Q:

Subsets of the boundary of a surface group

Consider the surface group $\Gamma=\langle a,b,c,d\mid [a,b][c,d]=1\rangle$: it is a Gromov hyperbolic group; its Gromov boundary $\partial\Gamma$ is homeomorphic to $S^1$ (the unit circle). 
I would like to define a family of subsets of $\partial\Gamma$ as follows: fix $x,y\in\Gamma$. Then
$$
U(x,y):=\left\{\xi\in\partial\Gamma\mid\text{ there exists a geodesic }g\text{ in }\Gamma\text{ starting from }x,\text{ passing through }y\text{ and s.t. } g(\infty)=\xi\right\}. 
$$

Does this definition make sense?  Is $U(x,y)$ trivial? 
Is $U(x,y)$ open? 
Is $U(x,y)$ connected? 

A:

You definition is closely related to the notion of the "cone type" introduced by Jim Cannon.  As Yves noted, $U(x,y)$ is compact.  It is also connected.  

Compactness part is immediate from the Arzela-Ascoli theorem: Take a sequence of rays $r_i: [0,\infty)\to X$ (where $X$ is the Cayley graph; here it does not matter for what hyperbolic group and what generating set) such that $r_i(0)=x$, $r_i(n)=y$ for all $i$ and a fixed $n$. Then the sequence $r_i$ subconverges to a geodesic ray $r$ from $x$ passing through $y$, such that $r(\infty)= \lim_i r_i(\infty)$. Hence, $U(x,y)$ is compact.   
To prove connectedness you need to use the fact that the Cayley graph $X$ in your case is planar, more precisely, is a subset of the hyperbolic plane $H^2$, so that the ideal boundary of $X$ is the boundary circle $S^1$ of $H^2$. In what follows, all the metric notions are with respect to the Cayley graph $X$. I will assume that $x\ne y$. Let $\xi_1, \xi_2$ be distinct points in $U(x,y)$. Consider geodesic rays $r_1, r_2$ from $x$ passing through $y$ and asymptotic to $\xi_1, \xi_2$. There exists the smallest $T\ge n=d(x,y)$ such that for all $t\ge T$, $r_1(t)\ne r_2(t)$. Thus, the union $A=r_1([T,\infty))\cup  r_2([T,\infty))$  is a topological line in $X\subset H^2$. This line splits $H^2$ in two components, one of then, call it $C$, does not contain $x$. The ideal boundary $\alpha$ of $C$ equals one of the two arcs of $S^1$ with the end-points $\xi_1, \xi_2$.  I claim that all points $\eta\in \alpha$ belong to $U(x,y)$. Indeed, consider a geodesic ray $r$ from $x$ asymptotic to $\eta$. This ray has to cross $A$ at some point $z=r(s)$. Say, $z\in r_1([0,\infty))$. Now, replace the portion $r[0,s])$ with $r_1([0,s])$. I will leave you to verify that the new ray $r_3$ is a geodesic ray from $x$ which passes through $y$. Hence, $\eta\in U(x,y)$. qed 

