Action of $SL_2\left(\mathbb{Z}\right)$ on $\mathbb{RP}^1$ is minimal 
Suppose that $SL_2\left(\mathbb{Z}\right)$ acts on $\mathbb{RP}^1$ by left multiplication i.e $A\cdot x=Ax$, where $A \in SL_2\left(\mathbb{Z}\right)$ and $x=[x_1:x_2] \in \mathbb{RP}^1$. Show that $O_x=\{A\cdot x: A\in SL_2\left(\mathbb{Z}\right)\}$ is dense for any $x \in \mathbb{RP}^1$.

Let $x=x=[x_1:x_2] \in \mathbb{RP}^1$. When $x_2 \ne 0$, we have $x=[x_1:x_2]=\left[\dfrac{x_1}{x_2}:1\right]$, otherwise $[x_1:0]=[1,0]$ . Hence, it is enough to look at the orbits of $[x:1]$ and $[1,0]$ in $\mathbb{RP}^1$. Let's look at $O_{[1:0]}=\left\{A[1:0]; A \in SL_2\left(\mathbb{Z}\right)\right\}$. Let $[y:1] \in \mathbb{RP}^1$. 
By Dirichlet's Theorem, there exists $P_N,Q_N \in \mathbb{Z}$ such that $ 1 \le Q_N \le N$ and $\left|Q_Ny-P_N\right| < \frac{1}{N}$. Let $A_N=\begin{bmatrix} P_N & P_NQ_N-1\\ 1 & Q_N \\ \end{bmatrix}$. Then $A_N[1:0]=[P_N:1]=\left[\frac{P_N}{Q_N}:  \frac{1}{Q_N}\right]$. 
The first component converges to $y$ the choice of $P_N$ and $Q_N$. How do I make the second component small? 
Thanks for the help!!
 A: Take a rational number $a/b$ close to $y$ with $(a,b)=1$. Say $ac-bd=1$. Which matrix sends $[1:0]$ close to $[y:1]$?
For the orbit of $[x:1]$, you can either adapt the above argument, or use that the action is continuous:
$\overline O_x$ contains $\infty=[1:0]$, so $O_\infty$ (closures of stable sets are stable), so $\overline O_\infty$.
A: As said by barto, the action is transitive on $\mathbf{Q}\mathbb{P}^1$.
Beware that a continuous action with a dense orbit on a topological is not always minimal. Here is, in this case, a way to conclude.
Assume by contradiction that $F$ is a closed invariant proper subset. Then $\mathbf{Q}\mathbb{P}^1$ is contained in its open complement $U$.
Let $V$ be the connected component of $U$ containing $\infty$. Then $V$ contains the complement of $[-n,n]$ for some $n$. Besides, $V$ is invariant by the stabilizer of $\infty$ in $\mathrm{SL}_2(\mathbf{Z})$, which contains the translation $x\mapsto x+1$. It follows that $V=\mathbf{Q}\mathbb{P}^1$ and hence $F=\emptyset$, proving minimality.
