discontinuity of functions

Question

got stuck on the next question,

Find the points of discontinuity of the function if they exist:

$f(x)={\begin{cases}\sin x&{\mbox{ for }}x \in\mathbb {Q} \\0&{\mbox{ for }}x\notin\mathbb{Q} \\\end{cases}}$

I would really appreciate a hint

Answer 1

Hint: Its continuous at those points $x$ such that $\sin x=0$ and only at those points.

Answer 2

$f(x)={\begin{cases}\sin x&{\mbox{ for }}x \in\mathbb {Q} \\0&{\mbox{ for }}x\notin\mathbb{Q} \\\end{cases}}$

we will choose $x_0 \notin \mathbb {Q}$ so that $\sin(x_0)=0$

so by definition $f(x_0) = 0$

${\displaystyle \lim _{x\to x_{0}}f(x)={\displaystyle \lim _{x\to x_{0}}\sin(x)= 0 = f(x_0)}}$

so for any ${\displaystyle x \not= \pi k }$ the function is discontinuous

Answer 3

If $q\ne0$ is rational, then take a sequence $(r_n)$ of irrationals converging to $q$. Then $$ \lim_{n\to\infty}f(r_n)=0\ne f(q)=\sin q $$ Thus $f$ is continuous at no rational point $q\ne0$.

Consider an irrational number $r$ and let $(q_n)$ be a sequence of rationals converging to $r$. Then $$ \lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}\sin q_n=\sin r $$ by continuity of the sine. In order that the function is continuous at $r$, you need $\sin r=f(r)=0$, that is, $r=k\pi$ for some integer $k\ne0$.

Therefore the function is not continuous at all irrational points not equal to $k\pi$, $k\ne0$ integer.

It remains to investigate whether the function is continuous at points of the form $k\pi$ ($k$ integer). Can you prove it?