P-adic balls are clopen and intersecting comparable balls are nested

If $x$ lies in the open ball $B(a,r)$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p<r$. For any $y$ with $\mathrm{\mid }y-x{\mathrm{\mid }}_p<r$, [L1] gives $\mathrm{\mid }y-a{\mathrm{\mid }}_p\le \max \{\mathrm{\mid }y-x{\mathrm{\mid }}_p,\mathrm{\mid }x-a{\mathrm{\mid }}_p\}<r,$ so $y\in B(a,r)$. Thus every point of an open ball is again a center, hence every open ball is open and every closed ball is open by the same argument with $\le r$.

If $\mathrm{\mid }y-x{\mathrm{\mid }}_p<\mathrm{\mid }x-a{\mathrm{\mid }}_p$, then [L1] forces $\mathrm{\mid }y-a{\mathrm{\mid }}_p=\mathrm{\mid }x-a{\mathrm{\mid }}_p$. Hence, if $x\mathrm{\notin }B(a,r)$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p\ge r$ and so $y\mathrm{\notin }B(a,r)$; and if $x$ lies outside the closed ball of radius $r$ around $a$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p>r$ and so $y$ lies outside that closed ball as well. Thus the complement of either the open or the closed ball around $a$ is a union of open balls. Therefore every open or closed ball is also closed.

If $B(a,r)$ and $B(b,s)$ intersect, choose $x$ in the intersection. For any $y\in B(a,r)$, $\mathrm{\mid }y-b{\mathrm{\mid }}_p\le \max \{\mathrm{\mid }y-x{\mathrm{\mid }}_p,\mathrm{\mid }x-b{\mathrm{\mid }}_p\}<\max \{r,s\}=s,$ so $y\in B(b,s)$. The same proof works for closed balls. Hence intersecting comparable balls are nested.