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Math Help - Function with open set properties

  1. #1
    Senior Member Pinkk's Avatar
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    Function with open set properties

    Suppose f: \mathbb{R}^{n} \to \mathbb{R}^{k} has the following property: For any open set U \subset \mathbb{R}^{k}, S = \{x : f(x)\in U\} is an open set in \mathbb{R}^{n}. Show that f is continuous.

    I am completely stuck on this. I know that there is a connection between the \epsilon - \delta definition of continuity and how if S is an open set, then given y\in S, there exists r>0 such that |x - y | < r \implies x\in S but I am missing the connection and can't really formalize any sort of proof. Thanks in advance.
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  2. #2
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    The balls B(x,r)={y : |x-y|<r} are open sets.
    Maybe this helps already?
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  3. #3
    Senior Member Pinkk's Avatar
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    Hmm, thanks, I think that does help. So here is my attempt.

    Given \epsilon > 0 and a\in \mathbb{R}^{n}, let U=B(\epsilon , f(a)), which is clearly an open set \{y : |y - f(a)| < \epsilon \}. Since f(a)\in U, we have that a\in S = \{x : f(x)\in U\}. So a is an interior point of S, which means there exists \delta > 0 such that |x - a|< \delta implies x\in S, which in turn implies f(x)\in U. Since f(x)\in U implies |f(x)-f(a)|<\epsilon, we have that given \epsilon > 0, there exists \delta > 0 such that |x - a| < \delta implies |f(x) - f(a)| < \epsilon, which means f is continuous at a. Since a was arbitrary, we have that f is continuous on \mathbb{R}^{n}. Q.E.D.
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Pinkk View Post
    Hmm, thanks, I think that does help. So here is my attempt.

    Given \epsilon > 0 and a\in \mathbb{R}^{n}, let U=B(\epsilon , f(a)), which is clearly an open set \{y : |y - f(a)| < \epsilon \}. Since f(a)\in U, we have that a\in S = \{x : f(x)\in U\}. So a is an interior point of S, which means there exists \delta > 0 such that |x - a|< \delta implies x\in S, which in turn implies f(x)\in U. Since f(x)\in U implies |f(x)-f(a)|<\epsilon, we have that given \epsilon > 0, there exists \delta > 0 such that |x - a| < \delta implies |f(x) - f(a)| < \epsilon, which means f is continuous at a. Since a was arbitrary, we have that f is continuous on \mathbb{R}^{n}. Q.E.D.
    I think it is perfectly rigorous
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