Let $f:[0,1] \to [0,1]$ be a continuous function such that $f \circ f=f$ on $[0,1]$ , then is it true that the set $\{x \in [0,1] : f(x)=x \}$ is a non-empty interval?
I can show that it is non-empty as for any $x\in [0,1]$,$f(f(x))=f(x)$ so every point of $f([0,1])$ is a fixed point. I think I can show that $\{x\in [0,1]:f(x)=x\}=f([0,1])$. But proving it is an interval is appearing difficult for me.
On
My proof : let $g(x)=f(x)-x$ which is clearly continuous , $g(0)=f(0)\geq 0$ and $g(1)=f(1)-1\leq 0$ hence by the IVT there is $c \in [0,1]$ such that $g(c)=f(c)-c=0$ hence $c\in E=\{x \in [0,1] | f(x)=x \}$ , now let $x,y \in E$ such that $x\leq y$ , for $t\in[0,1]$ let $z=(1-t)x+ty$ , we have $f(x)=x\leq z \leq y=f(y)$ by the IVT again there $h\in [0,1]$ such that $z=f(h)$ but since $f(f(h))=f(h)$ we get $f(z)=z$ hence $E$ is a convex part of $\mathbb{R}$ , and so $E$ is an interval .
The set $\{x \in [0, 1]: f(x) = x\}$ is just the image of $f$,which is an interval by the intermediate value theorem.