Editing Ultrametric space (section)

== Formal definition ==

An '''ultrametric''' on a [[Set (mathematics)|set]] {{mvar|M}} is a [[Real number|real]]-valued function
:<math>d\colon M \times M \rightarrow \mathbb{R}</math>
(where {{math|ℝ}} denote the [[real number]]s), such that for all {{math|''x'', ''y'', ''z'' &isin; ''M''}}:
# {{math|''d''(''x'', ''y'') &ge; 0}};
# {{math|1=''d''(''x'', ''y'') = ''d''(''y'', ''x'')}} (''symmetry'');
# {{math|1=''d''(''x'', ''x'') = 0}};
# if {{math|1=''d''(''x'', ''y'') = 0}} then {{math|1=''x'' = ''y''}};
# {{math|1=''d''(''x'', ''z'') &le; max {''d''(''x'', ''y''), ''d''(''y'', ''z'') }}} ('''strong triangle inequality''' or '''ultrametric inequality''').

An '''ultrametric space''' is a pair {{math|(''M'', ''d'')}} consisting of a set {{mvar|M}} together with an ultrametric {{mvar|d}} on {{mvar|M}}, which is called the space's associated distance function (also called a [[metric (mathematics)|metric]]). 

If {{mvar|d}} satisfies all of the conditions except possibly condition 4, then {{mvar|d}} is called an '''ultrapseudometric''' on {{mvar|M}}. An '''ultrapseudometric space''' is a pair {{math|(''M'', ''d'')}} consisting of a set {{mvar|M}} and an ultrapseudometric {{mvar|d}} on {{mvar|M}}.{{sfn | Narici|Beckenstein | 2011 | pp=1-18}} 

In the case when {{mvar|M}} is an Abelian group (written additively) and {{mvar|d}} is generated by a [[length function]] <math>\|\cdot\|</math>  (so that <math>d(x,y) = \|x - y\|</math>), the last property can be made stronger using the [[Wolfgang Krull|Krull]] sharpening to:
: <math>\|x+y\|\le \max \left\{ \|x\|, \|y\| \right\}</math> with equality if <math>\|x\| \ne \|y\|</math>.

We want to prove that if <math>\|x+y\| \le \max \left\{ \|x\|, \|y\|\right\}</math>, then the equality occurs if <math>\|x\| \ne \|y\|</math>. [[Without loss of generality]], let us assume that <math>\|x\| > \|y\|.</math> This implies that <math>\|x + y\| \le \|x\|</math>. But we can also compute <math>\|x\|=\|(x+y)-y\| \le \max \left\{ \|x+y\|, \|y\|\right\}</math>. Now, the value of <math>\max \left\{ \|x+y\|, \|y\|\right\}</math> cannot be <math>\|y\|</math>, for if that is the case, we have <math>\|x\| \le \|y\|</math> contrary to the initial assumption. Thus, <math>\max \left\{ \|x+y\|, \|y\|\right\}=\|x+y\|</math>, and <math>\|x\| \le \|x+y\|</math>. Using the initial inequality, we have <math>\|x\| \le \|x + y\| \le \|x\|</math> and therefore <math>\|x+y\| = \|x\|</math>.