Definition of a Tile I was given the following definition while trying to prove that 'if a group $G$ acts freely on a tree, then $G$ is a free group':
Choose an arbitrary vertex $v$
of a tree $T$, and consider the orbit of $v$ under the action of $G$. Since $G$ acts freely on $T$, it follows that
the points of the orbit are in bijection with the elements $G$. For each element $g$ of
$G$, let $T_g$ denote the set of points $x$ so that the distance $d(x,g\cdot v)$ is minimal over
all choices of $g$. In other words, the points of $T_g$ are the points of $T$ that are closer
to $g\cdot v$ than to any other point in the orbit of $v$.
What I do not understand is why such sets would not be singleton. 
 A: Perhaps the easiest way to explain this would be with an example. Let $T=(V,E)$ be the tree with vertex set the integers $V=\mathbb{Z}$ and edge set $E = \{(n,n+1)\mid n\in\mathbb{Z}\}$ - so $T$ is basically the real numbers as far as metric properties are concerned. Let $G = \langle \sigma\rangle$ where $\sigma$ is translation one to the right.
Then for any vertex, the orbit under the action of $G$ includes every vertex. Let $v=0$ and note that $\sigma^n(0) = n$. The tile $T_{\sigma^0} = \{x\in T \mid \forall n,\: d(x,0) \leq d(x,n)\}$. That is, all points which are closer to zero than any other vertex in the graph. Well that's just the interval $T_{\sigma^0} = [-\frac{1}{2},\frac{1}{2}]$. It's pretty easy to see that $T_{\sigma^n} = [n-\frac{1}{2},n+\frac{1}{2}]$ for every $n$.
If we replaced $G$ with the subgroup $H = \langle \sigma^2 \rangle$, then the tiles would get bigger (and be of the form $[n-1,n+1]$), because now the orbit of any vertex only hits half the points so there is more room between nearby vertices and so more points are closer to any one particular vertex in the orbit. In particular the tiles will also change depending on if our initial vertex is an even or odd integer.
