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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Ultrametric space ： ウィキペディア英語版 Ultrametric space In mathematics, an ultrametric space is a special kind of metric space in which the triangle inequality is replaced with $d(x,z)\backslash leq\backslash max\backslash left\backslash$. Sometimes the associated metric is also called a non-Archimedean metric or super-metric. Although some of the theorems for ultrametric spaces may seem strange at a first glance, they appear naturally in many applications. == Formal definition == Formally, an ultrametric space is a set of points $M$ with an associated distance function (also called a metric) :$d\backslash colon\; M\; \backslash times\; M\; \backslash rightarrow\; \backslash mathbb$ (where $\backslash mathbb$ is the set of real numbers), such that for all $x,y,z\backslash in\; M$, one has: # $d(x,\; y)\; \backslash ge\; 0$ # $d(x,\; y)\; =\; 0$ iff $x=y$ # $d(x,\; y)\; =\; d(y,\; x)$ (symmetry) # $d(x,\; z)\; \backslash le\; \backslash max\; \backslash left\backslash$ (strong triangle or ultrametric inequality). In the case when $M$ is a group and $d$ is generated by a length function $\backslash |\backslash cdot\backslash |$ (so that $d(x,y)\; =\; \backslash |x\; -\; y\backslash |$), the last property can be made stronger using the Krull sharpening〔Planet Math: (Ultrametric Triangle Inequality )〕 to: : $\backslash |x+y\backslash |\backslash le\; \backslash max\; \backslash left\backslash$ with equality if $\backslash |x\backslash |\; \backslash ne\; \backslash |y\backslash |$. We want to prove that if $\backslash |x+y\backslash |\; \backslash le\; \backslash max\; \backslash left\backslash$, then the equality occurs if $\backslash |x\backslash |\; \backslash ne\; \backslash |y\backslash |$. Without loss of generality, let us assume that $\backslash |x\backslash |\; >\; \backslash |y\backslash |$. This implies that $\backslash |x\; +\; y\backslash |\; \backslash le\; \backslash |x\backslash |$. But we can also compute $\backslash |x\backslash |=\backslash |(x+y)-y\backslash |\; \backslash le\; \backslash max\; \backslash left\backslash$. Now, the value of $\backslash max\; \backslash left\backslash$ cannot be $\backslash |y\backslash |$, for if that is the case, we have $\backslash |x\backslash |\; \backslash le\; \backslash |y\backslash |$ contrary to the initial assumption. Thus, $\backslash max\; \backslash left\backslash =\backslash |x+y\backslash |$, and $\backslash |x\backslash |\; \backslash le\; \backslash |x+y\backslash |$. Using the initial inequality, we have $\backslash |x\backslash |\; \backslash le\; \backslash |x\; +\; y\backslash |\; \backslash le\; \backslash |x\backslash |$ and therefore $\backslash |x+y\backslash |\; =\; \backslash |x\backslash |$. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Ultrametric space」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.