Take a function defined by
\begin{cases} 0 & x\in \mathbb{R} \smallsetminus \mathbb{Q}\\ \sin|x| & x\in\mathbb{Q} \end{cases} Find the points of continuity.
By solving $\sin|x|=0$, I get $x=n\pi$, where $n\in \Bbb N $, but $\pi$ is irrational.
So how do I find the points of continuity?
Note that points of continuity exists when the limit at the $x$-value exists and a point is defined on that $x$-value.
So inorder for
\begin{cases} 0 & x\in \mathbb{R}\setminus\mathbb{Q}\\ \sin|x| & x\in\mathbb{Q} \end{cases}
to have continuous points at $x=n\pi$, $\underset{x\in\mathbb{R/Q}}{\lim\limits_{x\to n\pi}}0=\underset{x\in\mathbb{Q}}{\lim\limits_{x\to n\pi}}\sin|x|$ and $\sin|n\pi|=0$
In real analysis textbooks a limit that is restricted to a subset of $\mathbb{R}$ may exist as long as the subset is dense in $\mathbb{R}$. This is because such a subset can "closely approximate any real value" or in formal terms every real number is a limit point of that subset. Hence you must prove that $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$.
Then we can say
$$\underset{x\in\mathbb{R\setminus Q}}{\lim\limits_{x\to n\pi}}0=\underset{x\in\mathbb{R}}{\lim\limits_{x\to n\pi}}0=0$$
and
$$\underset{x\in\mathbb{ Q}}{\lim\limits_{x\to n\pi}}\sin|x|=\underset{x\in\mathbb{R}}{\lim\limits_{x\to n\pi}}\sin|x|=0$$
Hence
$$\underset{x\in\mathbb{R \setminus Q}}{\lim\limits_{x\to n\pi}} 0 = \underset{x\in\mathbb{Q}}{\lim_{x\to n\pi}} \sin|x|$$
Finally since $\sin(n\pi)=0$ we know that all continous points are at $x=n\pi$.