Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 32A / b / i / Solution

Write the period-three points in increasing order and take the two intervals between consecutive points. For either possible cyclic ordering, the interval-covering transition graph is, after interchanging its vertices,The loop at the first vertex gives a fixed point. For every $n\ge 2$, the closed itineraryhas least period $n$. The interval-covering periodic-orbit theorem supplies a point with that itinerary and therefore a cycle of least period $n$. Thus $F$ has periodic orbits of every positive period. This is the period-three case of the Sharkovsky theorem.

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