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Can I have a metric space where the distance between two points is an arbitrary constant? Does this mean that there cannot be 'co-linear' points in the space? i.e. if A B and C are colinear, and B is not the same point as C, then the distance from A to B is my constant, from B to C is also my constant, and from A to C via B, is still my constant.

How does this make sense?

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It called the discrete metric. Colinear is a vague term. As it would depends on your definition and if it makes sense in a metric. – simplicity Oct 19 '11 at 0:43
So not all metric spaces can have something like co-linearity? – soandos Oct 19 '11 at 0:47
It depends what you call a straight line. If you take the taxicab metric the straigh lines are actually zig zags. In hyperbolic geometry the straight lines are arcs. – simplicity Oct 19 '11 at 0:51
But there is always some form of it no...? – soandos Oct 19 '11 at 0:52
soandos, there is no reason to expect any acceptable concept of collinearity in a metric space. – Will Jagy Oct 19 '11 at 1:09
up vote 4 down vote accepted

If $X$ is a set and $r$ is a positive number, we can define a function $d:X\times X\to\mathbb R$ so that for each $x$, $y\in X$, $$d(x,y)=\begin{cases}r, &\text{if }x\neq y;\\0&\text{if }x=y\end{cases}$$ and you can easily check that all the conditions for $d$ to be a metric are satisfied.

Therefore, the answer to your question is yes.

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I just made up a notion of metric space in which collinearity makes sense, see my answer. Wonder what these are like? – Will Jagy Oct 19 '11 at 1:11
You are correct about the 7 point plane, O wise one. – Will Jagy Oct 19 '11 at 2:46

As others have said, the discrete metric would meet your first requirement. It is often given a distance of 1 between distinct points, but there is no reason why you should not scale this.

You might define co-linearity as having equality in the triangle inequality, so saying that if $d(A,B)+d(B,C)=d(A,C)$ then we call $A$, $B$ and $C$ co-linear (and $B$ is between $A$ and $C$). The discrete metric does not have three co-linear points.

You could invent another metric where, for some point $O$, $d(A,O)=1$ whenever $A\not = O$ and $d(A,B)=2$ whenever $O\not =A\not =B\not =O$, so if $O$ is ignored then an example of a discrete metric. In this case $A$, $O$ and $B$ are always co-linear, but co-linearity is not a transitive property: you might have $A$, $O$ and $B$ co-linear, $A$, $O$ and $C$ co-linear, and $B$, $O$ and $C$ co-linear, but $A$, $B$ and $C$ not being co-linear.

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To expand on your other idea, for the case of a metric space we would want "collinearity" to resemble an equivalence relation. That is, three points $x,y,z$ will be called collinear if either (1) at least two of the points are actually the same point, or (2) if we rename the points so that the largest of the three distances is $d(x,z),$ then we require $d(x,z) = d(x,y) + d(y,z).$ The part about equivalence is just this: we require that if $(x,y,z)$ are collinear, and then $(x,y,t)$ are collinear, we require that the other two triples are also collinear: $$ (x,z,t), \; (y,z,t). $$

It is always possible that this kind of metric space has a name. If finite, it should be possible to isometrically imbed this in a Euclidean space. But I am not sure what happens if you define such an object and allow it to have infinitely many points. Hard to say, I just made it up.

Anyway, a metric space with Mariano's discrete metric cannot have this property. That is, there is no acceptable way to define "collinear" in that case.

P.S. Mariano and Fernando Q. Gouvea are the same person.

P.P.S. As pointed out by Mariano Suárez-Gouvea-Alvarez, even in the case of the 7 point plane with each uninterrupted segment having length 1, the result cannot be imbedded isometrically in a Euclidean space. Live and learn. The reprint by Wojtech Skaba does not include a metric space structure consistent with his collinearity structure, so this is slightly new territory, maybe there is something amusing there.

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Cute. It appears people have considered this: see You outed me! – Mariano Suárez-Alvarez Oct 19 '11 at 1:22
Consider the 1-skeleton of a barycentrically subdivided triangle, and put on it the combinatorial distance. Then collinearity is the usual collinearily, but one cannot isometrically embed it, no? (In an isometric embedding, the points distinct from the center have to be placed at the vertices and midpoints of an equilateral planar triangle, and one cannot put the 7th point anywhere) – Mariano Suárez-Alvarez Oct 19 '11 at 1:34

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