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What does diameter mean in the following sentence of Borsuk's conjecture?

Sentence: Can every set $S \subseteq \Bbb R^d$ of bounded diameter $\operatorname{diam}(S)>0$ be partitioned into at most $d+1$ sets of smaller diameter?

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If $m:\Bbb R^d\times\Bbb R^d\to\Bbb R$ is the relevant metric, then $$\mathrm{diam}(S):=\sup\{m(x,y):x,y\in S\}.$$ Intuitively, it is the least upper bound of the pairwise distances between points of $S$. Thus, $S$ is bounded if and only if it has a finite diameter (a good exercise).

For example, if $S$ is the unit ball (open or closed), one can see that $\mathrm{diam}(S)=2$. If $S$ is a singleton, then it has zero diameter. If $S$ is a finite, non-empty set, then the diameter is the maximum of the pairwise distances between its points. If $S$ is a right triangular plane region, then its diameter is the length of the hypotenuse.

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The diameter of a set $S \subset \mathbb{R}^n$ is defined as $$\operatorname{diam}(S) = \sup_{x,y \in S} \|x-y\|$$

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May you give an example please? – Victor Jul 20 '12 at 22:42
@Victor: Let $S=\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2<1$, the interior of the unit disk centred at the origin. Then $\sup\{\|x-y\|:x,y\in S\}=2$. – Brian M. Scott Jul 20 '12 at 22:45
Sure. For a ball, it is the "usual" diameter. For a cube, it's the length of the space diagonal. I.e it's the largest distance (or more precisely, the least upper bound of the distance) between two points in the set. – mrf Jul 20 '12 at 22:45

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