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The following theorem is given in Metric Spaces by O'Searcoid

Theorem: Suppose $V$ is a normed linear space. Then the function $d$ defined on $V \times V$ by $(a,b) \to ||a-b||$ is a metric on $V$

Three conditions of a metric are fairly straight-forward.

By the definitions of a norm, I know that $||x|| \ge 0$ and only holds with equality if $x=0$. Thus $||a-b||$ is non-negative and zero if and only if $a=b$.

The triangle inequality of a normed linear space requires: $||x+y|| \le ||x|| + ||y||$. Let $x = a - b$ and $y = b - c$. Then $||a - c|| \le || a - b || + || b - c||$ satisfying the triangle inequality for a metric space.

What I am having trouble figuring out is symmetry. The definition of a linear space does not impose any condition of a symmetry. I know from the definition of a linear space that given two members of $V$, $u$ and $v$ they must be commutative, however, I do not see how that could extend here.

Thus what I would like to request help with is demonstrating $||a - b|| = ||b - a||$.

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  • $\begingroup$ Well, which clause in the definition of a norm have you not used? $\endgroup$ – Chris Eagle Jun 29 '13 at 0:25
  • $\begingroup$ @PeterTamaroff I knew it had to be something really simple. Thanks! $\endgroup$ – GovEcon Jun 29 '13 at 0:26
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$$\|a-b\|=\|(-1)(b-a)\|=|-1|\cdot\|b-a\|=\|b-a\|$$

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Use $\lVert c\cdot x\rVert =|c|\lVert x\rVert$.

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