Prove there exists farthest points in a compact set.

Question

Let $K\subset\Bbb{R}^n$ be a non-empty compact set. Prove that there exists $a,b\in K$ so that $\|x-y\|\le\|a-b\|$ for all $x,y\in K$.

I thought it might be an application of extreme value theorem but I got to the point where for a fixed point $x_0\in K, \exists a,b\in K$ such that $\|x_0-a\|\le\|x_0-x\|\le\|x_0-b\|$ but I'm not sure where to proceed from here.

... or this one: diameter on a compact metric space. – Martin R Oct 17 '15 at 18:17 — Martin R, Oct 17 '15 at 18:17

score 2 · Answer 1 · answered Oct 17 '15 at 18:16

2

Use the product topology on $K×K$. This is still a compact set and thus $f(\,(x,y)\,)=\|x-y\|$ will have a maximum.

answered Oct 17 '15 at 18:16

Lutz Lehmann

126,666

score 2 · Answer 2 · answered Oct 17 '15 at 18:16

2

Hint: Consider the map $f:K\times K\to\mathbb{R}$ given by $f(x,y)=\|x-y\|$.

answered Oct 17 '15 at 18:16

Eric Wofsey

330,363

Prove there exists farthest points in a compact set.

2 Answers2