# At which points is the following function continuous?

From Royden and Fitzpatrick's Real Analysis, Fourth Edition (Chapter 1, Problem 48):

Let $f$ be the function defined by $$f(x) = \begin{cases} x & \text{if x is irrational} \\ p \cdot \sin \frac{1}{q} & \text{if x = \frac{p}{q} in lowest terms} \end{cases}$$ At what points is $f$ continuous?

This function is similar to the one considered here.

I was able to show that $f$ is continuous at $0$ and not continuous at any non-zero rational number. I believe $f$ is continuous on the irrationals, but am having trouble proving this conjecture. My idea was to pick a small enough neighborhood around $x$ that the only rationals in the neighborhood would have $q$ so large that $\left|1 - q \sin \frac{1}{q}\right|$ was arbitrarily small. The rest of the proof will follow nicely if I can show such a neighborhood exists, but I am having trouble constructing it.

• What is $\lim_{h\rightarrow0}{\sin h\over h}$? Commented Jul 20, 2016 at 22:29
• @DavidMitra I believe the limit is 1, but I don't quite see how this helps. Commented Jul 20, 2016 at 22:54
• If $x$ is irrational, then, given $N$, you can find an open interval $O$ containing $x$ so that any rational in $O$ in lowest terms has denominator exceeding $N$. This, and the limit above, gives you what you want ($q\sin(1/q) = \sin(1/q)/(1/q)$ ). Commented Jul 20, 2016 at 23:03
• @DavidMitra Right I guess my question was how to explicitly construct such an open interval. Can I just use the continued fraction expansion of $x$? Commented Jul 20, 2016 at 23:21

Let us first show that $f$ is continuous at 0. By construction, $f(0) = 0$. Now pick any $\epsilon > 0$. Since $\lim_{n \to \infty} n \sin \dfrac{1}{n} = 1$, there exists an index $N$ such that $$n \sin \dfrac{1}{n} < 1+\epsilon$$ for all $n \geq N$. Let $\delta = \min\left\{\dfrac{1}{N},\dfrac{\epsilon}{1+\epsilon}\right\}$. For any non-zero rational number $x = \dfrac{p}{q}$, where $\dfrac{p}{q}$ is in lowest terms, we have \begin{align*} \left|x\right| = \left|\frac{p}{q}\right| < \delta \qquad &\Rightarrow \qquad \frac{1}{\delta} \leq \frac{|p|}{\delta} < |q| \\ &\Rightarrow \qquad N \leq |q| \\ &\Rightarrow \qquad \left|q\right| \cdot \left|\sin \frac{1}{q}\right| < 1 + \epsilon \\ &\Rightarrow \qquad |f(x)| = \left|p \cdot \sin \frac{1}{q}\right| \leq \delta \cdot \left|q\right| \cdot \left|\sin \frac{1}{q}\right| < \delta \cdot (1 + \epsilon) \leq \epsilon \end{align*} If $x$ is irrational, then $$|x| < \delta \qquad \Rightarrow \qquad |f(x)| = |x| < \epsilon$$ Thus $|x - 0| < \delta$ implies $|f(x) - 0| < \epsilon$, so $f(x)$ is continuous at zero.
Let us now show that $f$ is discontinuous at all non-zero rational numbers. Let $x = \dfrac{p}{q}$ be a non-zero rational number and consider the sequence $$x_n = \frac{p}{q} \cdot \frac{(p\cdot q + 1)^n +q}{(p\cdot q + 1)^n} =: \frac{p_n}{q_n}$$ By construction, $p_n$ and $q_n$ are relatively prime and $\lim_{n\to \infty} x_n = x$. We therefore have $$\lim_{n\to \infty} f(x_n) = \lim_{n \to \infty} \frac{p_n}{q_n} \cdot q_n \cdot \sin \frac{1}{q_n} = \left(\lim_{n \to \infty} \frac{p_n}{q_n}\right) \cdot \left( \lim_{n \to \infty} q_n \cdot \sin \frac{1}{q_n}\right) = \frac{p}{q}$$ However $$f(x) = p \sin \frac{1}{q} \neq \frac{p}{q}$$ because the left-hand side is an irrational number for all integers $q$. Since $\lim_{n\to \infty} x_n = x$ but $\lim_{n\to \infty} f(x_n) \neq f(x)$, $f$ is not continuous at $x$.
Lastly, we can also show that $f$ is continuous at all irrational numbers. Let $x$ be an irrational number and fix $\epsilon > 0$. Let $x = [a_0; a_1, a_2, \cdots]$ denote the continued fraction expansion of $x$ and define the sequences \begin{align*} h_n = a_n h_{n-1} + h_{n-1} \qquad h_{-1} = 1 \qquad h_{-2} = 0 \\ k_n = a_n k_{n-1} + k_{n-1} \qquad h_{-1} = 0 \qquad h_{-2} = 1 \end{align*} Since the $a_n \geq 1$ for all $n \geq 1$, $k_n \to \infty$. It is possible to show $\left\{\dfrac{h_n}{k_n}\right\} \to x$ and $$\left|x - \frac{h_n}{k_n}\right| < \left|x - \frac{p}{q}\right|$$ for any rational $\dfrac{p}{q}$ satisfying $0 < q < k_n$, This means we can find an index $N$ such that \begin{gather*} \left|x - \frac{h_N}{k_N}\right| < \frac{\epsilon \cdot |x|}{\epsilon + 2 \cdot |x|} \end{gather*} and $$\left|1 - n \cdot \sin \frac{1}{n}\right| < \frac{\epsilon}{2 \cdot |x|}$$ for all $n \geq k_N$. Let $\delta = \left|x -\dfrac{h_N}{k_N}\right|$. Then for any rational $\dfrac{p}{q}$ satisfying $\left|x - \dfrac{p}{q}\right| < \delta$, we must have $q \geq k_N$. But this implies \begin{align*} \left|x - p \cdot \sin \frac{1}{q}\right| &\leq \left|x - \frac{p}{q}\right| + \left| \frac{p}{q}\right| \left| 1 - q \cdot \sin \frac{1}{q}\right| \\[8pt] &\leq \left|x - \frac{p}{q}\right|\left(1 +\left| 1 - q \cdot \sin \frac{1}{q}\right|\right) + \left|x\right| \left| 1 - q \cdot \sin \frac{1}{q}\right|\\ &< \epsilon \end{align*} For any irrational $x'$ satisfying $|x' - x| < \delta$, we have $|x' - x| < \epsilon$ because $\delta < \epsilon$. Thus $|x' - x| < \delta$ implies $|f(x') - f(x)| < \epsilon$, so $f$ is continuous at $x$.