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I'm having trouble proving this. I am able to prove other metrics, I think it is possibly the format of the railway metric that is confusing me...

Consider the function $d : \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow [0, \infty)$ given by

$d(x,y)=\begin{cases} d_2(x,y)~\text{if $x,y,0$ are collinear}\\d_2(x,0) + d_2(0,y) ~\text{otherwise}\end{cases}$

I have also tried to attach a photo of the question)

Cornman
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  • Should your function be $d$, not $\rho$? What is $d_2$? – Sambo Oct 08 '19 at 21:26
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    Try going through the definition of a metric, step by step. – Chris Culter Oct 08 '19 at 21:27
  • Yes it should sorry, and I'm guessing d2 is just like having a d and d'. –  Oct 08 '19 at 21:41
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    @mathsstudent101 I assume that $d_2$ is the ordinary Euclidean distance. – Chris Culter Oct 08 '19 at 22:24
  • yes I would assume so –  Oct 08 '19 at 22:28
  • There is something amiss in this and the linked question. For instance this question reads "Show that the railway metric is a metric". To prove this, one needs to prove 3 or 4 conditions (depending on the text one is reading). The answer given only proves partially the triangle inequality; the linked question does not(!) solve this question, since in there it is only asked to show the condition $x=y$ implies $\rho(x,y)=0$. Both questions are not duplicates, one requests the full proof and the other requests a partial proof. – user2820579 Oct 07 '20 at 16:28

1 Answers1

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Let d be the Euclidean metric for the real plane.
Let O be the point (0,0).
The railroad metric p for the real plane is
p(x,y) = d(x,y) when x,y,0 are collinear,
. . . . = d(x,0) + d(0,y) otherwise.

This is also called the Frence railroad metric.
All trains go through Pairs.
To get from a to b, take the train from a to Paris and then the train from Paris to b unless b is on the way from a to Paris or visa versa.

Clearly p(x,y) = p(y,x) and p(x,y) = 0 iff x = y.
For the triangle inequality, consider the case when x,y,0, y,z,0 and x,z,O are not collinear.
Then p(x,z) = p(x,0) + p(0,z)
<= p(x,0) + p(0,y) + p(y,0) + p(0,z)
= p(x,y) + p(y,z).
The other cases are left for the diligent reader.

William Elliot
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