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Let $(M, d)$ be a metric space and $A\subset M$ such that $$\mathrm{diam}(A)=\sup_{a,b\in A}d(a,b)=D<\infty.$$ How can I prove that for any $\varepsilon> 0$ there is $x\in A$ such that $A\subset\{y\in M: d(x,y)<(\varepsilon +D)\}$?

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Hint: by contradiction assume that the claim is false. So there exists an $\epsilon>0$ such that A is NOT in the set of points within $\epsilon+D$ of $x$. Notice that $x\in A$ by assumption. You can break it down into two cases, one where $A$ has just a single point, and when $A$ has at least two points. –  Alex R. Jul 8 '12 at 19:34
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I don't see why it says "there is" -- isn't this rather trivially true for all $x\in A$? –  joriki Jul 8 '12 at 19:36

1 Answer 1

Suppose that for some $\epsilon>0$, it is the case that for any $x\in A$, $A\not\subset\{y\in M:d(x,y)<(\epsilon+D)\}$. In other words, suppose that for any $x\in A$, there is some $y\in A$ such that $d(x,y)\geq\epsilon+D$. Then $$\operatorname{diam}(A)=\sup_{a,b\in A}d(a,b)\geq D+\epsilon>D,$$ which is a contradiction. In fact, this is a contradiction if for even a single $x\in A$, there is some $y\in A$ with $d(x,y)\geq \epsilon+D$. Thus, our assumption that there exists such an $\epsilon$ must be false.

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