Let $f,g : X\to X$ be real continuous functions such that $f(X)\cap g(X) = \emptyset$ and $f(X)\cup g(X) = X$. Which sets cannot be equal to $X$?

Question

Let $X \subset \mathbb{R}$ and let $f,g : X\rightarrow X$ 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. $\mathbb{R}$

Please explain all options.

My approach:

Simply we can see the conditions imply that $X$ is disconnected. Continuity implies that $f([0,1])$ and $g([0,1])$ are compact. From the above conditions. Compact set would not be connected so $X\ne [0,1]$.

Answer 1

If $X$ were compact, so would $f[X]$ and $g[X]$ be, and they'd be two closed disjoint non-empty subsets of $X$ that form a partition of $X$, so then $X$ is not connected.

So we know a compact $X$ cannot be connected.