Define the open balls of this metric The post office metric space, $P$ has the distance function defined as follows:
$$
d_P (\mathbf{x},\mathbf{y}) := \begin{cases}
0 & \mathbf{x} = \mathbf{y}\\
\Vert \mathbf{x-y}\Vert_2 & \mathbf{x}, \mathbf{y}, \mathbf{0} \mathbf{ \ are \ aligned } \\
\Vert \mathbf{x}\Vert_2+\Vert \mathbf{y}\Vert_2 & \mathbf{x}\neq \mathbf{y}
\end{cases}
$$
Where $\Vert \mathbf{x}\Vert_2 = \sqrt{x_1^2+x_2^2}$ is the Euclidean distance from $\mathbf{x}=(x_1,x_2) \in \mathbb{R}^2$ to the origin.
I am interested in drawing the balls of this metric centered at point $\mathbf{p}$, having a radius $r$:
$$B_r(\mathbf{p}) \triangleq  \{\mathbf{x} \in P\vert\ d(\mathbf{x},\mathbf{p})<r  \}$$
Does anyone have any tips on how to do this?
 A: The definition for the $\textit{post office metric}$ I know is slightly different than yours, namely:
$$d_P(\textbf{x}, \textbf{y}) = 
\begin{cases}  
0 & \textbf{x} = \textbf{y} \\
\|\textbf{x}\|_2 + \|\textbf{y}\|_2 & \textbf{x} \neq \textbf{y}
\end{cases}
$$
The way I've been told to think about this metric is, that a letter has to pass through the post office (the origin) in all cases, even if the house i want to send my letter to lies in a straight line with my own house. (In fact, when I learned about this metric it was called the 'French Railway Metric', since every train has to go through Paris, but that's the same concept...).
But nonetheless we can figure out how open balls in your metric look like! First the easy case, if $\textbf{p} = 0$ then the open ball around $\textbf{p}$ with radius $r$ is the same as the open ball in the euclidean metric, because any point is aligned with just 0.
Let's now look at $\textbf{p} \neq 0$. We have two cases, either $\|\textbf{p}\| \leq r$, or $\|\textbf{p}\| > r$. In the first case, we never reach the origin, so the open ball is just the straight line from $p$ in direction of the origin of length $r$, and also in the opposite direction (away from the origin, also of length $r$) not including the endpoints. So geometrically this open ball is just a straight line. If $\|\textbf{p}\|<r$, the open ball is the union of the line segment from $p$ to the origin, and the open disc centered at the origin of radius $r-\|\textbf{p}\|$ (with respect to the usual euclidean metric).
