3

I am trying to work just one case of showing that the French railways metric defined by a metric space $(\mathbb{R}^2,d)$ is actually a metric:

$$d(x,y) = \begin{cases} \|x-y\|, & \text{if $x,y,0$ are collinear;} \\ \|x\| +\|y\|, & \text{otherwise} \end{cases}$$

I am trying to show the case that if $x=y$, then $d(x,y)=0$. I have been able to show that if $x=y$, then $\|x-y\|=0$, but am having a hard time showing that if $x=y$, then $\|x\| +\|y\|=0$. Thank you!

Did
  • 279,727
user145416
  • 383

1 Answers1

5

If $x=y$ then we have that $x,y,0$ are collinear since $x,x,0$ i.e. $x,0$ are always collinear.

So in case $x=y$ we have $d(x,y)=\|x-y\|=\|x-x\|=\|0\|=0$

(There is no need to consider $d(x,y)=\|x\|+\|y\|$)

Michael Hardy
  • 1
usermath
  • 3,516