4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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
3
$\begingroup$

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: $d(\inf E, \sup E)=\delta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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