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I was wondering why metrics and norms are always defined to be real, rather than generalized to some other fields (or whatever). The best guess I have so far is:

Because every Archimedean ordered field is (up to unique isomorphism) a subfield of $\mathbb R$ anyway.

But is that actually true? And if it is, can it be strengthened to "every Achimedean ordered ring"? Or even semiring?

I know $\mathbb R$ is the only complete Archimedean field. But a priori, I suppose there could be non-complete examples that cannot be completed (without losing the Archimedean property).

hmakholm left over Monica
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  • Duplicate? http://math.stackexchange.com/questions/21435/equivalency-of-archimedian-fields-properties – t.b. Oct 27 '11 at 00:26
  • My guess is that we understand $\mathbb R$ pretty well, and it is nicely behaved. I had a discussion about this two years ago with two classmates. Eventually the teacher (measure theory) told us that you can define a valuation into any ordered Abelian group. My second guess would be that completeness is just very nice to have, just like complete Boolean-algebras are used for Boolean-valued models because incomplete ones could have critical holes that possibly remove well-definability. – Asaf Karagila Oct 27 '11 at 00:28
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    I guess this depends on your definition of Archimedean, but I think $\mathbb{C}$ is considered Archimedean although it is not ordered. So $\mathbb{R}$ is the only complete Archimedean ordered field. – Joel Cohen Oct 27 '11 at 00:37
  • @Joel, it's not clear to me what "Archimedean" would mean in the absence of an ordering. Please elaborate. – hmakholm left over Monica Oct 27 '11 at 01:06
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    @HenningMakholm : If $K$ is a field with an absolute value $|| : K \to \mathbb{R}^+$ (satisfying $|x| = 0 \Rightarrow x = 0$, $|ab| = |a| . |b|$, and $|a+b| \le |a| + |b|$), we say it is archimedean if integers (or precisely the image of the map $1_{\mathbb{Z}} \mapsto 1_K$) are not bounded. This definition implies an archimedean field with an absolute value is of characteristic $0$, so contains $\mathbb{Q}$. And it turns out all of then are contained in $\mathbb{C}$. Non archimedean complete fields with an absolute value include $p$-adic numbers $\mathbb{Q}_p$. – Joel Cohen Oct 27 '11 at 01:24

2 Answers2

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I defer to Proposition 12 (well... the second Proposition 12...) and Theorems 14 and 15 in this answer of mine.

It is not hard to construct an argument that $\mathbb{R}$ is a final object in the category of Archimedean fields from these results. For example see the notes that Pete L. Clark links to on the same page.

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kahen
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  • Your Theorem 14 takes nicely care of my principal question. I'm still wondering about possible strengthenings. I'm looking for a sort of minimal set of things one might want to do with the metric such that only substructures of the reals satisfy them. – hmakholm left over Monica Oct 27 '11 at 00:51
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    @kahen: We need a unique morphism from each Archimedean ordered field into $\mathbb{R}$ in order for it to be terminal. Unless I'm mistaken, there are two embeddings of $\mathbb{Q}[\sqrt{2}]$ (with its usual absolute value) into $\mathbb{R}$, so $\mathbb{R}$ cannot be terminal. – Zhen Lin Oct 27 '11 at 06:22
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    @Zhen Lin: one of these embeddings does not preserve order. (We're implicitly considering the category of Archimedean ordered fields with order-preserving field homomorphisms). – hmakholm left over Monica Oct 27 '11 at 11:37
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I wrote this answer to a related question. I think one can show as a consequence of this stuff that the Dedekind completion of a non-Archimedean ordered field is actually not a field.

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Michael Hardy
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