properties of nonarchimedean valuations
If $K$ is a field, and $\cdot $ a nontrivial nonarchimedean 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 nonarchimedean 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 \leftxy\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.
Proof. We start by proving (3). Suppose $y\in B(x,r)$. If $z\in B(x,r)$, then since the absolute value is nonarchimedean, 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 $$ is an open neighborhood of $y$ contained in $\overline{B}(x,r)$, which is therefore open.
Finally, to prove (6), we must show that given $\u03f5$, we can find $N>0$ sufficiently large such that $$ whenever $m,n>N$. Simply choose $N$ such that $$ for $i>N$; then
$$ 
Title  properties of nonarchimedean valuations 

Canonical name  PropertiesOfNonarchimedeanValuations 
Date of creation  20130322 18:01:11 
Last modified on  20130322 18:01:11 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  6 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 13F30 
Classification  msc 13A18 
Classification  msc 12J20 
Classification  msc 11R99 
Related topic  CompleteUltrametricField 