TITLE: French metro metric: difficulty to prove that $d(x, y) = 0\iff x = y$.
QUESTION [5 upvotes]: I think that it is related to the special definition of the metric in my book:
$$d(x, y) = \begin{cases}||x - y||,\mbox{ if }\exists \alpha\in\mathbb{R}: \alpha x + (1-\alpha) y = 0;\\ ||x|| + ||y||, \mbox{ otherwise.}\end{cases}$$
This way, for $x = y$, we have $\alpha x + (1 - \alpha) x = 0$, which is true only if $x = 0$, so we fall into the second case: $d(x, x) = ||x|| + ||x|| = ||x||^2 \neq 0$ if $x\neq 0$.
Seems like it doesn't satisfy the axioms of a metric. The case described in Woflram MathWorld is simpler, because the condition for the first case is: $x = \alpha y$. This way, for $d(x, x)$, we have $\alpha = 1$, and everything works fine!
Am I missing something or is there an error in the problem statement?
Thanks in advance!

REPLY [2 votes]: I agree. In some sense the actual condition they wanted is "$x,y$ lie on a common line through the origin", but the given condition breaks down as the points approach each other. If you divide the condition equation by $\alpha$ and then take $\alpha\to\infty$ it's clear that this would hold. Hence you should use Wolfram's definition which is almost the same.