euclidean-geometry
Given a set of Euclidean Vectors with $N$ dimensions, whose distance from a Euclidean Vector, $R$, is less than some Constant, $C$.
Can the max distance between any two vectors in the set be determined?
I have been searching for some sort of proof or rule but I can't seem to fine one, when I picture a sphere with $R$ at the center, I believe the max distance would be $2C$. However I am unsure if this is true for dimensions greater than $3$.
Maybe the following will help. It's not an answer, but it's too long for a comment.
Suppose the given distance is $d$. Then $L$ must be simultaneously tangent to three spheres of radius $d$, one centered at each of the given points.
Consider the case where the distances between some pair of the points is $D>2d$. The radius $d$ spheres can be nestled inside a ruled surface of this shape (so long as it's sized appropriately):
Any line of the ruled surface is tangent to both spheres.
You can also fit two spheres with centers distance $D$ apart into a ruled surface that's more (or completely) cone-like, or into one that's less cone-like and more (or exactly) cylindrical. Any of these ruled surfaces will yield an infinite number of lines $L$ tangent to these first two spheres.
If the sphere of radius $d$ about the third point intersects any of these ruled surfaces (hyperboloids of one sheet with axis containing the first two spheres' centers - the cone and the cylinder being degenerate ones), it seems like there will be either one or two lines $L$ from that ruled surface where $L$ equidistant from the three points.
I think that the third sphere will intersect one of these ruled surfaces if and only if it intersects either the cone or the cylinder, so maybe only the degenerate hyperboloids need to be considered.
If none of the points are more than $2d$ units apart, the first two spheres will intersect in a circle. The families of lines tangent to both will comprise a cylinder and some hyperboloids, but you can't twist them completely into a cone. Instead, there will be a different degenerate ruled family of tangent lines: those in the plane through the spheres' circle of intersection that are also tangent to that circle. Again, if the third sphere intersects any of these families of lines (and I think in this case it must), you'll get solutions.
In fact those n+1 vertex polytopes of n-dimensional space with equal pairwise distances are called simplices (singular: simplex). This is, as those are the most simple polytopes within each dimension. For obvious reasons there will be no polytope within n-dimensional space with fewer vertices. Moreover those simplices all are regular polytopes.
The described construction by @AsuraPath is that of continued pyramidisation: take a vertex atop the simplex of one dimension less.
--- rk
Best Answer
The triangle inequality gives $\|x+y\| \leq \|x+z\| + \|z+y\|$ for any points $x,y,z$.
In your case, you have a set $A$ such that $\|a-R\| < C$ for all $a \in A$. Hence if $a_1,a_2 \in A$, you have $\|a_1-a_2\| \leq \|a_1-R\|+\|R-a_2\| < 2C$.
Following a point made by Cameron below, the term $\max$ is used when the actual limit can be attained at some point. So it would be more correct to say that the supremum (or least upper bound) of the distance is $\sup_{a_1,a_2 \in A} \|a_1-a_2\| = 2C$.