ultrametric triangle inequality

Theorem 1.

Let $K$ be a field and $G$ an ordered group equipped with zero.  Suppose that the function  $|\cdot |:K\to G$  satisfies the postulates 1 and 2 of Krull valuation.  Then the non-archimedean or ultrametric triangle inequality

3.     $|x+y|\leqq \max \{|x|,|y|\}$

in the field is with the condition

(*) $|x|\leqq 1\Rightarrow |x+1|\leqq 1.$

Proof.  The value  $y=1$  in the ultrametric triangle inequality gives the (*) as result.  Secondly, let’s assume the condition (*).  Let $x$ and $y$ be non-zero elements of the field $K$ (if  $xy=0$  then 3 is at once verified), and let e.g.  $|x|\leqq |y|$.  Then we get  $|{\displaystyle \frac{x}{y}}|=|x|\cdot {|y|}^{-1}\leqq 1$,  and thus according to (*),

$$|x+y|\cdot {|y|}^{-1}=\left|\frac{x+y}{y}\right|=\left|\frac{x}{y}+1\right|\leqq 1.$$

So we see that  $|x+y|\leqq |y|=\max \{|x|,|y|\}$.

Theorem 2.

The Krull valuation (and any non-archimedean valuation (http://planetmath.org/Valuation))  $|\cdot |$  of the field $K$ satisfies the sharpening

$$|x+y|=\max \{|x|,|y|\}\mathit{\hspace{1em}}\mathrm{for}|x|\ne |y|$$

of the ultrametric triangle inequality.

Proof.  Let e.g.  $|x|>|y|$.  Surely  $|x+y|\leqq |x|$,  but also  $|x|=|(x+y)-y|\leqq \max \{|x+y|,|y|\}$;  this maximum is $|x+y|$ since otherwise one would have  $|x|\leqq |y|$.  Thus the result is:  $|x+y|=|x|$.

Note.  The metric defined by a non-archimedean valuation of the field $K$ is the ultrametric of $K$.  Theorem 2 implies, that every triangle of $K$ with vertices $A$, $B$, $C$ ($\in K$) is isosceles:  if  $|B-C|\ne |C-A|$,  then  $|A-B|=\max \{|B-C|,|C-A|\}$.

Theorem 3.

The valuation (http://planetmath.org/Valuation)  $|\cdot |:K\to ℝ$  of the field $K$ is archimedean if and only if the set

$$\{|1|,|1+1|,|1+1+1|,\mathrm{\dots }\}$$

of the “values” of the multiples of the unity is not bounded.

Proof.  If $|\cdot |$ is non-archimedean, then  $|n\cdot 1|=|1+\mathrm{\dots }+1|\leqq \max \{|1|\}=1$,  and the multiples are bounded.  Conversely, let  .  Now one obtains, when  $|x|\leqq 1$:

or     for all $n$.  As $n$ tends to infinity, this $n^{\mathrm{th}}$ root has the limit 1.  Therefore one gets the limit inequality  $|x+1|\leqq 1$,  i.e. the valuation is non-archimedean.

Title	ultrametric triangle inequality
Canonical name	UltrametricTriangleInequality
Date of creation	2013-03-22 14:54:15
Last modified on	2013-03-22 14:54:15
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	25
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13F30
Classification	msc 13A18
Classification	msc 12J20
Classification	msc 11R99
Related topic	MaximalNumber
Related topic	PAdicCanonicalForm
Related topic	UltrametricSpace
Related topic	MinimalAndMaximalNumber
Related topic	ExponentValuation2
Defines	non-archimedean triangle inequality