I need to prove that $\mathbb{R}/\mathbb{Z}^{2}$ is not a manifold when $\mathbb{Z}^{2}$ acts (continuously) on $\mathbb{R}$ by $t\mapsto t+m+n\alpha$ where $\alpha$ is a fixed irrational for all $t\in\mathbb{R}$ and $(m,n)\in\mathbb{Z}^{2}$. If I understand correctly, $\mathbb{R}/\mathbb{Z}^{2}$ is the space of orbits $O_{t}$ where $s,t\in O_{t}$, i.e. $\exists (m,n)\in\mathbb{Z}^{2}:s=t+m+n\alpha$, which we rewrite $s\sim t$.
I have some difficulties to handle a quotient space $M/G$ where $M$ is a manifold and $G$ is a group. I don't really understand how to intuitively visualize it. It is defined as the space of orbits equipped with the topology $\mathcal{T}_{M/G}:=\{U\subset M/G\vert\,\text{preimage of}\,\,U\,\,\text{is open in}\,\,\mathbb{R}\}$
In my particular example, we can see that, by the previously defined mapping, we can send $t$ as closer as we want to some $r\in\mathbb{R}$ (and sometimes, of course, reach it). We need to show that either it is not a topological space, or it is a topological space that is not a topological manifold, i.e. not locally homeomorphic to $\mathbb{R}^{n}$ for some $n$. I also understand that the preimage of any orbit is a countable set in $\mathbb{R}$, and that we have $\mathcal{T}_{\mathbb{R}/\sim}$ is an uncountable set.
I would like to know how I can intuitively visualize $M/G$ in a general way and to understand how to apply the definitions on my particular problem.
Thanks in advance
This action of $Z^2$ on $R$ is not proper. Since $\alpha$ is irrational, $n+m\alpha, n,m\in Z$ is dense, thus the orbit of $0$ is dense, thus $R/Z^2$ is not separated. Let $x\in R$ and $[x]$ the class of $x$ in $R/Z^2$ every neighborhood of $[x]$ is equal to $p(U)$ where $U$ is a neighborhood of an element $y\in p^{-1}([x])$, there exists $g\in Z^2$ such that $g(0)\in U$, thus $[0]\in p(U)$.