The real projective space $\mathbb{R}P^{n}$ is second countable. . The real projective space $\mathbb{R}P^{n}$ is second countable.
How to prove this. I have to use this proof in a solution of a question. But I cannot prove. Please help me. Write the proof clearly.
 A: You can show first $\mathbb{R}^n$ is 2nd countable, then show $\mathbb{RP}^n$ is 2nd countable, see: e.g. http://homepage.divms.uiowa.edu/~idarcy/COURSES/133/3_2rpns.pdf
A: $\Bbb R^{n+1}$ has a countable base (namely open balls of rational radii on rational points). 
Assume that $X$ has a countable base $\mathcal U$.


*

*Then any of its subsets $Y\subseteq X$ has a countable base (namely $\{U\cap Y\,\mid\,U\in\mathcal U\}$).

*If $Z=X/{\sim}$ is a quotient of $X$ by an equivalence relation $\sim$, then $\{\{[x]_\sim\,\mid\, x\in U\}\ \mid\, U\in\mathcal U\}$ will be a countable base for the induced topology of $Z$.


Now, apply 1. for $X=\Bbb R^{n+1}$ and $Y=\Bbb R^{n+1}\setminus\{0\}$, then apply 2. for $X=\Bbb R^{n+1}\setminus\{0\}$ and $a\sim b \iff (a=\lambda b$ for some $\lambda\in\Bbb R)$.
A: Use the follow proposition: The image of a continuous open map on a second-countable space is second countable.
Consider $\mathbb RP^n$ as $S^n$ with antipodal points identified. $S^n$ is second-countable as a subspace of a second-countable space. All that is left to prove is that the quotient map $\pi:S^n -> \mathbb RP^n$ is a open map, because it is continuous by definition.
The quotient map $\pi$ is a open map when for every open $V$ in the n-sphere, $\pi^{-1}(\pi(V))$ is also open in the n-sphere .
