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Let $f$ be a functions from reals to reals given by the following rule: $f$ = $\sin x$ if $x \in \mathbb{Q}$ and $f$ = $0$ if $x \in \mathbb{R} - \mathbb{Q}$. At what points if $f$ continous?

I think $f$ would only be continuous at $0$. Is that correct? Can someone help proving this using delta-epsilon machinery? thanks

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I apologize: my first comment was right. Consider what happens at $x=\pi$ for instance: $f(\pi)=0$, since $\pi$ is irrational, and $\lim_{x\to\pi}\sin x=0$ as well. The same thing happens at every integer multiple of $\pi$, so you get infinitely many points of continuity. – Brian M. Scott Feb 3 '13 at 5:56
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HINTS: If $\sin a\ne 0$, let $\epsilon=|\sin a|$, and show that no $\delta$ is small enough to guarantee that $|f(x)-f(a)|<\epsilon$ whenever $|x-a|<\delta$, thereby showing that $f$ is not continuous at $a$. (There will be two subcases, depending on whether $a\in\Bbb Q$ or not.)

For an $\epsilon$-$\delta$ argument showing continuity of $f$ at $0$, note that if $|\sin x|\le|x|$, so making $|x|$ small automatically makes $|f(x)|$ small.

Finally, if $a=n\pi$ for some non-zero integer $n$, use the fact that $|\sin(x-a)|=|\sin x|$ along with the idea of the previous paragraph.

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Hint: If $f$ is supposed to be continuos at $x_0$ something special needs to hold about $\sin(x_0)$. Namely, choose a sequence $\{a_n\}$ of rationals converging to $x_0$ and a sequence $\{b_n\}$ of irrationals converging to $x_0$. Then, on the one hand $f(x_0)=f(\lim a_n)=\lim f(a_n)=\lim \sin(a_n)=\sin(\lim a_n)=\sin(x_0)$ and on the other hand $f(x_0)=f(\lim b_n)=\lim f(b_n)=\lim 0=0$. Thus, $\sin(x_0)=0$.

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