How to interpret $\mathbb{R}\mathbb{P}^n=\mathbb{R}^{n+1}/\mathbb{R}^*\simeq S^n/\mathbb{Z}_2$?

Question

Problem: Show that the real projective space $\mathbb{R}\mathbb{P}^n$ is Hausdorff and compact(using the quotient topology).

I was able to look for a proof of this on a book that made use of a lemma. However it was pinted to me that $\mathbb{R}\mathbb{P}^n=\mathbb{R}^{n+1}/\mathbb{R}^*\simeq S^n/\mathbb{Z}_2$

I know that a map of the following kind is a diffeomorphism:

$f:\mathbb{R}^{n+1}\to S^n\\x\to \frac{x}{||x||}$

However I am do not know how to intepret $\mathbb{R}\mathbb{P}^n=\mathbb{R}^{n+1}/\mathbb{R}^*\simeq S^n/\mathbb{Z}_2$

I do not understand these equivalence classes. I do not even know how can I derive such results.

Question:

Can someone help me understand $\mathbb{R}\mathbb{P}^n=\mathbb{R}^{n+1}/\mathbb{R}^*\simeq S^n/\mathbb{Z}_2$?

Thanks in advance!

Update: As it as pointed out in the comment section, I leave here the definition of the projective space

$\mathbb{R}\mathbb{P}^n=\mathbb{R}^{n+1}\setminus\{0\}/\sim\\x\sim y$ if $y=\lambda x$ for some $\lambda\in \mathbb{R}\setminus{0}$

Answer 1

For $\left(\mathbb R^{n+1}\setminus\{0\}\right)/\mathbb R^* \cong S^n/\mathbb Z^2$, let us first settle what the two spaces are. Both are of the kind $X/G$ where $X$ is a topological space and $G$ is a group acting on $X$. Then $X/G$ denotes the space of orbits of this group action with the quotient topology. For $\left(\mathbb R^{n+1}\setminus\{0\}\right)/\mathbb R^*$ the multiplicative group of non-zero real numbers $\mathbb R^*$ acts by scaling vectors, so the equivalence classes are of the kind $[v]=\{\,\lambda v\,|\,\lambda\neq 0\,\}$ for any $v\neq 0$. For $S^n/\mathbb Z^2$ we identify the additive group $(\mathbb Z^2,+)$ with the multiplicative subgroup $(\{-1,+1\},\,\cdot\,)$ of $\mathbb R^*$ which then also acts by scaling vectors.

The idea behind the homeomorphism is that you take the quotient $\left(\mathbb R^{n+1}\setminus\{0\}\right)/\mathbb R^*$ in two steps. Every $\lambda\in\mathbb R^*$ is uniquely written as $$ \lambda = \underbrace{\operatorname{sgn(\lambda)}}_{\in\{-1,+1\}}\,\cdot\,\underbrace{\mathbb{|\lambda|}}_{\mathbb R_{>0}}. $$ Note that $\left(\mathbb R^{n+1}\setminus\{0\}\right)/\mathbb R_{>0} \cong S^n$ since each orbit has exactly one point of absolute value one, i.e., one point in $S^n$. Now all that is left is taking orbits with respect to the action of $\{-1,+1\}$ and you get the original statement.

Answer 2

In the $\Bbb{R}^{n+1}/\Bbb{R}^*$ case, the base space is $\Bbb{R}^{n+1}$ with the origin removed and an equivalence class is a line that would otherwise be through the (now missing) origin. So the class is two open rays, starting at the boundary at the origin, pointing in opposite directions. If the class were just one of these two rays, the quotient would be the deformation retraction of $\Bbb{R}^{n+1} \smallsetminus \{0\}$ onto $S^{n}$. However, since both rays are in the equivalence class, we identify antipodal points on this sphere. It is this last description, "$S^n$ with antipodal points identified" that is intended by $S^n/\Bbb{Z}_2$. ("$\Bbb{Z}_2$" because the involution given by negating all coordinates exchanges the two points in each equivalence class.)