properties of non-archimedean valuations

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).

Theorem 1.

Let $K$ be a field with a non-archimedean absolute value $\left\lvert\cdot\right\rvert$ . For $r>0$ a real number, $x\in K$ , define

	$\displaystyle B(x,r)=\{y\in K\ \mid\ \left\lvert x-y\right\rvert<r\},\ \text{% the open ball of radius $r$ at $x$}$
	$\displaystyle\bar{B}(x,r)=\{y\in K\ \mid\ \left\lvert x-y\right\rvert\leq r\},% \ \text{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

\left\lvert z-y\right\rvert=\left\lvert(z-x)+(x-y)\right\rvert\leq\max(\left% \lvert z-x\right\rvert,\left\lvert x-y\right\rvert)<r

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

\left\lvert\sum_{i=m}^{n}x_{i}\right\rvert\leq\max(\left\lvert x_{m}\right% \rvert,\ldots,\left\lvert x_{n}\right\rvert)<\epsilon

Title	properties of non-archimedean valuations
Canonical name	PropertiesOfNonarchimedeanValuations
Date of creation	2013-03-22 18:01:11
Last modified on	2013-03-22 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