Showing that a function has the intermediate value property

Question

Here's the question:

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.

Note that you can assume $a<0<b$. – paf Jul 02 '18 at 16:45 — paf, Jul 02 '18 at 16:45

José Carlos Santos · Accepted Answer · 2018-10-29T08:22:09.657

3

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 ($a<b\leqslant0$)?

edited Oct 29 '18 at 08:22

answered Jul 02 '18 at 16:44

José Carlos Santos

427,504

should not it be $n+2$ instead of $n+1$? – Intuition Oct 29 '18 at 07:31
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. – José Carlos Santos Oct 29 '18 at 08:01
Because of the discussion here https://math.stackexchange.com/questions/432647/showing-intermediate-value-property-holds – Intuition Oct 29 '18 at 08:04
I have a question does not sin n\pi for any n is zero..... so what is this interval look like? – Intuition Oct 29 '18 at 08:07
How will the remaining case differ from your second case I feel like they will be the same ..... am I correct? – Intuition Oct 29 '18 at 08:09
@hopefully I've edited my answer. Thank you. – José Carlos Santos Oct 29 '18 at 08:22
you are welcome :) ..... but can you answer my remaining questions please? – Intuition Oct 29 '18 at 08:25
I could not understand the intuition behind choosing the interval in that way ..... could you please explain this for me ? – Intuition Oct 29 '18 at 08:27
@hopefully Yes, the remaing case is like this one. The interval was chosen so that it is mapped by $f$ onto $[-1,1]$. – José Carlos Santos Oct 29 '18 at 08:29

Showing that a function has the intermediate value property

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