What can I say about the continuity of the function $f(x)=\sin(x)$ if $x$ is rational, and $f(x)=0$ otherwise?

Question

Let $$ f(x) = \begin{cases} \sin(x) & \text{if $x\in\mathbb{Q}$, and} \\ 0 & \text{otherwise.} \end{cases} $$

I need to study the continuity of the function described, but I don't remember how proceed to solve this.

Answer 1

$f(x) \; \text{is continuous at} \; x \Longleftrightarrow x = n\pi, \; n \in \Bbb Z; \tag 1$

for suppose $x \ne n \pi, n \in \Bbb Z$; then

$\sin x \ne 0, \tag 2$

so if $x \in \Bbb Q \setminus \{n\pi \mid n \in \Bbb Z \} = \Bbb Q \setminus \{ 0 \}$ (since $n\pi \in \Bbb Q$ if and only if $n = 0$), then

$f(x) = \sin x \ne 0, \tag 3$

and since there are irrationals $r$ arbitrarily close to $x$ where $f(r) = 0$, we cannot have

$\displaystyle \lim_{y \to x} f(y) = f(x) = \sin x \ne 0; \tag 4$

more formally, set

$\epsilon = \dfrac{\vert \sin x \vert}{2}; \tag 5$

for any $0 < \delta \in \Bbb R$, no matter how small,

$\exists y \; [[\vert y - x \vert < \delta] \wedge [\vert f(y) - f(x) \vert > \epsilon]], \tag 6$

which can't be true if

$\forall \epsilon \exists \delta [\forall y [[\vert y - x \vert < \delta] \Longrightarrow [\vert f(y) - f(x) \vert < \epsilon]]]; \tag 7$

the reader will recognize (7) as a statement of the continuity of $f(x)$ at $x$ thus we have shown that $f(x)$ cannot be continuous at $x \in \Bbb Q \setminus \{0\}$; likewise, if $x \in (\Bbb R \setminus \Bbb Q) \setminus \{n\pi \mid n \in \Bbb Z \}$, that is, a non-rational real which is not of the form $n \pi, n \in \Bbb Z$, then $\sin x \ne 0$ but $f(x) = 0$; there are thus rationals $y$ arbitrarily close to $x$ where $\vert f(y) \vert = \vert \sin y \vert > \vert \sin x \vert / 2$; since $f(x) = 0$, $f(x)$ is not continuous at such $x$.

So we have seen that $f(x)$ is not continuous at any real not of the form $n \pi$, which by contraposition is equivalent to the statement that

$f(x) \; \text{continuous at} \; x \Longrightarrow x = n \pi; \tag 8$

now if $x = n \pi$, then $\sin x = 0$ whether or not $n \pi$ is rational; thus as $y \to n \pi$ through rational values $\sin y \to \sin n \pi$; for irrational $y$ close to $x$, $\sin y = 0 = \sin x$ and in either case $f(y)$ is arbitrarily close to $f(x)$ for $y$ sufficiently close to $x$; thus $f(x)$ is continuous at $n\pi$; this completes the proof of (1).