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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?

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    $\begingroup$ $\mathbb Z[\frac 12]$ means the dyadic rationals (all rationals with denominator a power of 2). This is (presumably) not what you mean (but rather $\frac 12\mathbb Z$). $\endgroup$
    – Anthony Quas
    Commented Mar 2, 2018 at 17:24
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    $\begingroup$ @AnthonyQuas No. It's with $m\in\mathbf{Z}$. So it includes all iterated inverse images, that is, $\mathbf{Z}[1/2]/\mathbf{Z}$ (or intersection with $[0,1[$ for those who prefer fundamental domains). $\endgroup$
    – YCor
    Commented Mar 19, 2018 at 15:35
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    $\begingroup$ Anyway it's a bit weird to melt in the definition both the iterated inverse images (backwards orbit) and the iterated images (forward orbit). The orbit should then include all the inverse images of the forward images? For $1/2$ it does not matter but for $1/3$ one gets $1/3\to 2/3\leftarrow 5/6$, and $5/6$ is neither in the forwards nor backwards orbit of $1/3$. $\endgroup$
    – YCor
    Commented Mar 19, 2018 at 15:40

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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.

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  • $\begingroup$ Yes, it should be $[0,1)$, my mistake. Thanks $\endgroup$
    – Miguel Barata
    Commented Mar 2, 2018 at 18:56

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