Is the intersection of two annuli in a non-Archimedean space an annulus? Let $A$ be a ball in a non-Archimedean topological space, 
$A = \{x: |a-x| < r \}$ and let A contain two Balls that are disjoint from each other:
$B = \{x: |a-x| < s \}$ and $C = \{x: |b-x| < t \}$, so that $s \leq r$ and $t \leq r$, where $|\cdot|$ is a non-Archimedean absolute value.
Naturally, the complements of $B$ ( or $C$) in $A$ are annuli, but what about $A\setminus(B \cup C)$? I am trying to show that this is an annulus too, but I am not sure if I am right.
 A: It need not be an annulus; here’s a counterexample.
Let $X=\Bbb N^{\Bbb N}$, where $\Bbb N$ has the discrete topology; $X$ is homeomorphic to the irrationals. For distinct $x=\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ in $X$ let 
$$\delta(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}\;,$$
and let
$$d:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases}
0,&\text{if }x=y\\
2^{-\delta(x,y)},&\text{otherwise}\;;
\end{cases}$$
then $d$ is a non-Archimedean metric on $X$ that generates the product topology, and for each $x\in X$ and $n\in\Bbb N$ we have
$$B\left(x,2^{-n}\right)=\{y\in X:y_k=x_k\text{ for }k\le n\}\;,$$
where $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$. 
Note that if $0<\epsilon\le 1$, there is a unique $n\in\Bbb N$ such that $2^{-n-1}<\epsilon\le 2^{-n}$, and for that $n$ we have $B(x,\epsilon)=B\left(x,2^{-n}\right)$ for each $x\in X$, while if $\epsilon>1$, then $B(x,\epsilon)=X=B(x,2)$ for each $x\in X$. Thus, we need only consider radii of the form $2^{-n}$ for integers $n\ge -1$.
Now let $x$ be the sequence such that $x_n=0$ for each $n\in\Bbb N$, and let $y$ be the sequence such that $y_1=1$, and $y_n=0$ if $n\ne 1$: $x=\langle 0,0,0,0,\ldots\rangle$, and $y=\langle 0,1,0,0,\ldots\rangle$. Clearly $\delta(x,y)=1$, so $d(x,y)=2^{-1}$. Let $r=1$ and $s=t=2^{-1}$. Then
$$\begin{align*}
B(x,1)&=\{z\in X:z_0=0\}\;,\\
B\left(x,2^{-1}\right)&=\{z\in X:z_0=z_1=0\}\;,\text{ and}\\
B\left(y,2^{-1}\right)&=\{z\in X:z_0=0\text{ and }z_1=1\}\;,
\end{align*}$$
so $B\left(x,2^{-1}\right)$ and $B\left(y,2^{-1}\right)$ are disjoint open balls contained in $B(x,1)$. Let 
$$U=B(x,1)\setminus\left(B\left(x,2^{-1}\right)\cup B\left(y,2^{-1}\right)\right)=\{z\in X:z_0=0\text{ and }z_1>1\}\;;$$
I claim that $U$ is not an annulus.
For each $n\ge 2$ let $u^{(n)}\in X$ be such that $u_k^{(n)}=n$ if $k=1$, and $u_k^{(n)}=0$ otherwise; clearly
$$U=\bigcup_{n\ge 2}B\left(u^{(n)},2^{-1}\right)\;,$$
and the sets $B\left(u^{(n)},2^{-1}\right)$ are pairwise disjoint. If $U\subseteq B\left(u,2^{-m}\right)$ for some $u\in X$ and $m\in\Bbb N$. Then
$$B\left(u^{(2)},2^{-1}\right)\subsetneqq B\left(u,2^{-m}\right)=B\left(u^{(2)},2^{-m}\right)\;,$$
so $m<1$, i.e., $m=0$, or $m=-1$, and $B\left(u,2^{-m}\right)$ is either $B(u,1)$ or $X$. Now 
$$B(u,1)=B\left(u^{(2)},1\right)=\{z\in X:z_0=0\}=B(x,1)\;,$$ 
where the first equality holds because $u^{(2)}\in U$ and the metric is non-Archimedean, so $U$ is an annulus if and only if either


*

*$U=X\setminus B\left(w,2^{-n}\right)$ for some $w\in X$ and $n\ge -1$, or  

*$U=B(x,1)\setminus B\left(w,2^{-n}\right)$ for some $w\in B(x,1)$ and $n\ge -1$.


Either possibility requires that
$$B\left(x,2^{-1}\right)\cup B\left(y,2^{-1}\right)\subseteq B\left(w,2^{-n}\right)$$
and hence that $n<1$, so that $B\left(w,2^{-n}\right)$ is either $B(x,1)$ or $X$. In either case $U\subseteq B\left(w,2^{-n}\right)$ and hence $U=\varnothing$, which is false, so $U$ cannot be an annulus.
