($\mathbb{R},\mathcal{E}_1$) with equivalence relation $x\mathcal{R}y \Leftrightarrow a \neq 0$ y = ax I want to prove or disprove that ($\mathbb{R},\mathcal{E}_1$) with equivalence relation $x\mathcal{R}y \Leftrightarrow a \neq 0$ y = ax is homeomorphic to $(\{0,1\}, \tau)$ where $\tau$ is the discrete topology. 
To me it seems that there are two classes of equivalence: every point is equivalent to every point, and 0 is equivalent only to 0. It then seems plausible that they are homeomorphic. The function I tried is this one: f(x) = 0 if x is 0, and 1 in every other situation. Does that work?
Thank you!
EDIT: $\mathcal{E}_1$ is the euclidean topology. The question is: Is the quotient space homeomorphic to the space with two elements with discrete topology?
 A: You've observed that the quotient space $(\mathbb{R},\mathcal{E}_1)/\mathcal{R}$ consists of two points, the equivalence classes $C_0 = \{0\}$ and $C_1 = \mathbb{R}\backslash \{0\}$. Now what's the topology on the quotient space?
$(\mathbb{R},\mathcal{E}_1)/\mathcal{R}$ is homeomorphic to the two-point space with the discrete topology if and only if it has the discrete topology itself - that is, both of the singleton sets $\{C_0\}$ and $\{C_1\}$ are open. Are they? You just need to check the definition of the quotient topology... 
A: The equivalence relation has only two equivalence classes, as you pointed out:
$\{0\}$ and $\mathbb{R} \setminus \{0\}$.  
For the quotient space to a two-element discrete space it is necessary that $\{0\}$ is an open set in that quotient.
The open sets of the quotient are those whose inverse image by the map that sends the set to its quotient are open in the original space (see Wikipedia article).
Since $\{0\}$ is an equivalence class, it is its own inverse image.
But $\{0\}$ is not open in $\mathbb{R}$, with euclidean topology.
Hence they are not homemorphic.
