Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 14B / Solution

Let the three orbit points be $a<b<c$. Up to reflection of the interval, their images are $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b]$ and $J=[b,c]$. Continuity and the intermediate value theorem imply the covering relationsWe first prove the needed interval-covering principle using the permitted pullback fact. If a cyclic list of closed intervals $I_0,\dots ,I_{n-1},I_n=I_0$ obeys $F(I_j)\supseteq I_{j+1}$, pull back successively from the end to obtain a closed interval $K\subseteq I_0$ with $F^n(K)=I_0$ and $F^j(K)\subseteq I_j$. Since $F^n(K)\supseteq K$, the intermediate value theorem gives a fixed point of $F^n$ in $K$: choose preimages of the two endpoints of $K$, which give opposite weak signs of $F^n(x)-x$, irrespective of their order. This point realizes the cyclic itinerary.

For $n\ge 2$, use the word $I{J}^{n-1}$. It obeys the covering relations and therefore yields a point with period dividing $n$. A point away from the shared endpoint $b$ has an unambiguous itinerary, and this word has exactly one $I$ per cycle; it cannot be a repetition of a shorter word. Its least period is consequently $n$. If an orbit touches $b$, it is the original three-cycle. That realizes the case $n=3$ itself. It cannot realize any of the other words: for $n$ not divisible by three it is not fixed by $F^n$, and for a multiple of three larger than three it visits $a\in I\setminus J$ more than once, contradicting the itinerary. Thus the construction works for every $n\ge 2$. For $n=1$, $F(J)\supseteq J$ and the same fixed-point argument gives a fixed point. This proves period three implies all periods: there is an orbit of every positive least period.

For the two five-cycles forced by a three-cycle, use the two cyclic words $IJJJJ$ and $IJIJJ$. Both are admissible. Their points cannot be endpoints from the three-cycle, and because five is prime and the words use both intervals, each has least period five. Their itineraries have respectively one and two visits to the interior of $I$; cyclically shifting the starting point cannot change this count. Hence they give at least two distinct period-five orbits.