Cusp set is dense in boundary of hyperbolic plane Let $\Gamma$ be a Fuchsian group such that $\mathbb{H}/\Gamma$ is a finite-area hyperbolic surface with some cusps. How can we show that cusp set of $\Gamma$ is dense in the boundary of $\mathbb{H}$?
Whether we can prove it when $\mathbb{H}/\Gamma$ is a finite-area 2-orbifold?
Thank you in advance!
 A: The proof uses the limit set $\Lambda_\Gamma \subset \partial\mathbb H$ of $\Gamma$, so let me first review some basic facts about the limit set.
For any $x \in \overline{\mathbb H} = \mathbb H \cup \partial \mathbb H$, denoting its orbit in $\overline{\mathbb H}$ as $\Gamma \cdot x$, the limit set $\Lambda_\Gamma$ is equal to the set of all points $\xi \in \partial \mathbb H$ which are accumulation points of $\Gamma \cdot x$ in $\overline{\mathbb H}$. More precisely, $\xi \in \Lambda_\Gamma$ if and only if for every neighborhood $U \subset \overline{\mathbb H}$ of $\xi$ the set $\{\gamma \in \Gamma \mid \gamma \cdot x \in U\}$ is infinite. A key point about this definition is that $\Lambda_\Gamma$ is well-defined independent of the choice of $x \in \overline{\mathbb H}$. 
Another characterization of $\Lambda_\Gamma$ is that it is the unique minimal closed $\Gamma$-invariant subset of $\partial\mathbb H$. 
From the hypothesis that $\mathbb{H} / \Gamma$ has finite area, it follows that $\Lambda_\Gamma = \partial \mathbb H$. As a special case, it follows that $\Lambda_\Gamma$ is the set of accumulation points of $\Gamma \cdot x$ for every $x \in \partial\mathbb H$. To put it another way, every orbit of $\Gamma$ in $\partial\mathbb H$ is dense in $\partial\mathbb H$. In particular, the orbit of any cusp point is dense in $\partial\mathbb H$, and evidently every point in the orbit of a cusp point is also a cusp point. Therefore, the cusp set of $\Gamma$ is dense in $\partial\mathbb H$.
This proof uses only the hypothesis that $\Gamma$ acts properly discontinuously on $\mathbb{H}$ itself and has finite area quotient $\mathbb{H} / \Gamma$, so it works whether or not $\Gamma$ has torsion. In other words, the proof works whether $\mathbb{H} / \Gamma$ is a surface or an orbifold.
