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Math Help - advanced calculus, help!

  1. #1
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    advanced calculus, help!

    f: [-1, 1]---R and the graph of f is compact, prove that f is continuous?

    Any hint will be appreciated!!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Assume that the graph is compact. Assume for contradiction that  f is discontinuous at  x_0 \in [-1,1] . Then there exists an  \epsilon > 0 and a sequence  \{x_n \} \subset [-1,1] such that  \{x_n \} converges to  x_0 . Look at the sequence  \{(x_n, f(x_n)) \} . Since the graph is compact it has a subsequence \{(x_{n_{k}}, f(x_{n_{k}})) \}  which converges to a point  (x_1, f(x_1)) . So  \{x_{n_{k}} \} converges to  x_1 . But  \{f(x_{n_{k}}) \} converges to  f(x_0) which is a contradiction. This implies that  G(f) \to X is continuous and onto.
    Last edited by tukeywilliams; October 11th 2007 at 06:00 PM.
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  3. #3
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    Quote Originally Posted by violetsf View Post
    f: [-1, 1]---R and the graph of f is compact, prove that f is continuous?

    Any hint will be appreciated!!
    What about the Dirichlet function . Meaning rationals get mapped into 0 and irrationals into 1. So the image of [-1,1] is {0,1} which is certainly compact.
    --------
    Tukeywilliams, I think she is asking the converse.
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  4. #4
    Senior Member tukeywilliams's Avatar
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    I think I proved the right statement?
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  5. #5
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    I was thinking whether f(x0)=f(x1) ??
    Quote Originally Posted by tukeywilliams View Post
    I think I proved the right statement?
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  6. #6
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    Quote Originally Posted by tukeywilliams View Post
    I think I proved the right statement?
    This is my understanding. By "graph" the poster means "image".

    So the poster is asking: given a function mapping a compact set ([-1,1] is compact) into a compact set does it mean that the function is continous on the set?
    That is what I did.

    Now, I gave an example involving the Dirichelt (discontinous) function which shows it is false. You are saying you proved it, if so, then why do I have a conter-example?
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  7. #7
    Senior Member tukeywilliams's Avatar
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     f is continuous on  [-1,1] if and only if  G(f) is compact. I proved the  \Leftarrow direction.

    I defined  G(f) = \{x, f(x))| x \in [0,1] \}
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    Quote Originally Posted by tukeywilliams View Post
     f is continuous on  [-1,1] if and only if  G(f) is compact. I proved the  \Leftarrow direction.

    I defined  G(f) = \{x, f(x))| x \in [0,1] \}
    I am not going to read your proof now, but I think I see out difference. You use \{ (x,f(x)) \}, i.e. as a point in \mathbb{R}^2, while I used \{ f(x) \} as a point in \mathbb{R}. So maybe it is necessary for it to be continous.
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