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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    $\begingroup$ 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. $\endgroup$ Nov 5, 2011 at 22:09
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    $\begingroup$ 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. $\endgroup$ Nov 6, 2011 at 4:12
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    $\begingroup$ Bouncing off of robinhoode's comment, you could also try an orthonormal set of arbitrary cardinality $\endgroup$ Nov 6, 2011 at 4:23

3 Answers 3


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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    $\begingroup$ 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 $\endgroup$ Nov 10, 2011 at 12:36
  • $\begingroup$ @HernánEche, I do not quite understand why you do not like this answer. Were you looking for a subset of $\mathbf{R}^n$? $\endgroup$
    – hkBst
    Feb 13, 2016 at 17:56

With any metric in Euclidean space:
line endings in $\mathbb{R}^1$
vertices of equilateral triangles in plane $\mathbb{R}^2$
tetrahedrons in 3-space $\mathbb{R}^3$

These are made by adding one point, of two possible, to the previous. This seems extendable to higher dimensions.


In the discrete metric the whole plane is equidistant to any other point. How do we restrict that? This is one example I found inspired by the discrete metric. d0(x,y) is the metric I constructed.

\begin{equation} d_{0}(x,y) = \left\{ \begin{array}{lr} 0, & \text{if } x=y\\ 1, & \text{if } 0\leq d_{1}(x,y)< 1\\ 2, & \text{else } \end{array} \right\} \end{equation} $$d_{1}(x,y)= |x-y|$$ d1(x,y) can be any metric but, I will choose the euclidean one.

For the Euclidean Metric any subset of vertices in the open ball of radius 0.5 are equidistant. So is a square lattice of points spaced one or greater units away.

I came up with this construction myself so I don't know if it has a name.


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