Topological dynamics on the interval — Sharkovskii's order

Parent: orbit · Prereq: discrete-dynamical-systems

Setting: $I \subseteq \mathbb{R}$ an interval, $f: I \to I$ continuous (nothing more — no smoothness, no measure). The astonishing fact: mere continuity on a one-dimensional space already forces a rigid arithmetic of periods.

Sharkovskii's ordering

Order the positive integers as follows (⊳ = "precedes"):

3 \rhd 5 \rhd 7 \rhd 9 \rhd \cdots \ \rhd\ 2\cdot 3 \rhd 2 \cdot 5 \rhd \cdots \ \rhd\ 4\cdot 3 \rhd 4 \cdot 5 \rhd \cdots \ \rhd \cdots \rhd\ 2^3 \rhd 2^2 \rhd 2 \rhd 1.

Odd numbers first, then $2\times$ odds, then $4\times$ odds, …, and at the very end the powers of two, descending.

Sharkovskii (1964)

If a continuous $f: I \to I$ has a periodic point of period $m$ , it has periodic points of every period $n$ with $m \rhd n$ .

So period 3 forces all periods (3 is ⊳-maximal), and a map with only finitely many periods can only have periods that are powers of two. The ordering is sharp: for every $m$ there is a map whose set of periods is exactly $\{n : m \rhd n\} \cup \{m\}$ .

Li–Yorke (1975), "period three implies chaos"

If $f$ has a period-3 point, there is an uncountable scrambled set $S \subseteq I$ : for all $x \neq y$ in $S$ ,
$\limsup_{n\to\infty} |f^n(x) - f^n(y)| > 0 \quad\text{and}\quad \liminf_{n\to\infty} |f^n(x) - f^n(y)| = 0.$
Orbits of scrambled pairs approach each other arbitrarily closely, forever, without ever staying together.

The proof engine behind both: if intervals $J_0, J_1$ satisfy $f(J_0) \supseteq J_0 \cup J_1$ and $f(J_1) \supseteq J_0$ (a covering relation, which a period-3 orbit provides), then every infinite itinerary through the covering graph is realized by some point — an orbit for every symbol sequence. That is symbolic-dynamics entering through the back door.

Count along: a genuine period-3 orbit

Tent map $T(x) = 2\min(x, 1-x)$ on $[0,1]$ . Take $x = \tfrac{2}{7}$ :

T\!\left(\tfrac{2}{7}\right) = \tfrac{4}{7}, \qquad T\!\left(\tfrac{4}{7}\right) = 2\left(1 - \tfrac{4}{7}\right) = \tfrac{6}{7}, \qquad T\!\left(\tfrac{6}{7}\right) = 2\left(1 - \tfrac{6}{7}\right) = \tfrac{2}{7}. ✓

A 3-cycle — so by Sharkovskii the tent map has periodic orbits of every period, and by Li–Yorke it is chaotic.

The one-dimensionality is essential

On the circle, the rotation by $1/3$ has every orbit of period exactly 3 and no other periods at all. Sharkovskii's order is a theorem about the interval's topology (its proof lives on the intermediate value theorem), not about dynamics in general.