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I'm looking for a concrete example of a complete (in the sense that all Cauchy sequences converge) but non-archimedean ordered field, to see that these two properties are independent (an example of archimedean non-complete ordered field is obviously the rationals).

Thank you in advance.

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Complete in what sense? Cauchy? Non-standard Cauchy (that is Cauchy sequences which might be longer than $\omega$ but with $\epsilon$ from the field itself, and not real/rational as usual)? Is it complete in the form of order completeness, namely Dedekind complete? –  Asaf Karagila Jan 16 '11 at 11:36
    
I noticed that complete could mean different things and edited. I'm refering to (ordinary) Cauchy completeness. –  Charlie Jan 16 '11 at 11:54
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The completion of any non-archimedean ordered field is still non-archimedean. –  Qiaochu Yuan Jan 16 '11 at 14:15
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For a general ordered field, I don't think Cauchy sequences give the appropriate notion of completeness; I think you want Cauchy nets or filters. –  Harry Altman Jul 24 '11 at 6:29
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2 Answers

Consider the ring of formal Laurent series $R((x))$ with the ordering where $x$ is a positive infinitesimal.

That is, a rational function is positive if and only if its Laurent series has a positive leading coefficient.

What ($\omega$-indexed) sequences converge to zero?

Well, for some $n$, we must have $s_m < x^2$ for all $m > n$. In particular, this means the leading term cannot be of the form $a x^j$ with $j < 2$, because such a thing would be greater than x^2. So for all $m > n$, the coefficient on $x_j$ is 0 for all $j < 0$.

A similar argument can be used in each degree; so we have a simple characterization of sequences that converge to zero: they are the sequences bounded on the left for which the sequence of coefficients on each $x^i$ is eventually always zero. (Note that it's not enough to simply converge to zero!)

The bounded criterion rules out things like the sequence

$$ 1, x^{-1}, x^{-2}, x^{-3}, \cdots $$

which diverges to $+\infty$.

Correspondingly, there is a simple condition for Cauchy sequences: they are precisely the sequences which are bounded on the left and for which each the sequence of coefficients on each $x^i$ is eventually constant.

Therefore, $R((x))$ is Cauchy complete (in the sense that every Cauchy sequence converges), and it is non-Archimedean because it has positive infinite numbers, such as $x^{-1}$.

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In this context, complete means "order complete" rather than complete in the sense of metric spaces. A non-archimedian ordered field will necessarily have infinitely small elements, so I suspect that if you try to impose a metric you'll get something pathological like the closure of 0 containing all of the infinitely small elements.

You can construct an order-complete example by taking the field $\mathbb{R}(x)$ of rational functions over $\mathbb{R}$. Order it such that $x$ is less than every positive number but bigger than 0, and take its order completion.

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