Let the identity map map $(\mathbb{R}^{2}, d_f) \rightarrow (\mathbb{R}^{2},d^{2})$ be the French railway metric. Suppose the identity map $id:(\mathbb{R}^{2}, d_f) \rightarrow (\mathbb{R}^{2},d^{2})$ where $d_f$ is the French railway metric and $d^2$ is the Euclidean metric. Show that the inverse map is not continuous. 
French railway metric is defined: $$d_f(x,y) = 
   \left\{
     \begin{array}{lr}
       |x-y|& : \text{if}\ x,y\ \text{are on the same ray from the origin} \\
       |x| + |y| & : \text{otherwise} 
     \end{array}
   \right.
 $$
Where $|x| = \sqrt{x^2_1 + x^2_2}.$
Here is my attempt: 
I know that  $$d^{2}(f^{-1}(x),f^{-1}(z))=d^{2}(x,z)$$ since $f$ is the identity map. Let $d^{2}(x,y)\le \frac{\delta}{2}$ and $d^{2}(y,z) \le \frac{\delta}{2}.$
Then $$d^{2}(x,z)\le d^{2}(x,y) + d^{2}(y,z) \le \frac{\delta}{2} + \frac{\delta}{2} \le \delta$$ It is clear that $d_f(x,y) = d^{2}(x,y)$ when x,y are on the same ray from the origin. So that  $$ d^{2}(x,z) =d_f(x,z)< \epsilon. $$ But, if $x,y$ are not on the same ray from the origin $$d_f(x,z) > d^{2}(x,y)$$ so that $$d_f(x,y) > \epsilon$$ whenever $$d^2(x,y)< \delta.$$ 
 A: I’m afraid that your reasoning isn’t at all clear. I suggest taking either of the following approaches. I’ve spoiler-protected ways to carry out both approaches, but do try at least one of them yourself first.


*

*Find a set that is open in the $d_f$ metric but not in the $d^2$ metric.



 Consider a ray through the origin.



*

*Find a sequence that converges in the $d^2$ metric but not in the $d_f$ metric.



 Consider the sequence $\big\langle\langle 1,2^{-n}\rangle:n\in\Bbb N\big\rangle$.

A: Take the inverse id: $(\mathbb{R}^2,d^2) \rightarrow (\mathbb{R}^2,d_f)$. There exists an $x \in \mathbb{R}^2$ and $\epsilon > 0$ such that for all $\delta >0$ there exists $y \in \mathbb{R}^2$ with $d^2(x,y) < \delta$, but $d_f(x,y) > \epsilon$. Choose $x=(1,0)$ and $\epsilon =1$. For any $\delta$, let $y=(1,\delta)$.
Then $$d^2(x,y)= \sqrt{(x-y)^2}=\sqrt{(1-1)^2 + (0-\delta)^2}= \delta$$
We chose $x,y$ so that they do not like on the same ray through the origin so that $$d_f(x,y)= d^2(x,0)+d^2(0,y)= $$ $$d^2(x,0)=\sqrt{(x-0)^2+(0-0)^2}=1$$
$$d^2(0,y)=\sqrt{(0-1)^2+(0-\delta)^2}=\sqrt{1^2+\delta^2}=1$$ for sufficiently small $\delta$. This gives $$d_f(x,y)= 1+1 > \epsilon$$  $\square$ 
