Let $X⊂\mathbb R$ and let $f,g:X→X$ be a continuous functions? Let $X⊂\mathbb R$ and let $f,g:X→X$ be a continuous functions such that $f(X)∩ g(X)=∅$ and $f(X)∪g(X)=X$. Which one of the following sets cannot be equal to $X$?


*

*$[0,1]$

*$(0,1)$

*$[0,1)$

*$\mathbb R$



I think its $\mathbb R$, as $X⊂\mathbb R$ but I am not sure of how other options will behave here.
 A: Suppose $X=[0,1]$; $f(X)$ and $g(X)$ are compact intervals since the image of a compact set by a continuous map  is compact and the image of a connected set by a continuous map is connected. Write $f(X)=[a,b]$, $g(X)=[c,d]$ you may suppose $0\in f(X)$ so $f(X)=[0,b]$. $c>b$ otherwise, $[0,b]\cap [c,d]$ is not empty. This implies that $b+(c-b)/2$ is not in $f(X)\bigcup g(X)$. Contradiction.
A: I'm open to discussion if this is incorrect:
Let $X=(0,1)$.  Then $f(X)$ and $g(X)$ are both open sets.  Now suppose that $f(X) 
\cap g(X)=\emptyset$ and $f(X)\cup g(X)=X$.
Let $a=\sup\{(0,1)\setminus g(X)\}$ and $b=\inf\{(0,1)\setminus f(X)\}.$  Then $a <b$ and $f(X)=(0,a)$ and $g(X)=(b,1).$  It must be that $a<b$.  We arrive at a contradiction since $f(X)\cup g(X)\neq X$.
A: Since $[0,1]$ is compact, also $f([0,1])$ and $g([0,1])$ are compact and thus closed.
Let $[0,1]$ be the disjoint union of $f([0,1])$ and $g([0,1])$. Because $f([0,1])=[0,1] \setminus g([0,1])$ the set $f([0,1])$ must be open. This contradicts $f([0,1])$ being closed because the only sets in $[0,1]$ which are both open and closed are $[0,1]$ and $\emptyset$. Because $f([0,1])$ and $g([0,1])$ have at least one element, $f([0,1])$ cannot be $[0,1]$ or empty.
