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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