In $\mathbb{R}^{n+1}\setminus\{0\}$, let us introduce an equivalence relation by setting $x\sim y$ iff $y = \lambda x$ for some $\lambda \in \mathbb{R}\setminus\{0\}$. The set of all equivalence relations $[x] := \{y \in \mathbb{R}^{n+1}\setminus \{0\}: x\sim y\}$ is defined to be the real projective space $\mathbb{R}\mathbb{P}^{n}$.
Take $\pi: \mathbb{R}^{n+1}\setminus \{0\} \to \mathbb{R}\mathbb{P}^{n}$ to be the canonical projection map $\pi(x) := [x]$ and equip $\mathbb{R}\mathbb{P}^{n}$ with the quotient topology.
I want to prove that $\mathbb{R}\mathbb{P}^{n}$ is Hausdorff and second countable. I know that $\mathbb{R}^{n+1}\setminus \{0\}$ is Hausdorff and second countable because it is a topological subspace of $\mathbb{R}^{n+1}$, and I was trying to get the same properties by using pre-images of $\pi$. However, I got nowhere.
I know a proof of this fact, but it uses the real projection space built from the unit sphere $\mathbb{S}^{n}$ instead of $\mathbb{R}^{n+1}\setminus \{0\}$. I know both constructions are homeomorphic, but I wanted to prove it directly in my setting, using my construction. Can anyone help me?
Best Answer
Here's a proof I have that $\mathbb{RP}^n$ is Hausdorff.
First I will show that $\pi$ is an open map.
Suppose $U$ is an open subset of $\mathbb{R}^{n+1}\setminus\{0\}$. Define the set $$V = \bigcup\limits_{\lambda \in \mathbb{R} \setminus \{0\}}\lambda U$$ where $\lambda U$ means the set $\{\lambda x: x \in U\}$. For any nonzero $\lambda$, the map $x \to \lambda x$ is a homeomorphism of $\mathbb{R}^{n+1}\setminus\{0\}$ onto itself, so each $\lambda U$ is open and hence $V$ is open in $\mathbb{R}^{n+1}\setminus\{0\}$. It's not hard to see that $V$ is in fact the saturation of $U$ with respect to $\pi$.
So $V$ is a saturated open subset of $\mathbb{R}^{n+1}\setminus\{0\}$ and $\pi(U) = \pi(V)$ is open in $\mathbb{RP}^n$ with the quotient topology.
Now, I will use the following lemma (see Proposition 3.57 in John M. Lee's Introduction to Topological Manifolds, Second Edition):
In our case, $\pi$ was shown to be an open quotient map, so we just need to show that $\mathcal{R}$ is closed in $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$.
Now $\pi(x_1) = \pi(x_2)$ if and only if $x_1$ and $x_2$ lie on the same line crossing $0$, so $(x_1, x_2)$ lies in the complement of $\mathcal{R}$ if and only if $x_1$ and $x_2$ are linearly independent (regarded as vectors in $\mathbb{R}^{n+1}$ with $0$ as the starting point).
We may extend $\{x_1, x_2\}$ to a basis $\{x_1, x_2, \ldots, x_{n+1}\}$ for $\mathbb{R}^{n+1}$. Then $\det(x_1, x_2, \ldots, x_{n+1}) \neq 0$, and by continuity of $\det$, there is an open subset $W$ of $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$ on which $\det(x_1', x_2', x_3, \ldots, x_{n+1}) \neq 0$ for all $(x_1', x_2') \in W$. So all the pairs of vectors in $W$ are linearly independent, and hence do not lie in $\mathcal{R}$. This shows that the complement of $\mathcal{R}$ is open.