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I'm trying to prove:

If $f:A\subset \mathbb{R}^m \to \mathbb{R}^n$ and $A$ is open, then the following statements are equivalent:

1) $f$ is continuous on $A$.

2) $f^{-1}(V)$ is open in $\mathbb{R}^m$, for all $V\subset \mathbb{R}^n$ open.

Thanks for your help.

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    It's hard to know how to help in this situation without giving the whole game away. Could you say in a few words what you've tried?2011-09-05
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    @gary That's a good point. I was assuming that "continuous" means "satisfies the $\varepsilon$-$\delta$ condition at every point of $\mathbf R^m$", but it would be nice to spell that out in the question.2011-09-05
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    @Hiperion: As this is tagged "general topology", am I correct in guessing that your definition of "continuous function" is "the inverse image of an open set is open"?2011-09-05
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    My connection is slow today. Please be patient. Thanks2011-09-05
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    Of course. I hope that didn't come off as snooty. My point was only that you'll get much better answers and interactions here if you give your definitions and talk about your attempts to answer your questions!2011-09-05
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    @Dylan: I tried to follow your general suggestion from last time and I tried to do a generalization of the proof to maps between any two topological spaces X,Y.2011-09-05
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    What is the general definition of continuity you are using?2011-09-05
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    @gary the definition that I'm using is: Let $A\subset \mathbb{R}^n$, $f$ is continuous in $a\in A$ if and only if $\forall \varepsilon>0$ $\exists \delta >0$ such that $||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon$. Thanks.2011-09-09

2 Answers 2

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Assuming your using the definition of continuity that the inverse image of an open set is open, one way of doing the proof is:

i) Show that if the restriction of $f\colon \mathbb R^m \rightarrow \mathbb R^n:$ to any open subset of the domain is continuous, then f is globally-continuous, meaning f satisfies condition #2. For this, you can (Let X be the domain, mapping into Y to simplify; result generalizes to any two topological spaces X,Y anyway.) consider an open subset W of Y, and consider $f^{-1}(W)$ . By def. , f is continuous if $f^{-1}(W)$ is open in the subspace A of X .

2) If we use 2 for the definition of continuity, assume $f:X \rightarrow Y$ is continuous, so that $f^{-1}(W):=V$ is open in X. Can you show V is open in the subspace A?

3) Can you find an outlet, my battery is dy......

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    No problem, Hiperion.2011-09-05
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This is my answer:

1) $\to$ 2).

If $f^{-1}(V)=\emptyset$ the answer is trivial.

Let $a\in f^{-1}(V)$. Particularly $a\in A$. As $f$ continuous in $A$ then $\forall \varepsilon>0 \exists\delta>0$ such that, if $||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon$. i.e:

if $x\in \mathbb{B}_{\delta}(a) \implies f(x)\in \mathbb{B}_{\varepsilon}(f(a))$

$\implies f(\mathbb{B}_{\delta}(a))\subset \mathbb{B}_{\varepsilon}(f(a))$

$\implies \mathbb{B}_{\delta}(a)\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a)))$

$\therefore f^{-1}(V)$ is open.

2) $\to$ 1).

Let $\varepsilon>0$ and $a\in A$. $V$ is open then exists $\mathbb{B}_{\varepsilon}(f(a)))$ open. Now $a\in f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ is open. i.e:

$\exists \delta$ such that $\mathbb{B}_{\delta}(a))\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ (you can prove that this is true)

$\implies f(\mathbb{B}_{\delta}(a)))\subset \mathbb{B}_{\varepsilon}(f(a)))$

$||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon $

$\therefore f$ is continuous.