# 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 $X$?

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 $$X\ne [0,1]$$.

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.
• @JohnNash You know for sure that the option A is right: $X$ cannot be $[0,1]$, $X$ can be $\mathbb{R}$ or $[0,1)$ or $(0,1)$, but you have to come up with examples for that (just draw pictures). BCD are wrong. Nov 17, 2018 at 6:41
• @JohnNash B and D are homeomorphic, so an example for one can be turned into an example for the other. For $[0,1)$ it's easy: $f(x)=\frac{1}{2} x$ and $g(x) = \frac{1}{2}(x+1)$ will do, I think. Nov 17, 2018 at 6:44