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Math Help - Proof Closed set??

  1. #1
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    Proof Closed set??

    So I have f: R -> R be continuous. Let S={x in R: f(x)=0} be nonempty. Prove S is a closed set.

    So far I've started with:
    Let {x_n} be a sequence in S such that x_n -> x ?? yes? where to next??


    Any help would be greatly appreciated!
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  2. #2
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    That's how I would start. The next step would be to observe what f(x_n) equals for all n \in \mathbb{N}, and then conclude what the value of f(x) will be, using the continuity of f over \mathbb{R}.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by engineer5373 View Post
    So I have f: R -> R be continuous. Let S={x in R: f(x)=0} be nonempty. Prove S is a closed set.

    So far I've started with:
    Let {x_n} be a sequence in S such that x_n -> x ?? yes? where to next??


    Any help would be greatly appreciated!
    Problem: Let f:\mathbb{R}\mapsto\mathbb{R} be continuous. Define \mathcal{Z}\left(f\right)=\left\{x\in\mathbb{R}:f(  x)=0\right\}. Prove that \mathcal{Z}\left(f\right) is closed.

    Proof: Let \xi be a limit point of \mathcal{Z}(f). Then there exists some sequence \left\{\xi_n\right\} such that \xi_n\in\mathcal{Z}(f)\quad\forall n and \xi_n\to\xi. But then \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{n\to\in  fty}0=0. And since f was continuous as \xi this means that \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{x\to\xi  }f(x)=f(\xi)=0. Therefore \xi\in\mathcal{Z}(f). Thus \mathcal{Z}(f) contains all its limit points, which consequently means it's closed.
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    See now it makes sense. Thank you both so much!!!
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Problem: Let f:\mathbb{R}\mapsto\mathbb{R} be continuous. Define \mathcal{Z}\left(f\right)=\left\{x\in\mathbb{R}:f(  x)=0\right\}. Prove that \mathcal{Z}\left(f\right) is closed.

    Proof: Let \xi be a limit point of \mathcal{Z}(f). Then there exists some sequence \left\{\xi_n\right\} such that \xi_n\in\mathcal{Z}(f)\quad\forall n and \xi_n\to\xi. But then \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{n\to\in  fty}0=0. And since f was continuous as \xi this means that \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{x\to\xi  }f(x)=f(\xi)=0. Therefore \xi\in\mathcal{Z}(f). Thus \mathcal{Z}(f) contains all its limit points, which consequently means it's closed.
    Mathstud, is that you ?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    Mathstud, is that you ?
    You're like the third person that's said that?? Why do you think I'm 'Mathstud'?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Here is another way to do this problem


    Problem: Let \phi:\mathbb{R}\mapsto \mathbb{R} be continuous. Define \mathcal{Z}(\phi)=\left\{x\in \mathbb{R}:\phi(x)=0\right\}. Prove that \mathcal{Z}(\phi) is closed.

    Proof: Let \xi be a limit point of \mathcal{Z}(\phi). Then \forall \ell>0 there exists some y\in N_{\ell}(\xi) such that \phi(y)=0. Because \phi is continous at we know that \forall\varepsilon>0 there exists some \delta>0 such that if

    y\in N_{\delta}(\xi)\implies \phi(y)\in N_{\varepsilon}(\phi(\xi)). Now assume that \xi\notin\mathcal{Z}(\phi). Then \phi(\xi)\ne0\implies|\phi(\xi)|\ne0. But this is a contradiction, for since |\phi(\xi)|>0 one may let \varepsilon=|\phi(\xi)| in the above. But no matter

    what \delta>0 one picks there will be a point y\in N_{\delta}(\xi) such that \phi(y)=0, but clearly 0\notin N_{|\phi(\xi)|}(\phi(\xi)). Therefore, \xi\in\mathcal{Z}(\phi) and the conclusion follows.

    Remark: This can be generalized greatly to much more interesting forms, but I digress. Also, these are the times where one wishes they were a topologist. For topologists define a function continuous if whenever a set is open(closed) in the codomain it's inverse image is open(closed) in the domain. And since \left\{0\right\} is closed by definition \phi^{-1}\left(\left\{0\right\}\right)=\left\{x\in\mathbb  {R}:\phi(x)=0\right\}=\mathcal{Z}(\phi) is closed.
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