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Math Help - Continuity on rationals?

  1. #1
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    Continuity on rationals?

    If  g:R \rightarrow R is continuous and  f:R \rightarrow R is such that f(x)=g(x) for all rational numbers x, is f continuos?

    I'm not 100% sure about this because f is clearly continuos on the rationals, but what happens at the irrationals?

    EDIT: Being slow today... consider g(x) = 1 (hence cts) and f(x) = 1 for x in Q, and 0 for x in R\Q. f(x) is discontinuous everywhere.
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  2. #2
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    Let g(x) = 0 for all x and let f(x) = 0 when x is rational, f(x) = 1 when x is irrational. Clearly, f is not continuous.

    On the other hand, if both f and g are continuous and f(x) = g(x) for all rational x, then f(x) = g(x) everywhere.
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  3. #3
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    Quote Originally Posted by emakarov View Post
    On the other hand, if both f and g are continuous and f(x) = g(x) for all rational x, then f(x) = g(x) everywhere.
    How would I prove this bit?
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  4. #4
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    Given any irrational, x, there exist a sequence of rationals, \{r_i\} which converges to x. Since f is continuous, f(x)= \lim_{i\to\infty} f(r_i)= \lim_{i\to\infty} g(r_i)= g(x) because g is continuous.
    Last edited by Plato; May 16th 2011 at 05:03 AM. Reason: TeX Fix
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  5. #5
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    Re: Continuity on rationals?

    An even more interesting scenario on the properties f(x) can have is the following :
    Let g(x) = sin(x) which is continuous over R
    Let f(x) = g(x) for x rational (f agrees with g at rationals)
    = -g(x) for x irrational (reflection of graph of g along the X axis at irrational x values)
    It is easy to prove that f(x) is continouus only at the points K*pi, K belonging to Integers
    and discontinuous everywhere ! Similarly, we can construct counter-intuitive functions continuous, say at Integers and
    discontinuous elsewhere ! The old classic ,Hobson's "Theory of Functions-I,II" would be an illuminating read !!!!
    Last edited by LASLESH; September 1st 2012 at 09:56 AM.
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