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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!

Igor Deruga
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2 Answers2

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Let $x,y \in \mathbb{R}$, and suppose $$\exists \alpha\in\mathbb{R}: \alpha x + (1-\alpha) y = 0.$$ Then $$ \begin{align*} & \Rightarrow \alpha x + (1-\alpha) y = 0 \\ & \Rightarrow \alpha(x-y)+y=0 \\ &\Rightarrow \alpha=\frac{y}{y-x} \\ &\Rightarrow y \neq x. \end{align*} $$ Hence if $x=y$, no such $\alpha$ exists, and we must consider $d(x,y)=||x||+||y||$.

The bidirectional implication still poses a problem for the latter definition, as $$d(x,y)=||x||+||y||=0 \iff x=y=0,$$ and would not work for $x=y \neq 0$.

J. W. Perry
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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.

not all wrong
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