Show that if $S$ is compact, there are $x, y ∈ S$ so $|x − y| = d(S)$.
Where $ d(S)= \sup \{ \vec{x}-\vec{y}\}$ what i want to do is assume there is a sequence in S going to $\sup \{ \vec{x}-\vec{y}\}$ then use BW to choose a convergent sub-sequence of that sequence that converges to $x$ and a convergent sub-sequence of that sequence converging to y then since S compact this x, y are in S.
my concern is why can i say there a sequence $(x_n,y_n)$ to then pick a sub-sequence and how do i justify this?
Yes you can ! :P
Since S is compact it is also bounded , hence $d(S)<\infty$.
Remember the characterisation of sup : if $A\subset \Bbb R$ an upper bounded subset of the reals then the
following preposition holds $\alpha = supA \iff \forall \epsilon >0 $ there exist $b \in A$ such that
$\alpha - \epsilon < b \leq a$ $\ (*)$.
Now $d(S)$ by definition is the $sup$ of the set $\{||x-y||:x,y \in S\}$. So you start
using the $(*)$ consecutively for $\epsilon = \frac{1}{n}>0$.
1) for $\epsilon = 1>0$ there exist $x_1,y_1 \in S$ such that $d(S)-1 <||x_1 - y_1|| \leq d(S)$.
2) for $\epsilon=\frac{1}{2}>0$ there exist$x_2,y_2 \in S$ such that $d(S)-\frac{1}{2}<||x_2-y_2||\leq d(S)$.
So continuing with induction you can construct two sequences $(x_n,y_n)$ in $S$ such that
$d(S)-\frac{1}{n}<||x_n-y_n||\leq d(S)$ for all $n \in \Bbb N$.
Now $||x_n-y_n||\to d(S)$.
Since S is compact you can find a subsequence of $x_n$ lets say $x_{k_{n}}\to x \in S$.
Now again from the compacteness of $S$ you find a subsequence of $y_{k_{n}}$ lets say $y_{k_{s_{n}}}\to y \in S$.
Now since $||x_n-y_n||\to d(S)$ you also have that
$||x_{k_{s_{n}}}-y_{k_{s_{n}}}||\to d(S)$ $(1)$ and $||x_{k_{s_{n}}}-y_{k_{s_{n}}}||\to ||x-y||$ $(2)$.
Combining $(1),(2)$ you get $||x-y||=d(S)$ and $x,y \in S$. I hope this is better!