Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 7E (Dynamical Systems)

Choose binary expansions $x=0.b_1{b}_2\dots$ that do not end in an infinite string of ones. The doubling map deletes the first digit: $F(x)=0.b_2{b}_3\dots$. Given $x$ and any neighbourhood of it, choose $N$ sufficiently large that its binary cylinder of length $N$ is small enough, and choose a point $y$ sharing that prefix but with a prescribed tail. After $N$ shifts, the tail of $y$ may be chosen arbitrarily in $[0,1)$. Choose it close to zero or close to one, whichever is at distance more than $1/3$ from $F^N(x)$. Then $|x-y|\le 2^{-N}$ but $|F^N(x)-F^N(y)|>1/3$. Hence the map has sensitive dependence on initial conditions, with a uniform separation constant. At dyadic endpoints the cylinder sharing the chosen terminating expansion still contains points arbitrarily close to the endpoint.

Every nonempty relatively open interval contains a binary cylinder whose points share a finite prefix. Given nonempty open sets $U,V$, choose such a cylinder $C\subset U$ of length $N$. After $N$ binary shifts, $F^N(C)$ contains $[0,1)$ with the usual half-open cylinder convention. In particular, some point of $C$ maps into $V$. Thus $F$ is topologically transitive: $F^N(U)\cap V\ne \varnothing$. Equivalently, concatenate a prefix placing the point in $U$ with a tail placing its $N$th iterate in $V$.

Given a binary cylinder with finite prefix $b_1\dots b_N$, repeat a finite block beginning with that prefix indefinitely. The resulting binary sequence is periodic under the shift and represents a periodic point of an interval map of $F$ in the cylinder. If the given prefix consists entirely of ones, append a zero to the block so that its value is less than one; this still preserves the prefix. Since cylinders can be taken inside any nonempty open interval, periodic points are dense.

A point fixed by $F^4$ satisfies $(2^4-1)x\in \mathbb{Z}$, so $x=j/15$ for $0\le j\le 14$. The points with period dividing two are $0,1/3,2/3$, corresponding to $j=0,5,10$. All other twelve points have least period four. Multiplication by two modulo fifteen groups them into the three cyclesThis exhausts the requested four-cycles. The displayed order is the dynamical order, and cyclic changes of the starting point describe the same cycle.