Let $f:\mathbb{R}^2 \to \mathbb{R}$, and define $f^{-1}(A)=\{x \in \mathbb{R}^2\mid f(x) \subset A\}$ for any $A\subseteq \mathbb{R}$.
Prove: $f$ continuous iff for every open $A \in \mathbb{R}$, $f^{-1} \subset \mathbb{R^2}$ is open.
I have a problem adjusting to all these $\mathbb{R^n}$ definitions, and I'd like your guidance with solving this easy, but not that easy for beginners, question in "topology".
$(i)$ $f$ is continuous, so to all $\epsilon>0$ there's a $\delta>0$, for every $x \in B(x_0, \delta)$, $f(x) \in B(f(x_0), \epsilon)$.
$(ii)$ $A$ is open so for every $x \in A$ there's a radius $r$ such that $B(x,r) \subset A$
I need to prove that $f^{-1}(A)$ is open so let $x_0 \in f^{-1}(A)$ so $f(x_0) \in A$, Now from being A an open set I know there's an open ball s.t $B(f(x_0), \epsilon) \subset A$, what Do I do next? how do I apply the continuous?
Thanks!