Finding a two point scrambled set for the function $g:[0,1] \rightarrow [0,1], x \mapsto \min_{n\in \mathbb{Z}} |3x-2n|$?

Question

Let $I=[0,1]$ be the unit interval and $g$ as defined below.

Then $x \neq y$ with $x,y \in I$ are called "two point scrambled set"=$\{x,y\}$, if

1. $\lim\inf_{n \rightarrow \infty} | g^{(n)}(x) - g^{(n)}(y)| =0$

and

2. $\lim\sup_{n \rightarrow \infty} | g^{(n)}(x) - g^{(n)}(y)| >0$

It is known, that if $g$ has a period $3$ point, then it is chaotic and must have a two point scramble set as defined above. Since I am new to this topic of chaos theory, I am asking myself:

Q: How do I find in this specific situation for $g:[0,1] \rightarrow [0,1]$ a two point scrambled set?

Background for the question:

Let $ x $ be an infinite binary string. Define the function $ f(x) $ mapping $ x $ to the Cantor set of $ I = [0,1] $ as: $$ f(x) = \sum_{n=0}^{\infty} \frac{2 x_n}{3^{n+1}} $$ where $ x_n $ are the bits of $ x $. Define the function $ g(x) $ over the interval $ I $ as: $$ g(x) = \min_{n \in \mathbb{Z}} |3x - 2n| $$ which simplifies to: $$ g(x) = \begin{cases} 3x & \text{if } 0 \leq x \leq \frac{1}{3} \\ 2-3x & \text{if } \frac{1}{3} \leq x \leq \frac{2}{3} \\ 3x-2 & \text{if } \frac{2}{3} \leq x \leq 1 \end{cases} $$ This function $g$ maps the Cantor set to itself and has the property that for each infinite bitstring $ x $: $$ g(f(x)) = f(s(x)) $$ where $ s(x) $ is the Bernoulli shift of the bitstring $ x $, defined as $ s(x) = (x_{n+1})_{n \in \mathbb{N}_0} $.

Using this observation, and denoting $ \text{CB}(q) $ the Collatz-Bits / Parity sequence bits of infinite length of the rational number $ q $, then: $$ g(f(\text{CB}(q))) = f(\text{CB}(T(q))) $$ where $ T\left(\frac{a}{b}\right) = \frac{3(a/b)+1}{2} $ if $2$ occurs in the prime decomposition of $ a \cdot b $, $ \gcd(a, b) = 1 $ and $$ T\left(\frac{a}{b}\right) = (\frac{a}{b}) / 2 $$ otherwise.

Using this last observation and induction on $ n $, it is possible to map iterations of $ T $ to iterations of $ g $: $$ g^n (f(\text{CB}(q))) = f(\text{CB}(T^n(q))) $$

The interesting part is that the function $ g: [0,1] \to [0,1] $ defined over the whole interval shows a chaotic behaviour as defined by Li-Yorke in 1975, because of the simple fact that it has a point of period three: $ \frac{1}{13} $ with trajectory $ \frac{1}{13}, \frac{3}{13}, \frac{9}{13}, \frac{1}{13} $. Here are more examples : (Table 1).

x	alpha_x	f(x)	g(f(x))
[0]	0	0	[0, 0]
[1]	-1	1	[1, 1]
[1, 0]	1	3/4	[3/4, 1/4, 3/4]
[0, 1]	2	1/4	[1/4, 3/4, 1/4]
[1, 0, 0]	1/5	9/13	[9/13, 1/13, 3/13, 9/13]
[0, 1, 0]	2/5	3/13	[3/13, 9/13, 1/13, 3/13]
[0, 0, 1]	4/5	1/13	[1/13, 3/13, 9/13, 1/13]
[0, 1, 1]	-10	4/13	[4/13, 12/13, 10/13, 4/13]
[1, 1, 0]	-5	12/13	[12/13, 10/13, 4/13, 12/13]
[1, 0, 1]	-7	10/13	[10/13, 4/13, 12/13, 10/13]

The Collatz conjecture might be formulated as:

The natural number $n$ with infinte bitstring $x:=CB(n)$ has an eventually periodic sequence $g^{k}(f(x))$ which ends in a period $1/4,3/4$.

The Theorem of Li-Yorke (1975) states that in some specified sense:

If the function has a period three point, then it is chaotic.

This could maybe describe the observed "chaotic" behaviour of the Collatz $ T $ map as defined above.

For each finite $x$, let $\bar{x}$ denote the infinite bitstring with pure period $x$. We define $\alpha_x$, which is a 2-adic number, by the following expression:

$$ \alpha_x := -\sum_{n=0}^{\infty} \frac{x_n}{3^{x_0 + x_1 + \cdots + x_n}} \cdot 2^n $$

This formulation is borrowed from the discussion in Lagarias book where it is defined for finite strings. Here, we extend the concept using the 2-adic valuation for infinite bitstrings. Additionally one can use this function to prove that:

If $z = x.\bar{y}$ is an eventually periodic bitstring with period $y$, with lengths $r=|x|,s=|y|$, then

$$-\alpha_z = \frac{\rho_r(\alpha_x)}{\lambda_r(\alpha_x)}-\frac{\alpha_y}{\lambda_r(\alpha_x)}$$

and so, after some algebraic manipulations and since $\lambda_r(\alpha_x) = \lambda_r(\alpha_z),\rho_r(\alpha_x) = \rho_r(\alpha_z),$ :

$$\alpha_y = \alpha_z \lambda_r(\alpha_z) + \rho_r(\alpha_z) = T^{(r)}(\alpha_z)$$

where the last equality follows similar to the equation in Lagarias book, and $\lambda, \rho$ are copied as definitions from Lagarias $3x+1$ book.

But $y$ is periodic with period length $s$ so:

$$T^{(s)}(\alpha_y) = \alpha_y = \cdots = T^{(r)}(\alpha_z)$$

which proves that also in this case $\alpha_z$ must be eventually periodic.

The transformation $T$ on 2-adic numbers can be simply defined as:

$$ T(y) = \begin{cases} \frac{3y+1}{2} & \text{if } y_0 = (y \mod 2) = 1, \\ \frac{y}{2} & \text{otherwise}. \end{cases} $$

Here, $y_0$ is the 0-th bit of $y$. Thus, if $y_0 = 1$, then the transformation is $\frac{3y+1}{2}$; otherwise, it is $\frac{y}{2}$. This definition aligns with both the rational numbers' definition and the usual treatment for natural numbers, demonstrating its consistency across different numeric systems.

Answer 1

Having digested your definitions... Any number which has an aperiodic orbit (not a fixed point), and any number which has a periodic orbit (possibly a fixed point), will between them constitute a two-point scrambled set.

Let me elaborate first of all in a different domain, namely the domain of two points orbiting a unit circle:

Let $x_n=nq$ where $q\in\Bbb Q$

Then any rational orbit on the unit circle of the form $e^{x_n\pi i}$ will be periodic.

Now let $y_n=y_{n-1}+z$ where $z$ is an irrational real number.

Then any irrational orbit on the unit circle of the form $e^{y_n\pi i}$ will not be periodic, and it will come both arbitrarily close to the above rational orbit and arbitrarily close to the point on the circle exactly opposite the rational orbit, infinitely many times.

My interpretation of your definition is that these two orbits constitute a two-point scrambled set.

My claim is that in your above function a periodic orbit and an aperiodic orbit will constitute a two point scrambled set. My argument is that by Lagarias and Bernstein 2019, your map above is topologically conjugate to the bit shift map on the real unit interval, with periodic orbits corresponding to periodic orbits on the unit circle, and aperiodic orbits corresponding with irrational orbits on the unit circle.

It is a fairly obvious theorem of Lagarias and Bernstein 2019 that every irrational number has an aperiodic Collatz orbit therefore you can choose the orbit of any irrational number along with the orbit of the number $1$ or the fixed point $-1\cong\overline1_2$ as your two point scrambled set.

It is almost certainly the case that Lagarias' periodicity conjecture is true, in which case every rational number will have a periodic Collatz orbit, in which case you can pick any rational, irrational pair for your two-point scrambled set.