I am currently studying the dynamics associated with the function $f(x)=2x$ (mod 1). In particular, if we define the orbit of an element $y \in [0,1]$


$$ orb(y)= \{ f^m(y): m \in \mathbb{Z}\}$$


it is easy to see, for example, that $orb\big(\frac{1}{2}\big)=\mathbb{Z}\big[\frac{1}{2}\big] \cap [0,1)$. My question concerns what the orbits of elements of the form $\frac{1}{p}$, with $p$ prime would look like. It is easy to see (for example, for $p=3$) that 


$$ orb\Big(\frac{1}{3}\Big) \subset \Big(\frac{1}{3} \mathbb{Z} \Big[\frac{1}{2}\Big] \cap [0,1) \Big) \backslash \mathbb{Z} \Big[\frac{1}{2}\Big]$$


where $\frac{1}{3} \mathbb{Z} \Big[\frac{1}{2}\Big]= \Big\{\frac{m}{3\times 2^n}: m \in \mathbb{Z}, n \in \mathbb{Z}\Big\}.$ My problem is in proving (or disproving) the reverse inclusion. Is there any easy way to see that?



Orbits of the function f(x)=2x (mod 1)
To be precise, $orb\big(\frac{1}{2}\big)=\mathbb{Z}\big[\frac{1}{2}\big] \cap [0,1).$ 


Since you say $\mathbb{Z}$ rather than $\mathbb{N}$ you mean the orbit of $y$ to include the solutions $t$ of $f^k(t)=y$ for $k \in \mathbb{Z}.$


You say that the orbit of $\frac13$ is contained in $$ \Big\{\frac{m}{3\times 2^n}\mid n \in \mathbb{Z},\ 0 \leq m\lt 3\times2^n\text{ and }\gcd(m,3)=1\Big\}$$ and wish to show the reverse inclusion. Induction on $n$ goes pretty easily.


Any rational $r$ can be written uniquely as $r=2^eq$ where $e\in \mathbb{Z}$ and $q=\frac{2a+1}{2b+1}$ has odd numerator and denominator. To keep $r\in [0,1)$ there is an upper bound on $e$ but of course no lower bound.


