Let $E$ be a compact nonempty subset of $\mathbb R^k$, and let $\delta=\sup\{d(x,y): x, y\in E\}$. Show $E$ contains points $x_0,y_0$ such that $d(x_0, y_0) =\delta$.

I’ve managed to prove this by considering sequences $(x_n),(y_n)$ in $E$ such that for each $n\in\mathbb N,\delta-\frac{1}{n}< d(x_n,y_n)\leq\delta$. Then obviously $\lim_{n\to\infty}d(x_n,y_n)=\delta$. However, what I was wondering: is it true that $d(\inf E,\sup E)=\delta$? That was my original idea, and it seems correct to me, but I wanted to check here.

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    $\begingroup$ What is $\inf E$ for $E\subset \mathbb R^k$? $\endgroup$
    – celtschk
    Mar 5 '17 at 11:07
  • $\begingroup$ @celtschk I'm guessing we haven't assigned meaning to $\inf$ for k>1 dimensional Euclidean space. The answer below confirms that. In hindsight, I could have checked it myself - but thanks anyways! $\endgroup$
    – Sha Vuklia
    Mar 5 '17 at 11:38

There is an elegant approach: consider the function $f:E\times E \to R$, $f(x,y)=d(x,y)$. As it is a continous function (why?) on a compact set, it has a maximum, so that means there is a pair $(x_0, y_0)\in E$ so that $f(x_0, y_0)=\sup\{d(x,y): x, y\in E\}=\delta$.

Note that in $\mathbb R^k$, $k>1$, there's no order (usually), so the concepts of supremum and infimum have no sense. In $\mathbb R$, that would be true: $d(\inf E, \sup E)=\delta$.


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