We may use the explicit formulas of Cantor sets $$ \mathcal{C} = [0,1] \setminus\bigcup_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right) $$ to show that $C$ is invariant under the dilation Let $T:x \longmapsto T_3(x)=3x \quad \text{mod} \, \, 1$. That is Prove that $T(C) = C.$
We use the fact that for any set $A$ and a map $T:E \longrightarrow F$ we have $f(A^c)=f(A)^c$, where $A^c$ is the complement of $A$ in $E$. Let \begin{align*} A=& \bigcup_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right)\\ =& \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^{n}-1} \left(\left(\frac{3k+1}{3^{n+1}}\,,\, \frac{3k+2}{3^{n+1}}\right)\right) \quad (n=m-1)\\ =& \bigcup_{k=0}^{3^{0}-1} \left(\left(\frac{3k+1}{3^{1}}\,,\, \frac{3k+2}{3^{1}}\right)\right) \\ & = (\frac{1}{3}, \frac{2}{3}) \bigcup_{n=1}^{\infty} \bigcup_{k=0}^{3^{n}-1} \left(\left(\frac{3k+1}{3^{n+1}}\,,\, \frac{3k+2}{3^{n+1}}\right)\right)\\ \end{align*} So $$ C=\bigcap_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right) $$ Which gives $T_3(A)= (1/9;2/9) \cup A$, and by the Morgan law, we get \begin{align*} T_3(A^c)=(T_3(A))^c=(1/9,2/9)\cup A^c= A^c \end{align*} Question: I can't seem to find the power $n-1$ in the explicit formula of $C$?
Any help very much appreciated!
Perhaps not an answer to your specific question. Let $C_0 = [0,1], C_1 = [0,1/3] \cup [2/3,1],$ and so on, i.e removing the middle third of each interval at each iteration. Write
$$C = \bigcap_{k=1}^\infty C_k.$$
To show that $C$ is invariant under $T(x) = 3x \mod 1$ one observes that $T(C_k) = C_{k-1}$ for each $k = 1,2,...$ . We then have
$$T(C) = T\bigg(\bigcap_{k=1}^\infty C_k\bigg) = \bigcap_{k=1}^\infty T(C_k) = \bigcap_{k=1}^\infty C_{k-1} = [0,1] \cap \bigcap_{k=1}^\infty C_k = C.$$
You can of course use the construction used in your question but the one above is the most natural and intuitive one in my opinion.
Exercise: Show that the Sierpinski Carpet and the Sierpinski triangle are invariant under some $kx \mod 1.$