For every two elements $x,y$ of the disk $D^2$ let
$d(x,y)=\begin{cases}\|x-y\|,\text{if x and y are on the same line through (0,0)}\\ \|x\|+\|y\|, \text{else}\end{cases}$
(the 'french-railroad-metric)
What do the neighborhoods of $(0,0)$ and $(\frac12, 0)$ look like?
I want to figure out, what these neighborhoods look like. For $x=(0,0)$ it is easy, because every $y\in D^2$ can be on a line through (0,0).
So $d(x,y)=\|x-y\|=\sqrt{y_1^2+y_2^2}<\varepsilon$, which means that the neighborhoods of $(0,0)$ are just circles with centre (0,0) in $D^2$.
What do the neighborhoods of $x=(\frac12,0)$ look like?
For every point on the x-axis we get $d(x,y)=\sqrt{(y_1-\frac12)^2+y_2^2}<\varepsilon$. Which are again circles with centre $(\frac12, 0)$ on the disk.
If $y$ is not on the x-axis, we get:
$d(x,y)=\|x\|+\|y\|=\frac12+\|y\|<\varepsilon$
$\|y\|<\varepsilon-\frac12$.
Which are circles with centre (0,0) and a maximal radius of $1/2$?
So for $\varepsilon=1$ for example, we would get this picture? (The red colored area)
Thanks in advance, for your thoughts.
No, you wouldn't get that red-shaded area. The ball around $(\frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $\frac14$ centered at $(\frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.