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Math Help - Continuity

  1. #1
    Junior Member
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    Continuity

    For what value(s) of x is f continuous?


    f(x)= { 0 if x is rational
    2 if x is irrational
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  2. #2
    Senior Member Pinkk's Avatar
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    I am going to use the sequential definition of continuity and show the definition fails at rational and irrational points. It is not continuous anywhere. Let a be rational. Let x_{n} = \frac{1}{n\sqrt{2}} + a, which is a sequence of irrational numbers which converges to a. Since each x_{n} is irrational, f(x_{n}) = 2 for all positive integers n, so it trivially converges to 2, and since f(a) = 0 \ne 2, we have that  \lim x_{n} = a yet \lim f(x_{n}) \ne f(a), so the function is discontinuous on the rationals. Now let a be irrational. By the denseness of \mathbb{Q}, we can construct a sequence of rational numbers x_{n} that converges to a. Since each x_{n} is rational, we have that f(x_{n}) = 0 for all positive integers n. So f(x_{n}) trivially converges to 0, and since f(a) = 2 \ne 0, we have that \lim x_{n} = a yet \lim f(x_{n}) \ne f(a), so the function is discontinuous on the irrationals.

    If you need to find the values using the epsilon-delta definition of continuity, the basic idea behind the proof is that any interval contains infinitely many rationals and irrationals, so if the function were continuous then f(x) must get arbitrarily close to 0 and 2 simultaneously, which is clearly impossible.
    Last edited by Pinkk; September 21st 2010 at 06:19 AM.
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