Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.