Complete ordered field I'm trying to prove that;
If any Cauchy sequence is convergent in an ordered field F, every nonempty subset of F that has an upperbound has a sup in F.
Let A be a nonempty subset of F that is not a singleton and has an upperbound in F.
Let $a_0 \notin v(A)$ and $b_0 \in v(A)$.
It's written in my book that for every $e \in P_F$, there exists $N \in \omega$ such that $N≧(b_0 - a_0)/e$.
I think this is not accurate since it hasn't showed that such F is Archimedean.. Is such F archimedean? Or in such a condition does there exist such N?
-Definition of a Cauchy sequence;
For every $e\in P_F$, there exists $N\in \omega$ such that if $i,j≧N$, then $|x(i) - x(j)| < e$.
($x:\omega \to F$ is a sequence)
Least Upper Bound Property $\implies$ Complete

You’re right: there is a problem with the argument, because there is a Cauchy-complete non-Archimedean ordered field, and a non-Archimedean ordered field is not complete in the sense of having least upper bounds.
The standard example starts with the field $F$ of rational functions over $\Bbb R$, with positive cone consisting of those functions $f/g$ such that the leading coefficients of $f$ and $g$ have the same algebraic sign. Then form the Cauchy completion by extending this to equivalence classes of Cauchy sequences in $F$. This is Example 7 on page 17 of Gelbaum & Olmsted, Counterexample in Analysis; you may be able to see it here.
