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