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The problem is: For compact metric space $(X,d)$ prove that for every $r>0$ there exists a subset $S$ of $X$ such that $\{\mbox{Open balls of radius }r\mbox{ centered at }p \mid\mbox{ for all }p \in S\}$ forms a cover for $X$ and for every $p,q \in S$, $d(p,q) > r/2$.

I have an algorithm that would go like this: Form an open cover of $X$ by the set of open balls of radius $r/2$ around all points in $X$. By compactness there exists a finite number of these balls which cover $X$. Then for each point with a ball around it, "delete" the points which are inside of the ball and not the center of the ball. Then you will have a collection of points that are at least $r/2$ distance apart and the set of balls of radius $r$ around these points will cover $X$.

Does this "algorithm" work, and if so how do you denote such a set? I'm having problems figuring out exactly how to "delete" as I've used the word above.

Thanks!

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    your algorithm works perfectly; you just need to linearly order the (finitely many) balls, then see whether the center of the 1st ball is in another ball (if yes then remove the 1st ball), then the 2nd, etc.2012-11-01
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    Ok yeah I thought it would work. The issue is I need to construct a set of the correct points and I'm not sure how to denote such a set...2012-11-02

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