Geometric description of equivalence classes. For $X = R^2$ define the relation $R$ on $X$ by $(x_1, y_1)R(x_2, y_2)$ if $x_1 = x_2$.
a). Verify that $R$ is an equivalence relation on $X$.
I've already shown that this is reflexive, symmetric, and transitive. This isn't where my lack of understanding is.
b). Describe geometrically the equivalence classes of $R$.
My lack of understanding is here. The way that I see it, if we take any $x_i$ at a point  $(x_i, y_1)$, under the relation $R$, the point $(x_i, y_1)$ translates to $(x_i, y_2)$ and $x_i$ remains the same. So we have a vertical line segment. And this is the same for any $x$ in $X$.
Any help would be appreciated.
 A: Geometrically, I assume you meant $\mathbb{R}$, it is simply the set of verticle lines in the two dimensional plane. Another way you can also view it is that the equivalence relation collapses $\mathbb{R}^2$ to $\mathbb{R}$ and as such turn a plane into a line.
You can construct this to work forany line, having it point in any direction and not just verticle but the collapsing remains the same in all cases.
A: To make plane geometry, we place ourselves in the vector space $\mathbb{R}^2$.
Let $\mathcal{C}$ be a class, what we write formally $\mathcal{C} \in \mathbb{R}^2_{/R}$[hence the importance, as written by @Zelos Malum to differentiate well $\mathbb{R}$ and R]. By definition, there exists $(x_0,y_0) \in \mathbb{R}^2$ such that  $\mathcal{C}=\{(x,y)\in \mathbb{R}^2:(x,y)R(x_0,y_0)\}=\{(x,y)\in \mathbb{R}^2:x=x_0\}=\{(x_0,y): y\in \mathbb{R}\}=(x_0,0)+\mathbb{R}(0,1)$.
Thus, $\mathcal{C}$ is the "verticle line" of $\mathbb{R}^2$ passing by $(x_0,0)$.
We say that $(x_0,0)$ is a representative of $\mathcal{C}$. We can choose as representative any $(x_0,y_1)$ with $y_1 \in \mathbb{R}$, in other words any other point of the line $\mathcal{C}$.
