Let $U_0\subseteq\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$ defined as follows: $$U_0=\big\{(x_0,\dots,x_n)\in\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\;|\; x_0\ne 0\big\}.$$
I must prove that $U_0$ is an open set of $\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$
My attempt. Consider the function $$f\colon\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\to \mathbb{R}\quad\text{defined as}\quad f(x_0,\dots, x_n)=x_0.$$
We observe that $f$ is continuous, then since $$\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\setminus f^{-1}\big(\{0\}\big)=U_0,$$ we have that $U_0$ is open.
Question. It's correct? If yes, there is a more hasty way?
Thanks!
It's correct. Alternatively, observe $U_0=\left\{\Bbb R^{n+1}-\{0_{\Bbb R^{n+1}}\}\right\}\cap(\Bbb R-\{0\}\times\Bbb R^n)$, the latter is a product of open sets, hence open.