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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x)=\left\{\begin{array}{ll}\sin \pi x & \text { if } x \in \mathbb{Q} \\ 0 & \text { if } x \in \mathbb{R} \backslash \mathbb{Q} .\end{array}\right.$$ Prove that it's continuous only at integer points.

$\text{My attempt:}$ I was trying to prove this using sequential definition of continuity. Let $\{x_n\}$ be any sequence of rational converging to an integer $L$. We need to show, $f(x_n) \rightarrow f(L)$. $f(L)=0$, so we need to show $|f(x_n)-f(L)|=|f(x_n)|< \epsilon$ for $\epsilon >0$. But how do we proceed from here?

Anne Bauval
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Ellie_Wong
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  • By continuity of $\sin,$ $f(x_n)=\sin(\pi x_n)\to\sin(\pi L)=0.$ But you need to consider also any sequence $(y_n)$ of irrational numbers converging to $L$ (this will then be sufficient). And to prove integers are the only points of continuity. – Anne Bauval Jul 20 '23 at 21:45
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    This is a duplicate of https://math.stackexchange.com/q/3366544/ . – Xander Henderson Jul 20 '23 at 22:01
  • Any $x$ can be written as the limit of a rational sequence $q_k \to x$ or an irrational sequence $\alpha_k \to x$. To have a limit with these sequences, we must have $\sin (\pi q_k ) \to 0$. In particular this happens iff $x$ is an integer. Hence if $x$ is not an integer, $f$ is not continuous at $x$. If $x$ is an integer, since both $0 \to 0$ and $\sin (\pi q_k ) \to 0$ we see that $f$ is continuous at $x$. – copper.hat Jul 21 '23 at 02:40

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