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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

Vinny Chase
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    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 Nov 16 '17 at 01:10
  • ... 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. –  Nov 16 '17 at 01:12
  • @JonathanZ is it defined on $\mathbb{R}$? – Vinny Chase Nov 16 '17 at 01:15
  • @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 01:16
  • No. $\mathbb{R}$ is the range of $\rho$. You need to identify its domain. – JonathanZ Nov 16 '17 at 01:16
  • Well, what space is $\rho$ defined on? If you are unsure, take a quick look in the textbook, it won't hurt you! –  Nov 16 '17 at 01:17
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    @JonathanZ Is it $X$ x $X$? – Vinny Chase Nov 16 '17 at 01:18
  • Bingo! Now, it is a theorem that the product of two compact metric spaces is again compact. It is very likely that you are allowed to quote this (you are not asked to PROVE it from scratch). –  Nov 16 '17 at 01:19
  • @mathguy ok perfect! Thanks that was actually the problem before this. so now I have that $\rho$ is continuous on X x X so it takes its maximum on X x X and therefore its supremum so I'm done? – Vinny Chase Nov 16 '17 at 01:23
  • Oh, then it makes perfect sense. Yes, that's it. –  Nov 16 '17 at 01:25
  • It might seem obvious that $\rho$ is continuous on $X$ x $X$ (and it's true that it is), but you have to make an argument showing so. Then you can apply the EVT and you're done. – JonathanZ Nov 16 '17 at 01:25
  • @JonathanZ With a $\delta - \epsilon$ proof? – Vinny Chase Nov 16 '17 at 01:27
  • That's how I'd do it. Make sure to check in your text book what metric you should use on $X \times X$. There are a few possibilities, which all end up being equivalent to each other, but you should use the one specified in your text. – JonathanZ Nov 16 '17 at 01:30
  • Also https://math.stackexchange.com/q/124838 –  Nov 16 '17 at 02:48

1 Answers1

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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.

DanielWainfleet
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