Just to clarify the notation and the question:
Working in p-adic space $\mathbb{Q}_p$, we have the norm $|x|_p=p^{-ord_p(x)}$ and we define the metric over this space as $d(x,y)=|x-y|_p$. We are considering the open ball $B(0;r)=\{x\in \mathbb{Q}_p : |x|_p<r\}$. The question is to find the supremum of the metric $d$ over $B$.
My work so far:
Using the strong triangle inequality one can show that for any open ball $B(a;r)$ in p-adic space, $r\geq \sup(\{d(x,y) : x,y\in B\})$. So we get that $r$ is an upper bound. My guess is that $r$ actually is the $\sup$, since there are no other (obvious) candidates, but I am unsure how to go about proving this. One thought I had was to show that $r\leq\sup(\{d(x,y) : x,y\in B\})$, thereby getting equality. But I could not see how to show this.
Any help, hints, or solutions are greatly appreciated.
The $p$-adic absolute value takes values in $p^\mathbb{Z} \cup \{0\}$, which is discrete apart from the limit point $0$. Let $n \in \mathbb Z$ be such that $p^n < r \leq p^{n+1}$. Then for $x,y \in B(0,r)$ one has $|x-y|_p < r$, thus $|x-y|_p \leq p^n < r$. So $r$ can't be the supremum $\{ |x-y|_p : x,y, \in B(0,r) \}$. Rather the supremum is $p^n$ as $|p^{-n}|_p = p^n$ and $p^{-n} \in B(0,r)$.