Cloture of forward orbits of ternary dilation and Cantor set

Question

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.

Answer 1

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.

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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.

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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 $$ |T^n(x_0) -x|\leq |T^n(x_0) -x'| + |x'-x| < \varepsilon /2+\varepsilon /2=\varepsilon . $$