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

Please help me to get out these functions.

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    $\begingroup$ 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 $[0,1]$. $\endgroup$ Dec 3, 2013 at 6:34
  • $\begingroup$ I mean here typically the middle 1/3rd cantor set. Does the function you have mentioned a map from C to C? $\endgroup$
    – Fukuzita
    Dec 3, 2013 at 7:21
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    $\begingroup$ Well, there is also $f(x)=1-x$. $\endgroup$ Dec 3, 2013 at 17:01
  • $\begingroup$ 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 $\mu\gt2+\sqrt5$). $\endgroup$ Dec 3, 2013 at 23:10

1 Answer 1

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Let $f$ be such that

$ \begin{equation} f(x) = \cases{ x + \frac{2}{3} & \text{if $x<\frac{1}{2}$} \\ x - \frac{2}{3} & \text{if $x>\frac{1}{2}$} } \end{equation} $

Edit: as @studiosus points out, I left out the "rational" part of the requirement.

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.

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  • $\begingroup$ What makes you think that this functions is rational? Or that it preserves any nonempty compact set? $\endgroup$ Dec 3, 2013 at 13:35
  • $\begingroup$ I completely missed the "rational function" part... $\endgroup$
    – rewritten
    Dec 3, 2013 at 13:40
  • $\begingroup$ Do we have a non-linear kind of function from the classical Cantor set to itself? I mean can we define a function f(x)=P(x)/Q(x). where P and Q are quadratic polynomial of x. $\endgroup$
    – Fukuzita
    Dec 4, 2013 at 3:52

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