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

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

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 $\left\lvert\sum_{{i=m}}^{n}x_{i}\right\rvert<\epsilon$ whenever $m,n>N$. Simply choose $N$ such that $\left\lvert x_{i}\right\rvert<\epsilon$ for $i>N$; then