properties of non-archimedean valuations

If $K$ is a field, and $|\cdot |$ a nontrivial non-archimedean valuation (or absolute value) on $K$, then $|\cdot |$ 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 $\mathrm{|}\mathrm{\cdot }\mathrm{|}$. For $r\mathrm{>}\mathrm{0}$ a real number, $x\mathrm{\in }K$, define

$$\overline{B}(x,r)=\{y\in K\mid \left|x-y\right|\le r\},\mathit{\text{the closed ball of radius }}r\mathit{\text{ at }}x$$

Then

1. $B(x,r)$ is both open and closed;

2. $\overline{B}(x,r)$ is both open and closed;

3. If $y\in B(x,r)$ (resp. $\overline{B}(x,r)$) then $B(x,r)=B(y,r)$ (resp. $\overline{B}(x,r)=\overline{B}(y,r)$);

4. $B(x,r)$ and $B(y,r)$ (resp. $\overline{B}(x,r)$ and $\overline{B}(y,r)$) 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 ${lim}_{n\to \mathrm{\infty }}{x}_n=0$, then ${\sum }_{n=1}^{\mathrm{\infty }}{x}_n$ 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

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 $\overline{B}$ and by $\le$, we get equality of closed balls as well.

(4) is now trivial: If $B(x,r)\cap B(y,r)\ne \mathrm{\varnothing }$, 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 $r\le s$; then $B(z,r)=B_1$, and $B(z,r)\subset B(z,s)=B_2$, so that $B_1\subset B_2$. If $s\le r$, 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), $\overline{B}(x,r)$ is obviously closed; to see that it is open, take any $y\in \overline{B}(x,r)$; then $\overline{B}(x,r)=\overline{B}(y,r)$ and thus $B(y,s)\subset \overline{B}(y,r)$ for is an open neighborhood of $y$ contained in $\overline{B}(x,r)$, which is therefore open.

Finally, to prove (6), we must show that given $ϵ$, we can find $N>0$ sufficiently large such that whenever $m,n>N$. Simply choose $N$ such that for $i>N$; then