Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 7D / Solution

Use binary expansions ending in zeros at dyadic rationals. The doubling map shifts the expansion left:In ordinary coordinates it is $2x$ on $[0,1/2)$ and $2x-1$ on $[1/2,1)$. The branches are continuous; the left limit at $1/2$ is $1$ while its value and right limit are $0$. This is its only discontinuity in the half-open interval.

For every $N\ge 2$, the repeating binary block $0^{N-1}1$ givesIts least period is $N$: a shorter repeating block would make more than one digit equal to one in a length-$N$ period. The orbit avoids the discontinuity, and its multiplier is $(F^N{)}^{\prime }=2^N>1$, so these periodic orbits are repelling and unstable. The same multiplier argument applies to every periodic orbit, with the endpoint fixed point treated one-sidedly.

In the two-branch horseshoe formulation of Glendinning chaos, disjoint subintervals are each mapped onto the same interval by an iterate. Here $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$ already satisfy $F(K_0)=F(K_1)=J$. Their inverse branches $x\mapsto x/2$ and $x\mapsto (x+1)/2$ realize every binary itinerary. This is a horseshoe for a piecewise continuous map; on the circle the endpoint jump disappears. The symbolic explanation also shows dense periodic points, by repeating any prescribed finite prefix, and topological transitivity, because an iterate maps any sufficiently small dyadic cylinder onto $(0,1)$. Finally every neighborhood contains points sharing an arbitrarily long prefix but with different subsequent digits; an appropriate iterate separates their images by a fixed amount, for instance more than $1/4$. This proves sensitive dependence on initial conditions as well. The binary shift has full two-symbol chaotic dynamics.