Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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...

share|improve this answer
1  
$B(x_0, \delta)$? Isn't it just going backwards in the definition for the continuity? Am I allowed to? –  Jozef Feb 9 '12 at 17:18
1  
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
    
O.k Thank you arturo! –  Jozef Feb 9 '12 at 17:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.