1
$\begingroup$

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.

  • 0
    I don't think you want $f^{-1}$ to be the inverse. It might not exist.2012-12-16

2 Answers 2

2

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 $y\in B(x,\delta)\Rightarrow f(y)\in B(f(x),\epsilon)$?

  • 0
    Thanks to you too, I will try figuring out!2012-12-16
0

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. $f(B_{\delta}(x))\subseteq B_{\epsilon}(f(x))$.

  • 0
    uff, I see, topology seems hard to understand to me. Thanks again!2012-12-16