Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From http://en.wikipedia.org/wiki/Non-Archimedean_ordered_field:

The field of rational functions over $\mathbb{R}$ can be used to construct an ordered field which is complete (in the sense of convergence of Cauchy sequences) but is not the real numbers. Sometimes the term complete is used to mean that the least upper bound property holds. With this meaning of complete there are no complete non-Archemedian ordered fields. The subtle distinction between these two uses of the word complete is occasionally a source of confusion.

I had the idea that, in an ordered field, completeness in the sense of convergence of Cauchy sequences is equivalent to convergence in the sense that the least upper bound property holds. Now that I think about it, I know how to prove that the least upper bound property implies the convergence of Cauchy sequences, but I'm not so sure about the converse. According to this Wikipedia page, the converse is not true. Can anyone confirm this? (I need to be sure this is not an error on Wikipedia) If possible, provide a counter-example (perhaps the field of rational functions suggested in the quote, but how is the structure, i.e., ordering, operations, etc., defined?).

Thanks.

share|improve this question
    
In the wikipedia page there is a reference: Counterexamples in Analysis by Bernard R. Gelbaum and Bernard R. Gelbaum, Chapter 1, Example 7, page 15, where one should find the example of the rational functions over R, but I don't have that book at hand. –  becko Sep 5 '11 at 20:24
3  
Counterexamples must be non-Archimedean. The least upper bound property implies the Archimedean property, but convergence of Cauchy sequences does not. In an Archimedean ordered field, the 2 notions of completeness coincide. You will find that the example given in your reference is not Archimedean, and it is a good exercise to prove that an Archimedean ordered field in which all Cauchy sequences converge has the least upper bound property. –  Jonas Meyer Sep 5 '11 at 20:31
    
So a question becomes: How do you prove they're equivalent in Archimedean ordered fields? Would that be a good exercise in the right kind of analysis course? –  Michael Hardy Sep 6 '11 at 0:07
add comment

1 Answer

up vote 5 down vote accepted

The standard counterexample is the field $F$ of finitely-tailed Laurent series over $\mathbb{R}$: series of the form $\sum a_n t^n$ where $a_n \in \mathbb{R}$, exponents $n \in \mathbb{Z}$, but with only finitely many negative exponents (i.e., $a_n = 0$ for all but finitely many negative integers $n$.)

Addition and multiplication are defined just like for power series; you should verify that the "finitely many negative exponents" condition is essential to ensuring that multiplication of finitely-tailed Laurent series is well-defined. Check that non-zero elements of $F$ have multiplicative inverses: a series that starts with $a_n t^n$ has an inverse that starts with $a_n^{-1} t^{-n}$. (Here, and below, when I say that a nonzero element of $F$ starts with $a_n t^n$, I mean that $n$ is the lowest integer for which $a_n \ne 0$.)

Ordering is defined as follows: An element is positive if its starting term has positive coefficient. It's easy to check: (1) for every nonzero $x \in F$, exactly one of $x$ and $-x$ is positive; (2) the sum and product of 2 positive elements is positive.

So $F$ is an ordered field. Note that $F$ is not Archimedean: $t^{-1}$ is bigger than every integer. The field $F$ also does not satisfy the least upper bound property: $t^{-1}$ is an upper bound for the integers, but there is no least upper bound (check that if $x$ is such an upper bound, then $x/2$ is a smaller one.)

To discuss Cauchy completeness, we have to put a metric on $F$, as follows. Pick any positive real number $q>1$. If a nonzero element $x \in F$ starts with $a_n t^n$, then define $|x| = q^{-n}$ (and of course define $|0|=0$), and then define the distance between $x$ and $y$ to be $|x-y|$. Check that this is a metric space. (Caution: This doesn't extend the existing metric on $\mathbb{R}$; all real numbers have the same size under this metric.) In fact, it makes $F$ into a valued field; we have $|xy| = |x||y|$.

This is a "non-Archimedean" metric because it satisfies the strong triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$. For a non-Archimedean metric, a sequence is Cauchy iff $|x_n - x_{n+1}| \to 0$ as $n \to \infty$.

To prove that $F$ is complete under this metric, note that for a Cauchy sequence in $F$, the coefficients for a fixed exponent $t^n$ must be eventually constant; otherwise the difference between consecutive terms could never be smaller than $q^{-n}$. Call this constant $a_n$; then the Cauchy sequence converges to $\sum a_n t^n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.