Equivalence relation/Real projective space/Example

Let ${\displaystyle {}n\in \mathbb {N} }$, and set ${\displaystyle {}M=\mathbb {R} ^{n+1}\setminus \{0\}}$. ${\displaystyle {}\mathbb {R} ^{n+1}}$ is a real vector space, the scalar multiplication of ${\displaystyle {}\lambda \in \mathbb {R} }$ and ${\displaystyle {}x\in \mathbb {R} ^{n+1}}$ is denoted by ${\displaystyle {}\lambda \cdot x}$ Moreover, set

${\displaystyle R={\left\{(x,y)\in M\times M\mid {\text{ there exists some }}\lambda \in \mathbb {R} \setminus \{0\}{\text{ such that }}\lambda \cdot x=y\right\}}\,.}$

Therefore, two points are defined to be equivalent if they can be transformed to each other by scalar multiplication with a scalar ${\displaystyle {}\lambda \neq 0}$. We can also say that two points are equivalent if they define the same line through the origin.

This is indeed an equivalence relation. The reflexivity follows from ${\displaystyle {}x=1x}$ for every ${\displaystyle {}x\in M}$. To prove symmetry, suppose ${\displaystyle {}xRy}$, that is, there exists some ${\displaystyle {}\lambda \neq 0}$ such that ${\displaystyle {}\lambda x=y}$. Then also ${\displaystyle {}y=\lambda ^{-1}x}$ holds, as ${\displaystyle {}\lambda }$ has an inverse element. To prove transitivity, suppose that ${\displaystyle {}xRy}$ and ${\displaystyle {}yRz}$ holds; this means that there exist ${\displaystyle {}\lambda ,\delta \neq 0}$ such that ${\displaystyle {}\lambda x=y}$ and ${\displaystyle {}\delta y=z}$. Then ${\displaystyle {}z=\delta y=(\delta \lambda )x}$ with ${\displaystyle {}\delta \lambda \neq 0}$. The equivalence classes of this equivalence relation are the lines through the origin (but without the origin). The quotient set is called the real-projective space (of real dimension ${\displaystyle {}n}$), and is denoted by ${\displaystyle {}{\mathbb {P} }_{\mathbb {R} }^{n}}$.