Editing Triangle inequality (section)

==Metric space==
In a [[metric space]] {{mvar|M}} with metric {{mvar|d}}, the triangle inequality is a requirement upon [[Metric (mathematics)#Definition|distance]]:
:<math>d(A,\ C) \le d(A,\ B) + d(B,\ C) \ , </math>

for all points {{mvar|A}}, {{mvar|B}}, and {{mvar|C}} in {{mvar|M}}. That is, the distance from {{mvar|A}} to {{mvar|C}} is at most as large as the sum of the distance from {{mvar|A}} to {{mvar|B}} and the distance from {{mvar|B}} to {{mvar|C}}.

The triangle inequality is responsible for most of the interesting structure on a metric space, namely, convergence.  This is because the remaining requirements for a metric are rather simplistic in comparison.  For example, the fact that any [[limit of a sequence|convergent sequence]] in a metric space is a [[Cauchy sequence]] is a direct consequence of the triangle inequality, because if we choose any {{math|''x<sub>n</sub>''}} and {{math|''x<sub>m</sub>''}} such that {{math|''d''(''x<sub>n</sub>'', ''x'') < ''ε''/2}} and {{math|''d''(''x<sub>m</sub>'', ''x'') < ''ε''/2}}, where {{math|''ε'' > 0}} is given and arbitrary (as in the definition of a limit in a metric space), then by the triangle inequality, {{math|''d''(''x<sub>n</sub>'', ''x<sub>m</sub>'') ≤ ''d''(''x<sub>n</sub>'', ''x'') + ''d''(''x<sub>m</sub>'', ''x'') < ''ε''/2 + ''ε''/2 {{=}} ''ε''}}, so that the sequence {{math|{{mset|''x<sub>n</sub>''}}}} is a Cauchy sequence, by definition.

This version of the triangle inequality reduces to the one stated above in case of normed vector spaces where a metric is induced via {{math|''d''(''u'', ''v'') ≔ ‖''u'' − ''v''‖}}, with {{math|''u'' − ''v''}} being the vector pointing from point {{mvar|v}} to {{mvar|u}}.