Problem related to continuous functions on real numbers

Question

Let $X$ be a subset of $\mathbb R$ and let $f, g : X \to X$ be two continuous functions such that $f(X) \cap g(X) = \emptyset$ and $ f(X) \cup g(X) = X.$ Then which one of the following sets can't be equal to $X.$

1. $[0,1]$
2. $[0,1)$
3. $(0,1)$
4. $ \mathbb R.$

I am just feeling it too hard to start. Any suggestion will be highly appreciated. Thank you.

Update:

I was just giving a thought in the following way- If $X = [0,1]$ then if $f(x)$ is open(or closed) then from continuity of $f$, $X$ becomes open(or closed) and then from continuity of $g$, $X$ becomes closed(or open), which in either case a contradiction. So $X \neq [0,1].$ (Here both $f(X)$ and $g(X)$ can't be open(or closed) as $[0,1]$ is connected.)

Now if $X = (0,1)$ then no contradiction arises in the above argument so it can be possible. For $X=\mathbb R$ similarly it can be shown that it's a possible case. In this way I am struck at the case where $ X =[0,1).$

Am I thinking in a right direction ?

Answer 1

$X=[0,1]$ is indeed impossible, for otherwise $f[X], g[X]$ would both be compact (thus closed) and form a disconnection for $[0,1]$ which is connected. Contradiction.

Answer 2

We will use the fact that for a function $f: X \subset \mathbb{R} \to X$ defined on a compact domain $X$, there exists two points $x_-$ and $x_+$ such that $$ f(x_-) = \min_{x \in X}f(x) \quad f(x_+) = \max_{x \in X} f(x). $$

Case $X = [0,1]$: since $f$ and $g$ both are continuous, there exist four points $u_+,u_-$ (associated to $f$) and $v_+, v_-$ (associated to $g$) at which the extrema are reached.

Claim: We have $f(u_-) \leq f(u_+) < g(v_-) \leq g_+)$. The first and last inequality are trivial. The interesting one is the second one.

Assume it is not true. Then it holds $f(u_+) \geq g(u_-)$ and it follows that $\varnothing \neq [\max\{f(u_-),g(v_-)\}, \min\{f(u_+), g(v_+)\} \subset f(X) \cap g(X)$ which contradicts one of your assumptions. Therefore the claim holds.

Now, since let $y = \frac{f(u_+)+g(u_-)}{2}$. $y \in X$ (since $X$ is connected and contains pieces below and above this number) but $y \notin f(X) \cup g(X)$.

Note that, should $X$ not be connected, you could use this kind of thinking on each connected component.

Case 2: $X = [0,1)$: try to find examples, such that both functions are translated from one another.

Case 3: $(0,1)$: adapt case 4

Case 4: $\mathbb{R}$: try a downward facing parabola with an exponential function.

Hope this helps.