# Show a compact metric space $(X, \rho)$ has points $u,v$ such that diam $X = \rho(u,v)$

For a compact metric space $(X, \rho)$, show that there are points $u,v \in X$ for which $\rho(u,v) =$ diam $X$

I know that diam $X = sup$ {$\rho(u,v)|u,v \in X$} so does this just follow immediately from the Extreme Value Theorem?

Proof:

$\rho$ is continuous since it is a metric, so by EVT, $p$ takes its max value on X, therefore it takes its supremum on X so there exists points $u,v \in X$ such that $\rho(u,v) = sup$ {$\rho(x,y)|x,y \in X$} = diam $X$

Is this correct and is it missing any details? It feels way too easy so I think its not right

• Yeah, this isn't a valid proof (though you're going in the right direction to use the EVT). $\rho$ is not a function defined on $X$, so you can't apply it directly. Can you say what space $\rho$ is a function on? Can you show that space is compact, and can you show that $\rho$ is continuous? – JonathanZ supports MonicaC Nov 16 '17 at 1:10
• ... and it depends on what results you are allowed to use (as "known facts"). The fact needed here, to show the space on which $\rho$ is indeed a function, is compact, is not entirely trivial. – mathguy Nov 16 '17 at 1:12
• @JonathanZ is it defined on $\mathbb{R}$? – Vinny Chase Nov 16 '17 at 1:15
• @mathguy Then is there a better way to go about this? If this is the best way, could you explain how to go about showing the space $\rho$ is defined on is compact? – Vinny Chase Nov 16 '17 at 1:16
• @JonathanZ Is it $X$ x $X$? – Vinny Chase Nov 16 '17 at 1:18

Missing details: (1).$\rho$ is continuous from $X^2$ to $\Bbb R.$ (2). $X^2$ is compact because $X$ is compact. Error :$\rho$ does not take its maximum on $X.$ The domain of $\rho$ is $X^2.$ So $\rho$ takes its maximum on $X^2$ by EVT. Assuming $X\ne \phi.$
BTW , We can prove it briefly by entirely elementary means, using only the fact that any sequence in $X$ has a convergent sub-sequence. I will write it out on request, in return for the homeopathic essence of 24 beer.