let $f:\mathbb R^n\longrightarrow \mathbb R$ a continuous map and $\alpha\in \mathbb R$ . let $A=\{(x_1,\cdots,x_n)\in \mathbb R^n\;|\; f(x_1,\cdots,x_n) < \alpha\}$ How to show that $A$ is open in $\mathbb R^n$ and to what extent this result can be generalized?
I know that the complement $A^c=\bigsqcup_{k\geq \alpha} B_k$ where $B_k=\{(x_1,\cdots,x_n)\in \mathbb R^n\;|\; f(x_1,\cdots,x_n) =k\}$ and that $B_k$ is closed because $B_k=f^{-1}(\{k\})$ but we can't deduce that $A^c$ is closed since this is not a FINITE union.