Closed orbits are finite

 Let f:[a,b]\to [a,b] be continuous function and x_0\in [a,b]. Consider the sequence X=\{x_n\}_{n\ge 0} defined by x_n=f(x_{n-1}) for n\ge 1. Assume that X is a closed set. Prove that it consists of finitely many elements. (The problem is taken from IMC 2002. The set X is called the orbit of f starting from x_0)


\textbf{Proof.} Let us assume that X is infinite. So it has a limit point, which should belong to the set (since it is closed). Let n_0:=\min\{n\ | \ x_n \text{ is a limit point of } X \}.

Then define \hat X=\{x_n\}_{n\ge n_0+1}. Thus \hat X\cup \{x_{n_0}\} is compact and f:\hat X\cup\{x_{n_0}\}\to \hat X is a continuous function. Since the image of compact sets under continuous functions are compact set, we must have that \hat X is compact as well. But it is clearly not closed, since x_{n_0} is a limit point \hat X and does not belong to \hat X.

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