$f^{-1}$ and continuity I really need help concerning this task: let $f : X \to \mathbb{R}^n$, $X\subseteq \mathbb{R}^n$, be a function with $X$ open. 
Show: $f$ is continuous $\iff f^{-1} (W) $ is open for every open $W \subseteq\mathbb{R}^n$, when $f^{-1}$ is the inverse of $f$. 
Thanks for your help.
 A: My notation: $B(x,\epsilon)=\{y:\| x-y\|<\epsilon\}$ 
Here's the basic idea: 
$(\Rightarrow)$ Assume $f$ is continuous, and let $W\subset\Bbb{R}^n$ be open.  Now, let $x\in f^{-1}(W)$; we must find a $\delta$-ball such that $B(x,\delta)\subset f^{-1}(W)$.  How? Use the fact that $W$ is open and $f$ is continuous: $f(x)\in W$, $W$ is open so $\exists \epsilon>0$ such that $B(f(x),\epsilon)\subset W$.  Can you see how to "find" $\delta$ using continuity?
$(\Leftarrow)$ Suppose $f^{-1}(W)$ is open for every open set $W$, and let $\epsilon>0$.  Notice that for every $x\in X$, $B(f(x),\epsilon)$ is an open set.  Thus $f^{-1}(B(f(x),\epsilon))$ is an open set in $X$ which contains $x$.  Can you see how to find $\delta$ so that $y\in B(x,\delta)\Rightarrow f(y)\in B(f(x),\epsilon)$?
A: Let $x\in f^{-1}(W)$. Since $f(x)\in W$ exist $\epsilon>0$ s.t. $B_{\epsilon}(f(x))$$\subseteq W$. If $f$ is continuous at $x$ then exist $\delta>0$ s.t. $f(B_{\delta}(x))\subseteq B_{\epsilon}(f(x))$. Do you see now that $B_{\delta}(x) \subseteq f^{-1}(W)$ and $f^{-1}(W)$ is open?
For the converse let $x\in X$ and $\epsilon>0$. Using that $f^{-1}(B_{\epsilon}(f(x)))$ is open containing $x$ find $\delta>0$ s.t. $f(B_{\delta}(x))\subseteq B_{\epsilon}(f(x))$.
