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!