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Math Help - continuous function

  1. #1
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    continuous function

    suppose that the function g:R->R is continuous and that g(x)=0 if x is rational. Prove that g(x)=0 for all x in R.

    -- I am totally stuck on this one.
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by chutiya View Post
    suppose that the function g:R->R is continuous and that g(x)=0 if x is rational. Prove that g(x)=0 for all x in R.

    -- I am totally stuck on this one.
    Since you have g(x)=0 when x \in Q, suppose that x \in R-Q. You should know that the rationals are dense in R, so there is a rational x_n \in \left( x-\frac{1}{n}, x+\frac{1}{n}\right). so..

    Can you try to complete?
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  3. #3
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    Quote Originally Posted by harish21 View Post
    Since you have g(x)=0 when x \in Q, suppose that x \in R-Q. You should know that the rationals are dense in R, so there is a rational x_n \in \left( x-\frac{1}{n}, x+\frac{1}{n}\right). so..

    Can you try to complete?
    Do I have to take the limit? I still cannot figure out. Thanks for your help
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by chutiya View Post
    Do I have to take the limit? I still cannot figure out. Thanks for your help
    \lim_{n \to \infty} {x_n} = x since |x-{x_n}| < \frac{1}{n}.

    By continuity f(x) = \lim_{n \to \infty}f(x_n)= \lim_{n \to \infty}0 = 0
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chutiya View Post
    suppose that the function g:R->R is continuous and that g(x)=0 if x is rational. Prove that g(x)=0 for all x in R.

    -- I am totally stuck on this one.
    Quote Originally Posted by harish21 View Post
    \lim_{n \to \infty} {x_n} = x since |x-{x_n}| < \frac{1}{n}.

    By continuity f(x) = \lim_{n \to \infty}f(x_n)= \lim_{n \to \infty}0 = 0
    Maybe even more simply. We have by the continuity of g that g^{-1}(\{0\}) is closed. Thus, \mathbb{R}\supseteq g^{-1}(\{0\})=\overline{g^{-1}(\{0\})}\supseteq\overline{\mathbb{Q}}=\mathbb{R  } so that g^{-1}(\{0\})=\mathbb{R}
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  6. #6
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    For a beginner here is a different proof.
    If a continuous function is not zero at a point then there is an open interval containing the point on which the function has the same sign (either positive or negative throughout the interval). But every interval contains a rational number.
    So this function has to be zero everywhere.
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