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Thread: Function with open set properties

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

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

    I am completely stuck on this. I know that there is a connection between the $\displaystyle \epsilon - \delta$ definition of continuity and how if $\displaystyle S$ is an open set, then given $\displaystyle y\in S$, there exists $\displaystyle r>0$ such that $\displaystyle |x - y | < r \implies x\in S$ 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 $\displaystyle \epsilon > 0$ and $\displaystyle a\in \mathbb{R}^{n}$, let $\displaystyle U=B(\epsilon , f(a))$, which is clearly an open set $\displaystyle \{y : |y - f(a)| < \epsilon \}$. Since $\displaystyle f(a)\in U$, we have that $\displaystyle a\in S = \{x : f(x)\in U\}$. So $\displaystyle a$ is an interior point of $\displaystyle S$, which means there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a|< \delta $ implies $\displaystyle x\in S$, which in turn implies $\displaystyle f(x)\in U$. Since $\displaystyle f(x)\in U$ implies $\displaystyle |f(x)-f(a)|<\epsilon$, we have that given $\displaystyle \epsilon > 0$, there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a| < \delta $ implies $\displaystyle |f(x) - f(a)| < \epsilon$, which means $\displaystyle f$ is continuous at $\displaystyle a$. Since $\displaystyle a$ was arbitrary, we have that $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}^{n}$. Q.E.D.
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  4. #4
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    Quote Originally Posted by Pinkk View Post
    Hmm, thanks, I think that does help. So here is my attempt.

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