# Cloture of forward orbits of ternary dilation and Cantor set

I wonder if anyone can give me sensible proof.

The Cantor set $$C$$ is defined by \begin{align*} C_0&=[0,1],\\ C_1&=[0,\tfrac13]\cup [\tfrac23,1],\\ C_2&=[0,\tfrac19]\cup [\tfrac29,\tfrac13]\cup [\tfrac23,\tfrac79]\cup [\tfrac89,1],\\[-6pt] &\ \,\vdots\\[-12pt] C&=\bigcap_{n=0}^\infty C_n. \end{align*}

Observe that $$C=\left\{ x\in [0,1] \quad | \quad x=\sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} \quad \text{where} \quad \epsilon_k \in \{0,1\}^{\mathbb N}\right\}.$$

Consider the ternary dilation $$x\longmapsto T(x)=3x \quad \text{mod} \, \, 1$$.

Question: Show that there exists $$x_0 \in [0 ; 1)$$ with the closure of its forward orbit $$\{T^n(x_0) : n \in \mathbb N \}$$ are identical to $$C$$.

\begin{align*} T(x)=& 3x - \lfloor 3x \rfloor \\ T(T(x))= & T(3x - \lfloor 3x \rfloor) \\ = & 3(3x - \lfloor 3x \rfloor) - \lfloor 3(3x - \lfloor 3x \rfloor)\rfloor \\ = & 3^2 x - \lfloor 3^2 x \rfloor\\ \vdots \\ T^n(x) = & 3^n x - \lfloor 3^n x \rfloor. \end{align*}

So, by induction, \begin{align*} \overline{\{T^n(x_0)\}}=& \overline{\bigcup_{k=1}^{n}\{3^n - \lfloor 3^n x\rfloor}\} \\ & = \bigcup_{k=1}^{n}\{\overline{3^n - \lfloor 3^n x\rfloor}\}. \end{align*}

Another way consists to do some calculation from \begin{align*} x=& \sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} & \\ =& 2 \epsilon_1 + \sum_{k=}^\infty\frac{2\epsilon_{k+1}}{3^k} & \\ =& \sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} & \text{mod} \, \, 1 \end{align*} Thus $$T^n(x)= \sum_{k=1}^\infty\frac{2\epsilon_{k + n}}{3^k}$$, hence $$|T^n(x)- x| = \sum_{k=1}^\infty\frac{2(\epsilon_{k+n} - \epsilon_k)}{3^k}\leq \sum_{k=1}^\infty\frac{2}{3^k}=1$$ So it isn't clear that the claim is true.

Any help is welcome.

Consider the set $$W$$ of all finite words in the alphabet $$\{0,1\}$$. This is a countable set, so we may write $$W=\{w_n: n\in \mathbb N\}$$.

Now consider the infinite word $$w$$ obtained by concatenating the $$w_n$$ one after another.

If the digits of $$w$$ are used in the ternary expansion of $$x$$, then the orbit of $$x$$ under $$T$$ will be dense.

Notice that the collected works of Shakespeare, if encoded in a binary fashion, will appear as part of $$w$$, and so will all of the questions and answers in MSE.

EDIT: Let me try to be a bit more specific regarding the choice of the point with a dense orbit.

Take the universal word $$w$$ constructed above as the concatenation of all finite words $$w_n$$, and write $$w=\alpha _1\alpha _2\alpha _3\ldots ,$$ with each $$\alpha _i$$ in $$\{0, 1\}$$. Defining $$x_0=\sum_{k=1}^\infty\frac{2\alpha _k}{3^k},$$ I claim that $$\{T^n(x_0):n\in {\bf N}\}$$ is dense in the Cantor set $$C$$. To see this, let $$x$$ be any point in $$C$$, and write $$x=\sum_{k=1}^\infty\frac{2\epsilon _k}{3^k},$$ with each $$\epsilon _i$$ in $$\{0, 1\}$$. Given any $$\varepsilon >0$$ let us therefore find some $$n$$ such that $$|T^n(x_0)-x|<\varepsilon$$.

We first choose $$N$$ such that the truncated sum $$x'=\sum_{k=1}^N\frac{2\epsilon _k}{3^k}$$ satisfies $$|x-x'|<\varepsilon /2$$. By increasing $$N$$, if necessary, we may suppose that $$3^{-N}<\varepsilon /2$$.

Next, consider the finite word $$\epsilon _1\epsilon _2\epsilon _3\ldots \epsilon _N$$. Since all finite words appear in the set $$W$$ defined above, and since $$w$$ is the concatenetion of all words in $$W$$, there must be some $$n$$ such that $$\alpha _{n+1}=\epsilon _1, \quad \alpha _{n+2}=\epsilon _2, \quad \cdots , \quad \alpha _{n+N}=\epsilon _N.$$ As noted by the OP, to apply $$T$$ to any element in the Cantor set has the same effect as shifting its ternary digits. Therefore $$T^n(x_0) = \frac{2\alpha _{n+1}}{3} + \frac{2\alpha _{n+2}}{3^2} + \cdots + \frac{2\alpha _{n+N}}{3^N} + \sum_{k=N+1}^\infty \frac{2\epsilon _{n+k}}{3^k} =$$$$= x' + \sum_{k=N+1}^\infty \frac{2\epsilon _{n+k}}{3^k}.$$ This implies that $$|T^n(x_0) -x'|<3^{-N}$$, so $$|T^n(x_0) -x|\leq |T^n(x_0) -x'| + |x'-x| < \varepsilon /2+\varepsilon /2=\varepsilon .$$

• Although the information you provide is valuable to me, I believe you have not responded to my question. "Show that there exists $x_0 \in [0; 1)$ for which the closure of its forward orbit is equals $C_3$." – Furdzik Dec 31 '20 at 19:51
• I meant to describe the point $x$ whose orbit is dense in the third paragraph of my answer. – Ruy Dec 31 '20 at 20:41
• Thanks for your help and support. – Furdzik Jan 1 at 1:23
• You are welcome! – Ruy Jan 1 at 2:09