This question is about the proof of the following theorem by W.H.Schikhof in his book Ultrametric Analysis : An introduction to $p$-adic analysis:
Definition Let $(X,d)$ be a ultrametric space.
- $X$ is dense, if for any closed ball $B$ in $X$, the set $\{d(x,y)|x,y\in B\}$ is dense in $[0,\operatorname{diam}(B)]$.
- $X$ is separable, if it admits a countable dense subset.
In the proof,
the equivalence relation defined by the formula $d(x,y)\leq r_1$ divides $X$ into at least two balls; one of them, $B_1$ say, does not contain $a_1$.
This does make sense: the denseness allows us to find two points $x',y'\in B_0:= B(x,r_0)$ with $d(x',y')\in(r_1,r_0)$, thus $x'$ and $y'$ are not in the same equivalence class (i.e. the same ball with radius $r_1$ in $B_0$). Since different ball with same radius are disjoint in the ultrametric world, we know that we can always find a ball with radius $r_1$ that does not contain $a_1$.
My question is that why Schikhof make the conclusion that $\operatorname{diam}(B_1)=r_1$? It is clear that $\operatorname{diam}(B_1)\leq r_1$, but I don't see the converse inequality.
Actually, I don't think $r_1$ can always be realized as the diameter of a subball of $B_0$: the denseness can only guarantee that there exists a ball $B_1'$ in $B_0$ with diameter $r_1^+<r_0$ not less than (but very closed to) $r_1$. This seems to be harmless at the first glance, but now we have to show that $B_1'$ is not the only subball of $B_0$ with radius $r_1^+$, and this is what I don't know how to achieve. By the denseness of the metric, one can always find a ball $B_1''$ with larger diameter in $B_0$, but I don't know how to rule out the possibility that $B_1'\subset B_1''$.