Is a metrizable space $X$ compact $\iff \exists x,y$ such that $d(x,y) = \sup d(u,v)_{u,v \in X}$? Is a metrizable space $X$ compact $\iff \exists x,y$ such that $d(x,y) = \sup d(u,v)_{u,v \in X}$?
Is the condition that $x,y$ at least necessary for the space to be compact?
I have no idea  how to approach this using the definition of covers for $X$. Any help would be appreciated.
 A: No. Take $X = [-1,0)\cup (0,1]$. Then $d(-1,1) = 2 = \sup_{u,v\in X} d(u,v)$ but X is clearly non compact (for instance the open cover $ ([-1,1]\setminus[-\frac1n,\frac1n])_{n\in \mathbb N}$ doesn't have a finite subcover).
A: I think total boundedness is the way to frame the problem. The $\Longleftarrow$ direction would mean that you're assuming boundedness. In infinite dimensional metric spaces, total boundedness and boundedness are quite different animals. Consider $\ell^{\infty}$ and the subset $X = \{e_i\}_{i=1}^{\infty}$ where $e_i = (0,\ldots, 0,1,0\ldots)$ and the $1$ is in the $i$th place.
This subset $X$ is closed and bounded since the distance between any two elements of $X$ is $1$ (so the supremum is achieved), but what can you say about any ball of radius $\frac{1}{2}$ around any of the points in $X$? What does this say about compactness?
(This approach is equivalent to using an open cover, but I think explicitly framing the problem in terms of total boundedness makes it easier to see how things might play out.)
