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The long line is much longer than $\mathbb{R}$, and indeed many chains have this property. Thus, since metrics are usually assumed to be real-valued, this can be understood as an assumption that point-to-point distances are "short" in some sense.

Presumably, this has a variety of benefits. For example, every metrizable space is paracompact, a smallness condition that apparently follows from the shortness of $\mathbb{R}$.

However, I also think it has some drawbacks. There are many, many spaces that simply aren't metrizable. So I think that if we generalize the notion of a metric $d$ such that the codomain needn't be $\mathbb{R},$ this would make a larger class of spaces metrizable.

Reference request. Has the idea of generalizing the codomain of a metric been seriously considered? If so, a link or reference would be appreciated.

Remark. Its true that $\mathbb{R}$ is the unique Dedekind-complete linearly ordered field. So this would seem to put a kink in the idea. However, I think that $\mathbb{R}$ has much more structure than we really need to do metric space theory. In particular, every real number $x$ has an additive inverse $-x,$ and so long as $x$ is non-zero it has a multiplicative inverse $x^{-1}$. I don't think either of these properties are truly necessarily for the codomain of a metric.

goblin GONE
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  • paracompactness of metrizable spaces is not so much a consequence of the shortness of $\mathbb R$, as much as it is a consequence of it being separable. – Ittay Weiss Dec 04 '13 at 01:31

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I wrote a short note just recently on the precise question you are asking: A note on the metrizability of spaces.

The short answer is, yes, it was done and is very elegant and useful. My note is based on Flagg's work "quantales and continuity spaces".

Ittay Weiss
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  • It has been done in other ways, some of which are far more attractive from a purely topological point of view. – Brian M. Scott Dec 04 '13 at 01:34
  • @BrianM.Scott references please? – Ittay Weiss Dec 04 '13 at 01:38
  • and @BrianM.Scott if you have the time and if you feel like commenting on my note, I'd be very grateful. – Ittay Weiss Dec 04 '13 at 01:39
  • It’s been way too long since I looked at them for me to be able to come up with references; you could look for $\omega_\mu$-metrizable spaces and, more generally, linearly uniformizable spaces, with generalized metrics taking values in an ordered Abelian group. // I’m not a good person to comment on the note, I’m afraid: category theory and I don’t get along at all well. – Brian M. Scott Dec 04 '13 at 01:51
  • ok @BrianM.Scott, thanks! – Ittay Weiss Dec 04 '13 at 02:21
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    @BrianM.Scott, far more attractive than the usual "codomain = $\mathbb{R}$" approach, or far more attractive than Flagg's approach? Having spent the last half hour reading Flagg's stuff, I have to say it looks pretty elegant. – goblin GONE Dec 04 '13 at 02:49
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For concrete examples, if you have any ordered field, you can rewrite its order topology in terms of its "self-metric". In the largest (in some sense, most general) case, this was done for the surreals. Unfortunately, this doesn't really lead anywhere useful because of your remark: Everything bigger than the reals has gaps (and subfields of the reals have "holes":$\mathbb Q$-cauchy sequences that don't converge).

Mark S.
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    The requirement that whatever the codomain of a metric function should be, it should at least be a field is a strange one. Very little of the field structure of $\mathbb R$ is ever used in metric considerations. If one does not insist on having a field, but rather concentrates on the lattice properties one really needs, then every space is metrizable. – Ittay Weiss Dec 04 '13 at 01:36
  • @IttayWeiss I did not mean to imply that it was a good requirement, merely a source of a convenient class of examples one could consider, and that was considered. – Mark S. Dec 04 '13 at 01:43