Different Indicator

**manjohn12** · March 16th 2009, 10:24 AM

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 $\epsilon$ ?

**Showcase_22** · March 17th 2009, 11:22 AM

$<br /> f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ \frac{1}{q} \ \ \ \ \text{if} \ x \ \text{is rational} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}<br />$

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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