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A circle in Euclidean Geometry are all the point at same fixed distance (radius) from a single point called the center.

That fixed distance is only from every point to the center, but there are (obviously) different distances between point to point, by definition they are different points because they have a non-zero distance between them.

Well I wonder, (I've just imagine it and don't know if exist), an object of a special shape, of certain dimension, in a certain metric.. to have a single non zero distance between all points?

I mean, a shape/or metric, in wich points although perhaps they may have different distances between them, (as any known object in Euclides metric the circle, a plane, a sphere, a line, etc.), the special feature (or the special metric feature) would be that measuring those points from other metric, then all points are the same "distance" each other


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Any 2 point space, or the vertices of an equilateral triangle would do. You might want to look up "discrete metric" to see if it answers your question. – Jonas Meyer Nov 5 '11 at 22:09
Bouncing off of Jonas' answer, if you're interested in a geometric object that could "represent" a discrete metric space (of finite cardinality), you might try an n-simplex. – Robin Hoode Nov 6 '11 at 4:12
Bouncing off of robinhoode's comment, you could also try an orthonormal set of arbitrary cardinality – Jonas Meyer Nov 6 '11 at 4:23

Define a metric $d$ such that $d(x,y)=1$ if $x \not= y$ and $d(x,y)=0$ if $x=y.$ We should also check that $d$ satisfies the metric conditions:

i) $d(x,y) \geq 0$ for all $x$ and $y.$

ii) $d(x,y)=0$ iff $x=y.$ This is true by definition.

iii) $d(x,y)=d(y,x)$ which is obvious.

iv) $d(x,z) \leq d(x,y)+d(y,z)$ holds again by definition.

Therefore $d$ is a metric.

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It's indeed an answer, but it's also a formalization of my own question =), then question/answer remains almost open or incomplete because it still lack of a metric tensor and a way of transforming between it and a higher dimention discrete space, and this looks as a very hard task, like in continuous spaces with Nash Embedding Theorem, but I don't even know if there is some discrete equivalence on this. thanks – Hernán Eche Nov 10 '11 at 12:36
@HernánEche, I do not quite understand why you do not like this answer. Were you looking for a subset of $\mathbf{R}^n$? – hkBst Feb 13 at 17:56

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