Complex plane mod lattice spanned by two independent vectors with quotient topology is Hausdorff

Question

Below is an excerpt from an example from Forster's Lectures on Riemann Surfaces. In the example, I cannot figure out why the quotient space $\mathbb{C}/\Gamma$ is Hausdorff. How do we show this fact?

Answer 1

Take $u_1,u_2\in \mathbb{C}/\Gamma$. Consider the two representant $x_1,x_2 \in \mathbb{C}$ of $u_1,u_2$ respectively.

Now set $$r = \frac{1}{3} dist(\Gamma x_1, \Gamma x_2)$$ and consider $B_i$ the open ball of center $x_i$ and radius $r$. Obviously $B_1\cap B_2=\emptyset$. Note that $\pi(B_1)$ and $\pi(B_2)$ are open.

Let $u\in \pi(B_1)\cap \pi(B_2)$ be a point. Let $x\in\mathbb{C}$ be a representant of $u$. Then $x$ lie in $\gamma_1B_1\cap \gamma_2B_2$ for some $\gamma_1,\gamma_2\in\Gamma$ (two translations of the initial balls). Let $y_1,y_2$ be the centers of $\gamma_1B_1$ and $\gamma_2B_2$ respectively. Then \begin{eqnarray} dist(y_1,y_2) &\leq& dist(y_1,x)+dist(x,y_2) \\ &<&r+r\\ &\leq& \frac{2}{3} dist(\Gamma y_1,\Gamma y_2)\\ &\leq& \frac{2}{3} dist(y_1, y_2) \end{eqnarray} which is absurd. Hence $\pi(B_1), \pi(B_2)$ are disjoint open neighborhoods of $u_1$ and $u_2$.