Is $PSL_2(\mathbb Z)$ a Fuchsian group of the first kind? We know that as a discrete subgroup of $PSL_2(\mathbb R)$, $PSL_2(\mathbb Z)$ is a Fuchsian group. But how to prove/disprove that it is of the first kind. i.e. if every point on the extended real line is its limit point of some orbit?

Let's me add the following details to Moishe Kohan's answer and comments and please let me know if anything is wrong:
It is easy to see by Bezout theorem about coprime numbers that every rational point on $\mathbb R$ is $PSL(2,{\mathbb Z})$ equivalent to $\infty$. Since $\infty$ is a limit point of sequence $T^n z$ where $T:z\to z+1$. So $\mathbb Q \cup \infty$ is contained in the limit set of $PSL(2,\mathbb Z)$.
Now here is the harder part:
Any irrational point $x\in \mathbb R$ can be approximated by a squence of rational numbers $(r_n)$. Although $(r_n)$ don't belong to the orbit $PSL(2,\mathbb R).i$, for each $n$, there exists a sequence $(g_{nk}.i)$ converging to $r_n$ as $k \to \infty$. Now the diagonal sequence $(g_{nn}.i)$ converges to $x$ as $n\to \infty$.
