Prove $\langle x_0\rangle$ has only finitely many elements if and only if there exists $k_1$ and $k_2$ with $k_1 < k_2$ so that $x_{k_1} = x_{k_2}$

Question

Prove that the orbit $\langle x_0\rangle$ has only finitely many (distinct) elements if and only if there exists $k_1$ and $k_2$ with $k_1 < k_2$ so that $x_{k_1} = x_{k_2}$

I know this to be true but how does one go about proving this?

Answer 1

If the orbit has finitely many elements, there exists $n<m$ with $x_n=x_m$ (otherwise there would exists infinitely many elements).

If there exists $n<m$ with $x_n=x_m$, then $x_{n+k}=x_{m+k}$ for all $k\ge0$, which means that the orbit keeps repeating the elements that were already there. In other words, the orbit is (possibly with repetitions, since $n$ may not be minimal) $x_0,\ldots,x_{m-n},x_0,\ldots,x_{m-n},\ldots$.