Let $X\subset \mathbb{R}$ and let $f,g:X \to X$ be continuous functions such that $f(X)\cap g(X)=\emptyset$ and $f(X)\cup g(X)=X$. Which one of the following sets cannot be equal to $X$?
A) $[0,1]$
B) $(0,1)$
C) $[0,1)$
D) $\mathbb{R}$
I don't know how I go through this problem. I think I have to choose some function to get the correct option but I can't. Thank you for help.
It is A) because then both $f[0,1]$ and $g[0,1]$ are compact and hence closed, and they are disjoint, nonempty, and $f[0,1]=[0,1]-g[0,1]$, so $f[0,1]$ is open in $[0,1]$, so is $g[0,1]$, this shows that $[0,1]$ is disconnected, a contradiction.
The continuous image of an interval is again an interval (connectedness). The continuous image of a closed and bounded interval is a closed and bounded interval (compactness and connectedness).
Thus $X$ cannot be $[0,1]$, because there's no way to write $X$ as the disjoint union of two closed and bounded intervals: without loss of generality, you can assume that the maximum of $f(X)$ is less than the minimum of $g(X)$ and a contradiction is reached.
All other cases can happen. For instance, consider $X=\mathbb{R}$ and $f(x)=x^2$, whereas $g(x)=-e^{-x}$. The cases B and C are similarly dealt with.
