If $K$ is a field, and $\Abs{\cdot}$ a nontrivial non-archimedean valuation (or absolute value) on $K$ , then $\Abs{\cdot}$ has some properties that are counterintuitive (and that are false for archimedean valuations).

Theorem 1   Let $K$ be a field with a non-archimedean absolute value $\Abs{\cdot}$ . For $r>0$ a real number, $x\in K$ , define

the open ball of radius $r$ at $x$
the closed ball of radius $r$ at $x$

Then

1. $B(x,r)$ is both open and closed;
2. $\bar{B}(x,r)$ is both open and closed;
3. If $y\in B(x,r)$ (resp. $\bar{B}(x,r)$ ) then $B(x,r)=B(y,r)$ (resp. $\bar{B}(x,r)=\bar{B}(y,r)$ );
4. $B(x,r)$ and $B(y,r)$ (resp. $\bar{B}(x,r)$ and $\bar{B}(y,r)$ ) are either identical or disjoint;
5. If $B_1 = B(x,r)$ and $B_2 = B(y,s)$ are not disjoint, then either $B_1\subset B_2$ or $B_2\subset B_1$ ;
6. If $(x_n)$ is a sequence of elements of $K$ with $\lim_{n\to\infty} x_n=0$ , then $\sum_{n=1}^{\infty} x_n$ is Cauchy (and thus if $K$ is complete, a sufficient condition for convergence of a series is that the terms tend to zero)

Proof. We start by proving (3). Suppose $y\in B(x,r)$ . If $z\in B(x,r)$ , then since the absolute value is non-archimedean, we have $$ \Abs{z-y} =\Abs{(z-x)+(x-y)} \leq \max(\Abs{z-x},\Abs{x-y}) < $$ so that $z\in B(y,r)$ . Clearly $x\in B(y,r)$ , so reversing the roles of $x$ and $y$ , we see that $B(x,r) = B(y,r)$ . Finally, replacing $B$ by $\bar{B}$ and $<$ by $\leq$ , we get equality of closed balls as well.

(4) is now trivial: If $B(x,r)\cap B(y,r)\neq\emptyset$ , choose $z\in B(x,r)\cap B(y,r)$ ; then by (3), $B(x,r) = B(z,r) = B(y,r)$ . An identical argument proves the result for closed balls.

To prove (5), choose $z\in B_1\cap B_2$ . Assume first that $r\leq s$ ; then $B(z,r) = B_1$ , and $B(z,r)\subset B(z,s) = B_2$ , so that $B_1\subset B_2$ . If $s\leq r$ , then we have identically that $B_2\subset B_1$ . (Note that (4) is a special case when $r=s$ ).

(1) and (2) now follow: for (1), note that $B(x,r)$ is obviously open; its complement consists of a union of open balls of radius $r$ disjoint with $B(x,r)$ and its complement is therefore open. Thus $B(x,r)$ is closed. For (2), $\bar{B}(x,r)$ is obviously closed; to see that it is open, take any $y\in \bar{B}(x,r)$ ; then $\bar{B}(x,r)=\bar{B}(y,r)$ and thus $B(y,s)\subset \bar{B}(y,r)$ for $s<r$ is an open neighborhood of $y$ contained in $\bar{B}(x,r)$ , which is therefore open.

Finally, to prove (6), we must show that given $\epsilon$ , we can find $N>0$ sufficiently large such that $\Abs{\sum_{i=m}^n x_i}<\epsilon$ whenever $m,n>N$ . Simply choose $N$ such that $\Abs{x_i}<\epsilon$ for $i>N$ ; then $$ \Abs{\sum_{i=m}^n x_i}\leq \max(\Abs{x_m},\ldots,\Abs{x_n}) < \epsilo $$