Define the projective quotient space as follows:
$\mathbb{R}P^n$ is the quotient of $\mathbb{R}^{n+1}-\{0\}$ by the quotient equivalence relation $(x_0,\cdots,x_n)\sim (\alpha x_0,\cdots,\alpha x_n)$ for all $\alpha\neq 0$. Denote the quotient map as $r$.
Consider the set $V_i=\{x\in\mathbb{R}^{n+1}:x_i=0\}$. Denote $U_i$ as $r(\mathbb{R}^{n+1}-V_i)$.
Question: Prove $\mathbb{R}P^n$ is Hausdorff.
My approach:
I have proved that $U_i$ is open and Hausdorff for all $i$. Thus, for two distinct points $x$, $y$ in $\mathbb{R}P^n$, if there are in the same $U_i$, we can find two distinct open sets to separete them. But there are also points that are not in the same $U_i$. How do we deal with this kind of points?
Update:
Consider the diagonal set $\Delta(\mathbb{R}P^n)=\{(x,x):x\in\mathbb{R}P^n\}=\cup_{i}\{(x,x):x\in U_i\}=\cup_{i}\Delta(U_i)$. (The second equality is because $\{U_i\}$ forms a cover for $\mathbb{R}P^n$.)
Because $U_i$ is Hausdorff, each $\Delta(U_i)$ is closed. This implies $\mathbb{R}P^n$ is Hausdorff.
Is my proof correct? Thank you!
Best Answer
Hint: (I think the sets $U_i$ and $V_i$ are a bit of a distraction). Each point in $\Bbb{RP}^n$ is represented by a unique pair of points $\pm x \in \Bbb{R}^{n+1} \setminus \{0\}$ with $\|x\| = 1$.Given two distinct pairs $\pm x$ and $\pm y$ with $\|x\| = \|y\| = 1$ you can find disjoint open cones $C_x$ and $C_y$ such that $\pm x \in C_x$ and $\pm y \in C_i$. The images of $C_x$ and $C_y$ in $\Bbb{RP}^n$ are open sets that separate $r(x)$ and $r(y)$.