there is a natural number $n\ge 1$ such that

$\frac{1}{n\cdot 1_F}<\epsilon ,$

where $n\cdot 1_F$ is the canonical natural of $F$ (Every complete ordered field is Archimedean) and $1\mathrm{/}(n\cdot 1_F)$ is its multiplicative inverse (Field). As is standard we abbreviate $n\cdot 1_F$ to $n$ and write the conclusion $1\mathrm{/}n<\epsilon$.

This is the reciprocal form of the Archimedean property. Every complete ordered field is Archimedean on its own delivers only the assertion that the canonical naturals are cofinal, $x<n\cdot 1_F$; the form actually used in analysis, that the reciprocals of the naturals get below every positive bound, is the statement above, and it is recorded separately so that no proof has to reconstruct the inversion step in passing.

Complete ordered field (least-upper-bound property)

Every complete ordered field is Archimedean

Field

• Monotonicity gives the eventual form for free. If $m\ge n\ge 1$ then $m\cdot 1_F\ge n\cdot 1_F>0$, because $k\mapsto k\cdot 1_F$ is strictly increasing on the naturals $\ge 1$ (Canonical naturals are positive and strictly increasing), and so $1\mathrm{/}(m\cdot 1_F)\le 1\mathrm{/}(n\cdot 1_F)<\epsilon$ by Inverses of positives are positive, and reciprocation reverses order again. So the corollary yields not merely one index but a threshold: every $m\ge n$ satisfies $1\mathrm{/}m<\epsilon$. That one extra line is what a convergence proof needs, and it is left to the caller rather than folded into the statement, because the caller usually has a threshold of its own to combine it with.

• Completeness is used only through Every complete ordered field is Archimedean. Nothing here needs the least-upper-bound property directly. The corollary therefore holds verbatim in any Archimedean ordered field, in particular in $\mathbb{Q}$, and it fails in a non-Archimedean ordered field, where an infinitesimal $\epsilon >0$ is below every $1\mathrm{/}n$ by construction.

• The equivalence is exact: the reciprocal form implies the cofinal form back again, since given $x>0$ one applies it to $\epsilon =1\mathrm{/}x$. The two are the same property written on the two sides of the inversion, and only the direction proved above is used in this library.