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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
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user145416
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2 Answers2

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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
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usermath
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While it seems your question has already been answered by @usermath and @Daniel Fischer, your question is linked to a couple closed questions (this and that) that ask specifically about proving the triangle inequality of this metric. Similar questions also appear here and here, so I wanted to give a proof that the above is a metric to serve as a reference for this and the other questions.


The hardest property to show is the triangle inequality, so that's where we'll start. To cut down our casework, we can observe $||a-b|| \le ||a|| + ||b||$. We consider two cases:

  1. Suppose x is colinear to y through the origin. Then, $$ dx,y) = ||x-y|| = ||x-z+z-y|| \le ||x-z|| + ||z-y|| \le d(x,z) + d(z,y)$$

  2. Suppose x is not colinear to y through the origin. Notice that we cannot have both (i) x being colinear to z and (ii) y being colinear to z through the origin (otherwise, x would be colinear to y through the origin). Then, it suffices to show the triangle inequality holds for at most one point being colinear to z and the origin. We assume x is colinear to z through the origin without loss of generality: $$d(x,y) = ||x|| + ||y|| = ||x|| - ||z|| + ||y|| + ||z|| $$ Using the reverse triangle inequality, $$ d(x,y) \le ||x - z|| + ||z|| + ||y|| \le d(x,z) + d(z,y)$$


Now, we proceed to positivity. By definition, it is easy to see $d(x,y) \ge 0$ so we focus on the equality condition:

$$ \text{If } x = y \text{, then they are colinear through the origin, so} \\ d(x,y) = ||x - x|| = 0 $$

$$ \text{If } x \ne y \text{, then by properties of the norm, } \\ || x - y || > 0 \text{, and} \\ ||x|| + ||y|| > 0 \text{, given it is not possible for both } x = 0, y = 0 $$


Finally, symmetry $d(x,y) = d(y,x)$ follows almost immediately from the definition and properties of norms.

Duke Earcham
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