# Canonical projection is open - Projective Space

Show that canonical projection $\pi:\mathbb{R}^{n+1}\setminus \{0\}\rightarrow \mathbb{RP}^{n}$ is open.

I tried to prove it, but I have a hard time pulling the aberts. I search the internet but only find the projection using the sphere $\pi:\mathbb{S}^{n}\rightarrow \mathbb{RP}^{n}$.

• You wrote "$\mathbb{R}^{n+1}\backslash$ {0}" Then someone corrected that so that it said "$\mathbb{R}^{n+1}\backslash \{0\}$". Then I corrected that further so it said "$\mathbb{R}^{n+1}\setminus \{0\}.$ So note proper MathJax usage. – Michael Hardy Apr 7 '18 at 17:32
• "pulling the aberts" means "manipulating the open sets" ? – fredgoodman Apr 7 '18 at 17:32
• @freegoodman In portuguese, "open set" translates as "conjunto aberto". I'd guess OP is not used to certain math terms in English. – Ivo Terek Apr 7 '18 at 17:34
• I've never seen the word "abert" before and Google isn't helping with that. – Michael Hardy Apr 7 '18 at 17:35
• Portuguese "aberto" = Spanish "abierto" = English "open". – fredgoodman Apr 7 '18 at 17:39

$\def\R{\mathbb R}$ $\def\Rnp{\mathbb R^{n+1}}$ Let $U \in \Rnp$ open. We need to show that $\pi^{-1}(\pi(U))$ is open. But $$\pi^{-1}(\pi(U)) = \{r x : r \in \R \setminus \{0\}, x \in U\} = \bigcup_{r\in \R \setminus \{0\}} r U,$$ which is a union of open sets, hence open.