# How to define a “metric” whose range is not the reals?

This may sound a very stupid question. Why do we need to restrict a metric from a general set $X$ to map to the positive real numbers? I try to be clearer.

We are given a set $X$ and a totally ordered set ($Y,\succeq$) with least element $0$ and "an addition-like operation on it" denoted by $+$. A metric $d$ is a function $d:X\times X \rightarrow Y$ satisfying the following axioms $\forall x,y,z\in X$: (1) $d(x,y)\succeq 0$ if $x\neq y$ and $d(x,y)=0$ if $x=y$; (2) $d(x,y)=d(y,x)$; (3) $d(x,z)\preceq d(x,y)+d(y,z)$.

Does this definition make any sense? If yes, has work been done on this subject?

Thank you

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It does make sense. I might make the distinction "$Y$-valued metric" in this case. –  Cameron Buie Oct 16 '12 at 20:12
... and why totally ordered and not partially? The notion of "uniform space" is often used as a generalization of metric space. encyclopediaofmath.org/index.php/Uniform_space –  GEdgar Oct 16 '12 at 20:17
@ CameronBuie Thank you. Does any work exist on it? Would it be a prolific definition or is too exotic to be useful (in any sense)? –  dado Oct 16 '12 at 20:18
See mathoverflow.net/questions/10870/… for an attempt with a particular $X$ ("the" nonstandard reals). –  Jason DeVito Oct 16 '12 at 20:23
@GEdgar Would a poset fine as well? I thought we need the elements to be comparable. –  dado Oct 16 '12 at 20:23
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There is quite some work done exactly along the lines you suggest. Two sources that answer your question in somewhat different ways are: Flagg's "Quantales and continuity spaces" and Heckmann's "Similarity, Topology and Uniformity".

In more detail, given a quantale (that is a complete lattice together with a binary operation (taken to be commutative, associative, and whose unit is the bottom element of the lattice)) one can define a V space to be a category enriched in V. That amounts to a set $X$ and a function $d:X\times X\to V$ satisfying $d(x,x)\ge \bot$ and $d(x,z)\le d(x,y)+d(y,z)$. In that setting one can do quite a lot of the usual constructions, perhaps most notably the interpretation of Cauchy completeness (see http://ncatlab.org/nlab/show/Cauchy+complete+category). Particular choices of $V$ will recover familiar cases such as: ordinary (non-symmetric) metric spaces, posets, probabilistic metric spaces, as well as Lawvere's fundamental work on generalized metric spaces.

With any V-space one can associate (two) topologies in a way that extends the familiar one from ordinary metric spaces. Varying $V$ yields in this way familiar classes of topologies, most notably the Scott topology. This latter observation explains why V-spaces are studied in Domain Theory.

A nice early result is due to Flagg (and is related to earlier result of Kopperman): Every topological space is V-metrizable if one is allowed to choose $V$ based on the given topology.

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Thank you a lot. I will definitely look into some of the things you just suggested. –  dado Oct 16 '12 at 22:04

Yes, it makes sense, and yes, work has been done on such things. You’ll find work on it under the 2000 AMS Mathematics Subject Classification 46A19, though that does include a few other topics as well.

Non-Archimedean metrics lend themselves to a similar generalization that doesn’t even require a group operation on the linear order in which they take values. Let $\kappa$ be a regular cardinal, and suppose that $d:X\times X\to\kappa+1$ has the following properties:

1. For all $x,y\in X$, $d(x,y)=\kappa$ iff $x=y$.
2. For all $x,y\in X$, $d(x,y)=d(y,x)$.
3. For all $x,y,z\in X$, $d(x,z)\ge\max\{d(x,y),d(y,z)\}$.

For $x\in X$ and $\alpha<\kappa$ let $B(x,\alpha)=\{y\in x:d(x,y)\ge\alpha\}$. Then $\{B(x,\alpha):x\in X\text{ and }\alpha<\kappa\}$ is a base for a topology on $X$. If $\kappa=\omega$, this topology is generated by a non-Archimedean metric $\rho$ on $X$ given by $\rho(x,y)=2^{-d(x,y)}$. A closely related idea from algebra is the notion of a valuation.

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