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Math Help - Different Indicator

  1. #1
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    Different Indicator

    Define  f: [0,1] \to [0,1] by  f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ \frac{1}{q} \ \ \ \ \text{if} \ x \ \text{is rational} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}

    (a) Show that  f is discontinuous at every rational number.

    So there exists a sequence  (x_n) converging to  a \in [0,1] ,  a \in \mathbb{Q} such that  (f(x_n)) does not converge to  f(a) . So maybe come up with a sequence that has both irrational and irrational terms? Thus it would oscillate? For example, we can approximate  \frac{1}{2} by a sequence with both rational and irrational terms?

    (b) Fix  t \in \mathbb{R} and  n \in \mathbb{N} . Show that  f maps only finitely many elements of  (t-\frac{1}{2}, t+ \frac{1}{2}) to  \frac{1}{n} .

    Suppose  t = 1 and  n =2 . In other words, we want to show that  f maps only finitely many terms from  \left(\frac{1}{2}, \frac{3}{2} \right) to  \frac{1}{2} . So given a sequence  (x_n) that approximates  \frac{1}{2} , it will have a finite number of rational terms from part (a). Is this correct?


    (c) Show that  f is continuous at every irrational number.

    Given any sequence  (x_n) in  [0,1] converging to  a \in [0,1] ,  a \in \mathbb{R} \ \backslash \mathbb{Q}, (f(x_n)) converges to  f(a) . So we can always find an "irrational" sequence that approximates  a ? Thus  (f(x_n)) converges to  0 ?

    (d) Let  \epsilon >0 . Prove that  f maps only finitely many elements of  [0,1] to a value greater than  \epsilon .

    This is the same as part(b) right?

    (e) Prove that  f is Riemann integrable on  [0,1] and that  \int_{0}^{1} f = 0 .

    Here we just consider  |\mathcal{R}(f,P)| ? And bound it and show it is less than  \epsilon ?
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  2. #2
    Super Member Showcase_22's Avatar
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    <br />
f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ \frac{1}{q} \ \ \ \ \text{if} \ x \ \text{is rational} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}<br />
    A). (I'm not that good at these so it would be dandy if someone could check this over).

    Let C be our rational number. Hence C=\frac{p}{q}.

    Let's construct a sequence a_n such that a_n=C+\frac{1}{n} and a sequence b_n=C-\frac{\pi}{n} (where n \in \mathbb{N}).

    Clearly, a_n approaches C from the right and b_n approaches C from the left. Our goal is to show that f(\lim_{n \rightarrow \infty} a_n) and f(\lim_{n \rightarrow \infty} b_n) are different.

    \frac{1}{n} is always rational so a_n=C+\frac{1}{n} is always rational. Hence f(\lim_{n \rightarrow \infty}a_n)=\frac{1}{q}.

    \frac{\pi}{n} is always irrational so b_n=C+\frac{\pi}{n} is always irrational. Hence f(\lim_{n \rightarrow \infty}b_n)=0.

    Therefore:

    f(\lim_{n \rightarrow \infty}a_n) \neq f(\lim_{n \rightarrow \infty} b_n) so the function f is not continuous at every rational number.
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