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.
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.