closed subset of $\mathbb R^n$ defined by a continuous map 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.
 A: Hints:
1) The preimage of an open set under a continuous map is open.
2) Look at the set $[\alpha,\infty)$ instead of $\{k\}$.
A: Here $A= f^{-1}(  (-\infty,a) ) .$ It can be shown that for maps $ f:\mathbb{R}^n \to \mathbb{R}^m $ the $\epsilon - \delta$ definition of continuity is equivalent to the statement that for every set $O$ open in $ \mathbb{R}^m $, the preimage $f^{-1} ( O) $ is open in $\mathbb{R}^n.$ Since $ (-\infty, a) $ is open in $\mathbb{R}$ and $f$ is continuous, $A$ is also an open set.
This formulation of continuity in terms of open sets becomes the definition in more general spaces where open sets are still defined but concepts of distance between points is not. Consider the generalized definition: 
Let $f: X \to Y $ be a map between topological spaces $X$ and $Y.$ Then $f$ is called continuous if for every set $O$ open in $Y$ we have $f^{-1} (O)$ open in $X.$ 
Thus if another similar problem gives again $f$ is continuous and $A$ is the preimage of an open set, we can automatically conclude $A$ is open.
