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For the following, we need the extra structure of a topology on $X\cup X(\infty)$ (see [Ballmann '95]).

**Lemma 1.** In this topology, $(x_n)_n\subset X$ converges to $\xi\in X(\infty)$ if and only if $d(x,x_n)\to\infty$ for some (and hence any) $x\in X$ and the geodesic segments $\sigma_{x,x_n}$ converge to $\sigma_{x,\xi}$.

Given $x\in X$, denote by $\Lambda(\Gamma,x)$ the set of accumulation points of the orbit $\Gamma x=\{\gamma x\colon\gamma\in\Gamma\}$ in $X\cup X(\infty)$.

**Lemma 2.** 	$\Lambda(\Gamma,x)=\overline{\Gamma x}\cap X(\infty)$.

**Proof.** 	Since the action of $\Gamma$ is properly discontinuous, for every $K\subset X$ compact, the set $\{\gamma\in\Gamma\colon \gamma(K)\cap K\ne\emptyset\}$ is finite. This implies the claim.

**Lemma 3.** 	$\Lambda(\Gamma,x)$ is independent of $x\in X$.

**Proof.** Let $\xi\in \Lambda(\Gamma,x)$. Let us show that $\xi\in\Lambda(\Gamma,y)$ for any $y\in X$. As $\xi\in\Lambda(\Gamma,x)$, there exists $(\gamma_n)_n\subset\Gamma$ such that $\gamma_nx\to\xi$ as $n\to\infty$. By the above, $d(x,\gamma_nx)\to\infty$ and $\sigma_{x,\gamma_nx}\to\sigma_{x,\xi}$. Recall that $\Gamma$ is a group of isometries acting on $X$. Hence, given $y\in X$, for every $n\in\bN$ it holds $d(\gamma_ny,\gamma_nx)=d(x,y)$.
Hence, 
\[
	d(x,\gamma_nx)
	\le d(x,y)+d(y,\gamma_ny)+d(\gamma_ny,\gamma_nx)
	= 2d(x,y)+d(y,\gamma_ny)
	\to\infty.
\]	
It follows that $\sigma_{x,\gamma_nx}$ and $\sigma_{y,\gamma_ny}$ are asymptotic and hence $\sigma_{y,\gamma_n y}\to\sigma_{y,\xi}$. Hence $\xi\in\Lambda(\Gamma,y)$, proving the lemma.

Denote $\Lambda(\Gamma)=\Lambda(\Gamma,x)$, $x\in X$ arbitrary.

**Lemma 4.** (See [Knieper '97]) 	If $X/\Gamma$ is compact, then $\Lambda(\Gamma)=X(\infty)$.

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