Comparable elements and classes. Linear Algebra - Shilov. 
*

*Given a subspace $\mathbf L$ of a linear space $\mathbf K$, an element $x\in\mathbf K$ is said to be comparable with an element $y\in\mathbf K$ (more exactly, comparable relative to $\mathbf L$) if $x-y\in \mathbf L$. Obviously, if $x$ is comparable with $y$, then $y$ is comparable with $x$, so that the relation of comparability is symmetric. Every element $x\in \mathbf K$ is comparable with itself. Moreover, if $x$ is comparable with $Y$ and $y$ is comparable with $z$, then $x$ is comparable with $z$, since
$$x-z=(x-y)+(y-z)\in \mathbf L$$

*The set of all elements $y\in \mathbf K$ comparable with a given element $x\in \mathbf K$ is called a class, and denoted by $\mathbf X$. As just shown, a class $\mathbf X$ contains the element $x$ itself, and every pair of elements$y\in \mathbf X$, $z\in \mathbf X$ are comparable with each other. Moreover, if $u\notin\mathbf X$, then $u$ is not comparable with any element of $\mathbf X$. Therefore two classes either have no elements in common or else coincide completely. The subspace $\mathbf L$ itself is a class. This class is denoted by $0$, since it contains the zero element of the space $\mathbf K$.


Although the above lines seem to me quite cryptic, I can follow the entire reasoning. Unfortunately, I can't understand why the subspace $\mathbf L$ is itself a class. And why it is denoted by $0$, since it contains the zero element of the space $\mathbf K$.
Furthermore, what does $\mathbf L$ stand for? Is it a general subspace of $\mathbf K$?
Since the whole space $\mathbf K$ can be partitioned into a set of nonintersecting classes, how can I visualise a class? Could you make an example with $\mathbb R^3$?
Could you help me? Could you try to explain me the above concepts in a less obscure way? I am studying on Linear Algebra by Shilov, chapter 2 section 2.48.
Thank You.
 A: Not really an expert on this topic, but your question made me revisit this indeed obscure part of my old Shilov copy!
I will give an example on $\mathbf{K} = \mathbb{R}^2$. Let 
\begin{equation}
\mathbf{L} \triangleq \{\mathbf{x} \in \mathbb{R}^2 : \mathbf{x} = (\alpha, 0), \alpha \in \mathbb{R}\}
\end{equation}
that is, $\mathbf{L}$ the set of points along the x-axis of the x-y plane (see figure below). Now consider a vector $\mathbf{x} = (\beta, \gamma)$. Then the class of $\mathbb{x}$ is
\begin{equation}
\mathbf{X} \triangleq \mathbf{X}_\gamma=\{\mathbf{x} \in \mathbb{R}^2 : \mathbf{x} = (\alpha, \gamma), \alpha \in \mathbb{R}\},
\end{equation}
that is, $\mathbf{X}_\gamma$ is the set of points along the line $y=\gamma$ on the x-y plane (see figure). 
Note that the class of the zero vector $(0,0)$ is $\mathbf{X}_{\gamma=0}= \{\mathbf{x} \in \mathbb{R}^2 : \mathbf{x} = (\alpha, 0), \alpha \in \mathbb{R}\} = \mathbf{L}$, hence the naming convention $\mathbf{L} = \mathbf{0}$. Furthermore, note that it holds $\mathbf{X}_\alpha \cap \mathbf{X}_\beta= \emptyset$, for $\alpha \neq \beta$ and $\mathbb{R}^2 = \cup_{\alpha \in \mathbb{R}} \mathbf{X}_\alpha$.

