Let $f: I \rightarrow I$ be a continuous one-dimensional map of an interval $I \subset \mathbb{R}$. Explain what is meant by the statements (i) that $f$ has a horseshoe and (ii) that $f$ is chaotic (according to Glendinning's definition).

Assume that $f$ has a 3-cycle $\left\{x_{0}, x_{1}, x_{2}\right\}$ with $x_{1}=f\left(x_{0}\right), x_{2}=f\left(x_{1}\right), x_{0}=f\left(x_{2}\right)$ and, without loss of generality, $x_{0}<x_{1}<x_{2}$. Prove that $f^{2}$ has a horseshoe. [You may assume the intermediate value theorem.]

Represent the effect of $f$ on the intervals $I_{a}=\left[x_{0}, x_{1}\right]$ and $I_{b}=\left[x_{1}, x_{2}\right]$ by means of a directed graph, explaining carefully how the graph is constructed. Explain what feature of the graph implies the existence of a 3-cycle.

The map $g: I \rightarrow I$ has a 5-cycle $\left\{x_{0}, x_{1}, x_{2}, x_{3}, x_{4}\right\}$ with $x_{i+1}=g\left(x_{i}\right), 0 \leqslant i \leqslant 3$ and $x_{0}=g\left(x_{4}\right)$, and $x_{0}<x_{1}<x_{2}<x_{3}<x_{4}$. For which $n, 1 \leqslant n \leqslant 4$, is an $n$-cycle of $g$ guaranteed to exist? Is $g$ guaranteed to be chaotic? Is $g$ guaranteed to have a horseshoe? Justify your answers. [You may use a suitable directed graph as part of your arguments.]

How do your answers to the above change if instead $x_{4}<x_{2}<x_{1}<x_{3}<x_{0}$ ?