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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).

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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 $\mathbb{Q}_p$.2011-10-27

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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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    @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).2011-10-27
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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.