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

  • $\begingroup$ What are $P_F$, and $\nu(A)$? $\endgroup$ – William Jun 19 '12 at 9:56
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    $\begingroup$ I assume that $P_F$ is the positive cone of $F$ and $v(A)$ is the set of upper bounds of $A$? Also, what exactly is your definition of a Cauchy sequence in $F$? $\endgroup$ – Brian M. Scott Jun 19 '12 at 9:56
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    $\begingroup$ Brian's right. My definition of a cauchy sequence is 'for every $e\in P_F$, there exists $N\in \omega$ such that if $i,j ≧ N$, then |$x(i) - x(j)$| < $e$. ($x:\omega →F$ is a sequence) $\endgroup$ – Katlus Jun 19 '12 at 10:02

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.


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