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Thread: 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 $\displaystyle a$ be rational. Let $\displaystyle x_{n} = \frac{1}{n\sqrt{2}} + a$, which is a sequence of irrational numbers which converges to $\displaystyle a$. Since each $\displaystyle x_{n}$ is irrational, $\displaystyle f(x_{n}) = 2$ for all positive integers $\displaystyle n$, so it trivially converges to $\displaystyle 2$, and since $\displaystyle f(a) = 0 \ne 2$, we have that $\displaystyle \lim x_{n} = a$ yet $\displaystyle \lim f(x_{n}) \ne f(a)$, so the function is discontinuous on the rationals. Now let $\displaystyle a$ be irrational. By the denseness of $\displaystyle \mathbb{Q}$, we can construct a sequence of rational numbers $\displaystyle x_{n}$ that converges to $\displaystyle a$. Since each $\displaystyle x_{n}$ is rational, we have that $\displaystyle f(x_{n}) = 0$ for all positive integers $\displaystyle n$. So $\displaystyle f(x_{n})$ trivially converges to $\displaystyle 0$, and since $\displaystyle f(a) = 2 \ne 0$, we have that $\displaystyle \lim x_{n} = a$ yet $\displaystyle \lim f(x_{n}) \ne f(a)$, 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 $\displaystyle f(x)$ must get arbitrarily close to 0 and 2 simultaneously, which is clearly impossible.
    Last edited by Pinkk; Sep 21st 2010 at 05:19 AM.
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