Periodic points of image of Cantor set by ternary dilation Let $C$ be the Cantor. The set is built up by cutting all intervals into thirds and removing the middle third step one by one and to infinity.
\begin{align*}
 C_0 = &  [0,1]\\
  C_1 =&  [0, \frac 13] \cup [\frac 23, 1]\\
  C_2 =&  [0, \frac 19]\cup [\frac 29, \frac 13]\cup[\frac 23,\frac 79]\cup [\frac 89, 1].\\
\vdots \\
C= & \cap_{i=0}^{\infty} C_i 
\end{align*}
The Cantor set corresponds to the resulting "infinite number of times" ( observe that $T(C)=C$ ).
I wonder how to get prove that the periodic points of  $T(C)$ are dense in $C$, where
$T$ is the transformation $T: C \to C$ defined by $x \longmapsto 3x \quad \mathrm{mod}\, \,  1$.
A periodic point is a point $x_0$ such that $f^k(x_0)=x_0$ for some  integer $k \geq 1$.
So
\begin{align*}
T(x)= x \quad \mathrm{mod} \, 1    \\
2x -  \lfloor 2x \rfloor =& 0 \\
2x =  \lfloor 2x \rfloor &.
\end{align*}
How to finish this exercise.
Many thanks for your help.
 A: Notice that every $x$ in $[0, 1]$  may be expressed as
$$
  x=\sum_{n=1}^\infty  a_n3^{-n},
  $$
with $a_n\in \{0,1,2\}$, and that $x$ lies in the Cantor set iff it has an expression as above with every  $a_n\neq 1$.
It should be stressed that $x$ may have two different such expressions, e.g.
$$
  \frac 13 = 1\times 3^{-1} = 2\times 3^{-2} + 2\times 3^{-3} + 2\times 3^{-4} + \cdots 
  $$
and that it is enough for one of these expressions to satisfy   $a_n\neq 1$ in order  to qualify.
This said,  observe that  ternary dilation corresponds to shifting the digits, namely
$$
  T\left(\sum_{n=1}^\infty  a_n3^{-n}\right) =
  \sum_{n=1}^\infty  a_{n+1}3^{-n}.
  $$
This implies that $x$ is periodic for $T$ if its  digits are periodic,  that is,  if there exists $p$ such that
$$
  a_{n+p}=a_n, \quad\forall n\in \mathbb N.
  $$
Given any $x$ in $C$ we may therefore approximate it by a periodoc point $y$ whose digits $a_n$ coincide with those of $x$ up
to a big enough $n$,  and are then repeated periodically.
