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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?

bof
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Wolfgang
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  • The points in Cantor's $C$ are those with digits only $0$ and $2$ in base $3$. Assume first that $x$ has a $1$ in its $3$-ary representation. If the first digit is a $0$, then $x<1/3$. Therefore $g(x)=3x$, i.e. the digits shift to the left. If the first digit is a $2$, then $g(x)=3-3x$, i.e. the digits shift to the left and $2$'s turn into $0$'s and $0$'s become $2$'s. If there were no digit $1$ you keep applying $g$ and the number is still in $C$. Keep applying $g$ until the first digit of $g^n(x)$ is a $1$. Divide in two cases, $g^n(x)\leq 1/2$ and $g^n(x)>1/2$. But in both $g^{n+1}(x)>1$. – orole Feb 06 '18 at 22:24
  • Thanks for your fast response orole. I really like the idea of tackling the problem in base $3$. I see how it has to be true that an element $x$ of $\Gamma$ can't have a digit $1$ and that this can be proven as you described. However I am confused as to how this proves that this is the Cantor set. Is this something that I will have to prove seperately? – Wolfgang Feb 06 '18 at 22:36
  • That the Cantor set consists of the numbers that in base $3$ have only digits $0$ and $2$? You can prove it from the construction that eliminates the middle thirds. In the $n$-th step the intervals removed are precisely those numbers that have a $1$ in the $n$-th digit in base $3$. This last assertion is pretty much by definition of base $3$ expansion. – orole Feb 06 '18 at 22:44
  • Intuitively it does feel correct that a digit $1$ signifies that it is in a removed interval. I will try and work on the problem for a bit to see if I can link the concept together with the definition of the Cantor middle-thirds set that the book provided. – Wolfgang Feb 06 '18 at 22:47
  • $x=(0.a_1a_2a_3...)_3$ means: Divide $[0,1)$ in $3$ equal parts, the number is in the $a_1$-th subinterval. Why? Because if $a_1=0$, then $x<1/3$, if $a_1=1$, then $1/3<x<2/3$, if $a_1=2$, then $x\geq 2/3$. Divide that subinterval into $3$ equal parts. Then $x$ in the $a_2$-th subinterval. ... – orole Feb 06 '18 at 22:53

1 Answers1

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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 $\deg (g(x'))=\deg (x')-1.$

DanielWainfleet
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