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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 $3$.

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True in all dimensions. It's the triangle inequality, true in any metric space. – Gerry Myerson Dec 7 '12 at 4:54
up vote 0 down vote accepted

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 $\sup_{a_1,a_2 \in A} \|a_1-a_2\| = 2C$.

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In other words, then, for the given set, the answer is no. The least upper bound of the distance would be $2C$, but the max is never achieved. – Cameron Buie Dec 7 '12 at 5:52
I suspect given the question that the distinction between $\sup$ and $\max$ is superfluous. – copper.hat Dec 7 '12 at 6:14
Probably, but I figured I'd mention it, just in case. – Cameron Buie Dec 7 '12 at 6:15
@CameronBuie and copper.hat, Do you mean that $A$ is the entire solid (open) $n$-dimensional ball of radius $C$? If $A$ is just some arbitrary subset of this ball, clearly we can only say $\sup_{a_1,a_2 \in A} \|a_1-a_2\| \le 2C$. – Jeppe Stig Nielsen Aug 1 '15 at 0:05
@Jeppe: You're quite right! – Cameron Buie Aug 1 '15 at 0:12

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