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 $A\subset\{y\in M: d(x,y)<(\varepsilon +D)\}$?
How to show this property for the diameter of a set
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2I don't see why it says "there is" -- isn't this rather trivially true for all $x\in A$? – 2012-07-08
1 Answers
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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.