continuity problem

Question

Let $f:\Bbb R \to\Bbb R$ be defined by setting $f(x) = \sin x$ if $x$ is rational, and $f(x) = 0$ otherwise. At what points is $f$ continuous?

is this true?

Let $c$ be irrational. Then there exists a sequence of rational numbers $x_n$ such that $\lim x_n=c$. Then $f(x_n)=\sin x_n$, $\lim f(x_n)=\sin c$. However $f(c)=0$. Thus $f$ is discontinuous at $c$.

score 0 · Accepted Answer · answered Dec 16 '13 at 22:47

0

HINT: You’re thinking in the right direction, but you’ve overlooked the possibility that $\sin c=0$. For what irrationals $c$ is this true?

answered Dec 16 '13 at 22:47

Brian M. Scott

616,228

so c differant 2n*pi ? – user115722 Dec 16 '13 at 22:51
@user115722: It’s continuous at $2n\pi$ for integer $n$, but it’s also continuous at infinitely many other points. Where is $\sin x=0$? – Brian M. Scott Dec 16 '13 at 22:54
i'm not native english speaker.also sin(n*pi)=0 am ı wrong? – user115722 Dec 16 '13 at 23:01
@user115722: Now you’re right: $\sin x=0$ if and only if $x=n\pi$ for some integer $n$, and your argument shows that those are exactly the points at which $f$ is continuous. – Brian M. Scott Dec 16 '13 at 23:03
so answer is x=nπ.am ı need to show c is rational case ? – user115722 Dec 16 '13 at 23:08
@user115722: You need to show that $x=0$ is the only rational number at which $f$ is continuous. Use the same idea: if $c$ is rational, there’s a sequence of irrationals converging to it. – Brian M. Scott Dec 16 '13 at 23:09
thanks brian i understand.i guess :) – user115722 Dec 16 '13 at 23:15
@user115722: I hope so! You’re welcome. – Brian M. Scott Dec 16 '13 at 23:16

continuity problem

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