Is the Post Office Metric applicable in $\Bbb{R}^n$ for all $n$? I was required to provide a metric space $(X,d)$ with $x,y\in X$ and $0<r<R$ such that $B_R(x)\subsetneq B_r(y)$. After a lot of thinking and reading, I came by a metric function called the "Post Office Metric", always attributed to $\Bbb{R}^2$, in particular when giving examples for a metric space such as the above. I constructed a metric function (defined on $\Bbb{R}$) of the same concept as follows: $d(x,y)=
\begin{cases}
|x|+|y|,  & \text{if } x\ne y \\
0, & \text{otherwise}
\end{cases}$, 
and to what I checked it seems like a legitimate metric function. Then I looked for it and came by nothing like that, and I begin to feel like I am doing something wrong. I would appreciate your thought on the matter.
 A: I think the easiest route would be to take $X$ to be a finite graph and let $d(x,y)$ be length of the shortest path between vertices $x,y$. So for instance, you might take $X$ to be the graph

a----b----c----d----e

Note that $B_r(c)=X$ for any $r>2$. On the other hand, $B_R(e) \neq X$ when $R<4$. 
Added: In fact, come to think of it, you can also achieve this with the interval $X=[0,1]$ in the ordinary absolute value metric. 
$$B_{0.8}(0) \subsetneq B_{0.6}(0.5)$$
A: The idea is fundamentally similar to Mike F's answer, but perhaps simpler or at least more common in analysis.
Consider $X=[0,1]$ with the induced Euclidian metric. Let $x=0$; then $B_R(x)=[0,x)$. Take say, $R=1/2$.
Then take, say, $y= 1/3$ and $r=2/5$, we have that $B_r(y)=[0,11/15)\simeq [0,0.73)$.
A: UPDATE: Have fun with the standard interval $[0, 1]$! I think that's what the other answer did. (saw the picture of the graph answer and that idea popped up at the same time the third answer came)
What about something like $d(x, y)=||x||/2+||y||$? Should be pretty easy to come up with the values of $x, y$ and $0<r<R$...assuming you meant for all your parameters to be existential.
