Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $ g(x) = \left\{ \begin{array}{ll} 3x & \mbox{if } x \leq\frac{1}{2} \\ 3-3x & \mbox{if } x>\frac{1}{2} \end{array} \right. $
Show that the set $\Gamma=\{x\in[0,1] \mid g^n(x)\in[0,1] \forall n\}$ is the Cantor middle-thirds set.
I have been attempting to solve this sixth problem from chapter 8 of A First Course in Discrete Dynamical Systems. It is quite easy to see that if $x>1$ or $x<0$ the orbit diverges to $-\infty$. The only approach I have been able to come up with is that therefore any element of $\Gamma$ has to be in $[0,1]=C_0$ of the Cantor Set. If I can then prove through induction that if I assume that all $x\in\Gamma$ are in $C_n$ implies that $x\in C_{n+1}$ it seems I would have enough from there to finish a proof. However I have not been getting anywhere with this. Would anyone be so kind to give me any pointers or help me understand the question better?
Let $C$ be the Cantor set.
(I). If $x\in C$ then $x$ can be written in base-$3$ without using the digit $1,$ from which it is easily seen that $x\in C\implies g(x)\in C.$
(II). For $x\in [0,1]$ \ $C:$ There are unique $a,b\in \Bbb N$ such that $$x\in \left(\frac {3a-2}{3^b}, \frac {3a-1}{3^b}\right) \quad (\bullet)$$ so let $b=\deg (x).$
....(IIa). If $\deg (x)=1$ then $x\in (1/3,2/3)$ and $g(x)>1.$
....(IIb). If $\deg (x)>1$ then $g(x)\in [0,1]$ \ $C$ by $(\bullet),$ and $\deg (g(x))=\deg (x)-1,$ so $\deg (\;g^{\deg (x)-1} (x) \;)=1.....$ So $g^{\deg (x)}(x)>1.$
(III). If $x\not \in [0,1]$ then obviously $g(x)\not \in [0,1].$
Remark. In (IIb), by $(\bullet)$ we have $x\in [0,1] \setminus C \implies [\;1-x\in [0,1] \setminus C \land \deg (x)=\deg (1-x)\;]....$
....And $g(x)=g(1-x).$ So take $x'$ to the member of $\{x,1-x\}$ which is less than $1/2.$ By $(\bullet)$ applied to $x'$ we have $g(x)=g(x')=3x' \in [0,1]$ \ $C$ and $\deg (g(x'))=\deg (x')-1.$