Equivalence relation/Real projective space/Example

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

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

Therefore, two points are defined to be equivalent if they can be transformed to each other by scalar multiplication with a scalar ${}\lambda \neq 0$. 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 ${}x=1x$ for every ${}x\in M$. To prove symmetry, suppose ${}xRy$, that is, there exists some ${}\lambda \neq 0$ such that ${}\lambda x=y$. Then also ${}y=\lambda ^{-1}x$ holds, as ${}\lambda$ has an inverse element. To prove transitivity, suppose that ${}xRy$ and ${}yRz$ holds; this means that there exist ${}\lambda ,\delta \neq 0$ such that ${}\lambda x=y$ and ${}\delta y=z$. Then ${}z=\delta y=(\delta \lambda )x$ with ${}\delta \lambda \neq 0$. 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 ${}n$), and is denoted by ${}{\mathbb {P} }_{\mathbb {R} }^{n}$.