# Different Indicator

• Mar 16th 2009, 10:24 AM
manjohn12
Different Indicator
Define $\displaystyle f: [0,1] \to [0,1]$ by $\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ \frac{1}{q} \ \ \ \ \text{if} \ x \ \text{is rational} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$

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

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

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

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

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

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

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

This is the same as part(b) right?

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

Here we just consider $\displaystyle |\mathcal{R}(f,P)|$? And bound it and show it is less than $\displaystyle \epsilon$?
• Mar 17th 2009, 11:22 AM
Showcase_22
Quote:

$\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ \frac{1}{q} \ \ \ \ \text{if} \ x \ \text{is rational} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$
A). (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 $\displaystyle C=\frac{p}{q}$.

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

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

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

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

Therefore:

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