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I was reading (in a real analysis book) about this topological (I think it's topological) concept "diameter of a set of points". And there's a side note there which says: the diameter of a triangle (in $\mathbb{R}^2$) is equal to the length of its longest side.

This made me thinking... if we take two points $P,Q$ inside a closed triangle ABC with max side length $m$, how do we prove that $PQ \le m$. It seems to me it's not so easy, is it?

Also, is there an analogous statement for $\mathbb{R}^3$? Is it maybe about a tetrahedron (that the diameter of a tetrahedron is the length of its longest edge)? How does one prove that?

Sorry for the confusing tags, I am just not quite sure what tags to put here.

peter.petrov
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  • Thanks. But the question I am directed to obviously doesn't answer all the questions I asked. What about the analogue in the $R^3$ space? – peter.petrov Jan 05 '22 at 21:55
  • The proof uses that $\Vert x-y \Vert^2$ is a convex function of both arguments, and therefore attains its maximum at extremal points on a convex set. The same argument works for a tetrahedron in $\Bbb R^3$. – Martin R Jan 05 '22 at 22:01
  • @MartinR OK, thanks, I will reread it, see if I can understand it in full. I was hoping for some more elementary proof but anyway. OK, thanks. – peter.petrov Jan 06 '22 at 14:03

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