Function from Cantor Set to itself.

I am stuck in getting rational functions (except identity) defined from Cantor set to itself.

• Are you talking about the Cantor set, or a Cantor set? The function $f(x)=\mu x(1-x)$, for various values of $\mu\gt4$, is defined on a Cantor-type subset of $[0,1]$. Dec 3, 2013 at 6:34
• I mean here typically the middle 1/3rd cantor set. Does the function you have mentioned a map from C to C? Dec 3, 2013 at 7:21
• Well, there is also $f(x)=1-x$. Dec 3, 2013 at 17:01
• For $\mu\gt4$, the function $f(x)=\mu x(1-x)$ is a map from a Cantor set $C$ to itself. A Cantor set is defined to be a closed, bounded, perfect, totally disconnected subset of the reals. The middle-thirds set is an example of a Cantor set. The Cantor set $C$ is a subset of $[0,1]$. A proof is given in Elaydi's textbook, Discrete Chaos (but only for the case $\mu\gt2+\sqrt5$). Dec 3, 2013 at 23:10

Let $f$ be such that
$$$f(x) = \cases{ x + \frac{2}{3} & \text{if x<\frac{1}{2}} \\ x - \frac{2}{3} & \text{if x>\frac{1}{2}} }$$$
In this case, I would try to prove that there is no such function. Start by saying that rational functions are $C^\infty$ in their domain, and then relate the distances between points in your "middle-third" Cantor set to the continuiti of the derivative of the function.