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Let $f(x)=\sin \left( \frac{1}{x} \right)$ for $x\ne 0$ and let $f(0)=0$. Show that $f$ is discontinuous on $\mathbb{R}$ and still has the intermediate value property on $\mathbb{R}$.

It's easy to show that $f$ is discontinuous at $x=0$. Now, I go on to prove that $f$ has the intermediate value property. Let $a,b \in \mathbb{R}$ with $a<b$. Assume that $y$ be some real such that $f(a) < y < f(b)$. If $y=0$ we are done for $f(0)=y$. If $y\ne 0$, I was hoping to use the continuity of sine on $\mathbb{R}$ but nothing seems work out.

Hints would be appreciated. I do not need full solution.


marked as duplicate by José Carlos Santos real-analysis Oct 29 '18 at 8:23

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    $\begingroup$ Note that you can assume $a<0<b$. $\endgroup$ – paf Jul 2 '18 at 16:45

If $0<a<b$ or $a<b<0$, then you just use the fact that $f$ is continuous in $\mathbb{R}\setminus\{0\}$.

If $a\leqslant0<b$, you use the fact that there's some natural $n$ such that$$\frac1{n\pi}<b$$and, since $y\in[-1,1]$, there's a $c\in\left[\frac1{(n+2)\pi},\frac1{n\pi}\right]$ such that $f(c)=y$.

Can you do the remaining case ($a<b\leqslant0$)?

  • $\begingroup$ should not it be $n+2$ instead of $n+1$? $\endgroup$ – hopefully Oct 29 '18 at 7:31
  • $\begingroup$ Why? The range of the restriction of $f$ to the interval that I mentioned is already $[-1,1]$. There is no need to consider a larger interval. $\endgroup$ – José Carlos Santos Oct 29 '18 at 8:01
  • $\begingroup$ Because of the discussion here math.stackexchange.com/questions/432647/… $\endgroup$ – hopefully Oct 29 '18 at 8:04
  • $\begingroup$ I have a question does not sin n\pi for any n is zero..... so what is this interval look like? $\endgroup$ – hopefully Oct 29 '18 at 8:07
  • $\begingroup$ How will the remaining case differ from your second case I feel like they will be the same ..... am I correct? $\endgroup$ – hopefully Oct 29 '18 at 8:09

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