# Prove a function $\mathbb{R}^2\to \mathbb{R}$ is continuous if and only if inverse image of open sets is open

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!

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Remember that (ii) is actually an "if and only if": a subset $A$ is open if and only if for every $x\in A$ there is an $r\gt 0$ (which may depend on $x$) such that $B(x,r)\subseteq A$.
You start well: take $x_0\in f^{-1}(A)$. We want to show that there exists $r\gt 0$ such that $B(x_0,r)\subseteq f^{-1}(A)$.
So we look at $f(x_0)\in A$. Since $A$ is open, there exists $\epsilon\gt 0$ such that $B(f(x_0),\epsilon)\subseteq A$. Now, by your definition of continuity, there exists $\delta\gt 0$ such that for every $x\in B(x_0,\delta)$, $f(x)\in B(f(x_0),\epsilon)\subseteq A$. So, for every $x\in B(x_0,\delta)$, $f(x)\in A$. So $B(x_0,\delta)$ is contained in...
$B(x_0, \delta)$? Isn't it just going backwards in the definition for the continuity? Am I allowed to? –  Jozef Feb 9 '12 at 17:18
What do you mean, going backwards? The definition of continuity says: for every $x_0$, for every $\epsilon\gt 0$, there exists $\delta\gt 0$ such that $f(B(x_0,\delta))\subseteq B(f(x_0),\epsilon)$. Here, we have $B(f(x_0),\epsilon)$, and we apply the definition. You aren't going backwards, because it doesn't matter where $\epsilon$ came from, all that matters is that you have an $x_0$ and an $\epsilon\gt 0$. –  Arturo Magidin Feb 9 '12 at 17:21