$U_{0}=\{[a^{0},a^{1},…,a^{n}]\in \mathbb{R}P^{n}\,:\,a^{0}\neq 0\}$. Is it open in $\mathbb{R}P^{n}$?

Question

Is $U_{0}$ an open neighborhood of a $[a^{0},a^{1},…,a^{n}] \in \mathbb{R}P^{n}$ ? How can I prove that? It tried to see if $\pi^{-1}(U_{0})$ is open in $\mathbb{R}^{n+1}\smallsetminus\{0\}$. I see that $$\pi^{-1}(U_{0}) = (\mathbb{R}^{n+1}\setminus\{0\})\setminus\left(\{a^{1}\neq0\}\cup\{a^{2}\neq0\}\cup\dots\cup\{a^{n}\neq0\}\right)$$Where $\{a^{j} \neq 0\}$ is just the axis of the j-st coordinate. Is this set open in $\mathbb{R}^{n+1}\smallsetminus\{0\}$?

Answer 1

$\pi^{-1}(U_0)=\{(x_0,\cdots,x_n)\in\Bbb R^{n+1}\mid x_0\ne 0\}$

Which is open in $\Bbb R^{n+1}$ by definition of product topology. Hence, $\pi^{-1}(U_0)$ is open in $\Bbb R^{n+1}\setminus \{0\}$. Hence, by definition of quotient topology, $U_0$ is open.