Relative compactness and sequences such that $|x_p-x_q|\geq c$ Let $X$ be a Banach space and $B$ is a bounded subset of $X$.
If there exist a constant $c>0$ and a sequence $(x_n)_n\in B$ such that $$|x_p-x_q|\geq c,$$ for all $p,q$ with $p\neq q$, then $B$ is not relatively compact (its closure is not compact), because this sequence does not have a convergent subsequence.
Now if $B$ is not relatively compact , do we have the existence of such a sequence ?
If it's true, how can we construct it ?
 A: I'll assume you know the Heine-Borel theorem for metric spaces, which states that a subset $B$ of a metric space $X$ is compact if and only if it is complete and totally bounded, i.e. for every real $\epsilon > 0$, there is a finite set $\{x_1,\dots,x_n\}$ from $X$ such that $B\subseteq \bigcup_{i = 1}^n B_\epsilon(x_i)$, where $B_\epsilon(x)$ is the open ball of radius $\epsilon$ around $x$.
Now if $B$ is not relatively compact, then $\overline{B}$ is not compact. But since $X$ is complete, and closed sets contain all their limit points, $\overline{B}$ is also complete. Hence $\overline{B}$ is not totally bounded. Unpacking the definition, there exists $\epsilon > 0$ such that for all $\{x_1,\dots,x_n\}$ in $X$, there exists $y\in \overline{B}$ such that $y\notin \bigcup_{i = 1}^n B_\epsilon(x_i)$, so $|y - x_i|> \epsilon$ for all $i\leq n$. 
Taking $c = \epsilon/2$, we can inductively construct a sequence in $B$ with the desired property. Start with any $x_1$ in $B$. Given $x_1,\dots,x_n$, we can find $y\in \overline{B}$ such that $|y - x_i|> \epsilon$ for all $i\leq n$. But $y$ is a limit point of $B$, so we can find $x_{n+1}\in B$ such that $|y - x_{n+1}| < \epsilon/2$. By the triangle inequality, $|x_{n+1} - x_i| > \epsilon/2$ for all $i \leq n$.
