Normed vector space/Metric space/Fact/Proof/Exercise/Solution

1. We have $d(u,v)=\Vert u-v\Vert =0$ if and only if $u-v=0$, that is, $u=v$.
2. We have

   $\begin{array}{cc}d(u,v)& =\Vert u-v\Vert \\ & =\Vert -1(v-u)\Vert \\ & =|-1|\cdot \Vert v-u\Vert \\ & =d(v,u).\end{array}$

3. For arbitrary $w\in V$, we have, due to the definition of a norm

   $\begin{array}{cc}d(u,v)& =\Vert u-v\Vert \\ & \le \Vert u-w\Vert +\Vert w-v\Vert \\ & =d(u,w)+d(w,v).\end{array}$

To the exercise