Topology of the completed upper-half plane

Question

Define the topology on $\mathbb{H}^* : = \mathbb{H} ∪ \mathbb{Q} ∪\infty$ by taking a basis of open sets around $\infty$ to be $S_{\epsilon} : = \{ z ∈ H : Im ( z ) > 1 /\epsilon \}∪\infty$ , and taking $Γ( 1 )$ -transforms to get bases of open sets around the points in $\mathbb{Q}$ (along with the usual topology on $\mathbb{H}$ ).

Sketch neighbourhoods of open sets around some points in $\mathbb{Q}$

(here, $\mathbb{H}$ is the upper half plane and $Γ(1)$ is $Γ(1):=Γ(1)′/\{±I\})$, where $Γ(1)′:=SL(2,\mathbb{Z})$ )

Super stuck, any help greatly appreciated!

Answer 1

If $a/b \in \mathbf Q$ with $(a,b) =1$, there exist integers $m, n$ such that $am-bn=1$ (Bezout's identity). The modular transformation $z \mapsto (az + n)/(bz + m)$ belongs to $\Gamma$, and it sends $\infty$ to $a/b$. What does it send the neighborhood $S_\epsilon$ to? (Look at where it sends the line $\text{Im } z = 1/\epsilon$.)

Answer 2

To see how the neighbourhoods of $x\in\mathbb{Q}$ in $\mathbb{H}^\ast$ look, it is helpful to first get a picture of how the neighbourhoods $S_\varepsilon$ of $\infty$ look, when viewed as subsets of the Riemann sphere. The straight line $\operatorname{Im} z = 1/\varepsilon$ becomes a circle tangent to the great circle corresponding to $\mathbb{R}\cup \{\infty\}$, touching it in the north pole. Möbius transformations map circles on the sphere to circles on the sphere, so the neighbourhood of an $x\in\mathbb{Q}$ corresponding to $S_\varepsilon$ is bounded by a circle tangent to the same great circle, touching it in the point corresponding to $x$. In the plane, it becomes a circle in the upper half plane tangent to $\mathbb{R}$, touching it in $x$. So the neighbourhood of $x$ corresponding to $S_\varepsilon$ is the union of an open disk in the upper half plane whose centre has real part $x$ and whose boundary circle is tangent to $\mathbb{R}$, with $\{x\}$.