A function defined on
$X=(0,1) \cup(2,3)$ is open set in $\mathbb{R}$ such that $f'(x)=0$ for all $x\in X$.
Then the range of function is
$1)$ uncountable number of point
$2)$ Countably infinite number of points
$3)$ atmost 2 points
$4)$ atmost 1 point
Here in this question The co-domain is not given rather it is $\mathbb{R}$ or something else ,
If we consider the co domain $\mathbb{R}$ then ,The function is continous if the range is a singleton set then its inverse image should be closed (singleton set is closed in $\mathbb{R}$) but the set is open so it can't be true ,same will be for two point option so option 3th and 4th is not true.
But if i consider $f$ as a constant function then 4th option is true ,but this condition works on Connected set while the given domain is not connected , next i am not getting how to solve
Please help
Best Answer
Since $f'$ is the null function, the restriction of $f$ to any connected subset of $X$ is constant. So, $f|_{(0,1)}$ and $f|_{(2,3)}$ are constant functions and therefore the range of $f$ consists of $2$ points, at most. And, of course, in some cases, it consists of exactly $2$ points.