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I would like to show the following implication.

Let $f$ be a function from $\mathbb{R}^n$ to $\mathbb{R}^m$. If $f^{-1}(U)\subset\mathbb{R}^n$ is open for every open $U\subset\mathbb{R}^m$, then $f$ is continuous.

Here is my attempt:

Let $B$ be the $m$-ball of radius $\epsilon>0$ centered at $f(a)$, where $a$ is a point in the domain of $f$. By hypothesis, $f^{-1}(B)$ is open. Thus, there is an $n$-ball of radius $\delta$ for some $\delta>0$ centered at $a$ that lies entirely within $f^{-1}(B)$. Call this $n$-ball $B^{-1}$. Note, $f(B^{-1})=B$ is not necessarily true, as $f$ isn't necessarily surjective.

enter image description here

But we know that, by the definition of $B$, for every point $x$ within $f^{-1}(B)$, we have that $\left|f(x)-f(a)\right|<\epsilon$. We also know that $B^{-1}\subset f^{-1}(B)$. Thus, we can say that there exists a $\delta>0$ such that $\left|f(x)-f(a)\right|<\epsilon$ whenever $0<\left|x-a\right|<\delta$.

Is this fine? I have tried not to look at any other proofs of this either here on MSE or anywhere else.

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  • $\begingroup$ This proof works. Good job. $\endgroup$ – Crostul May 21 '15 at 21:32
  • $\begingroup$ It looks ok. You should mention your $n$-ball of radius $\delta$ has center $a$. This $n$-ball is the red disc. Your (otherwise nice) picture is mislabelled. The red disc is $B^{-1}$?) $\endgroup$ – David Mitra May 21 '15 at 21:35
  • $\begingroup$ @DavidMitra I have added that fact and relabeled the picture. $\endgroup$ – Arturo don Juan May 22 '15 at 0:14
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You're very close. In particular, there is an open $n$-ball centered at $a$ contained entirely in $f^{-1}(B),$ having radius $\delta$ for some $\delta>0.$

You should probably then use the definition of preimage to finish the proof.

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