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Let $X \subseteq \mathbb{R}$ and $f,g : X \rightarrow X $ be continuous functions such that $f(X) \cap g(X) = \emptyset$ and $f(X) \cup g(X) = X$. Then which of the following cannot be $X$ ?

A. $[0,1]$
B. $(0,1)$
C. $[0,1)$
D. $\mathbb{R}$

Now I can see B. , C. , D. can be $X$ through some examples. Then A. must be the answer. Also $[0,1]$ being compact and connected its image under $f$ and $g$ has to be closed intervals and no to closed sub-intervals can be found of $[0,1]$ satisfying all the above criteria. But this is my intuition. How to begin if I have to write a solid proof for option A. ?

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user118494
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1 Answers1

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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.

Stephan Kulla
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