Let $X$ be a set and $f:X\to X$ a map. An element $x\in X$ is said to be preperiodic if $\exists n>m\geq 0$ such that $f^{n+m}(x)=f^m(x)$, where $f^k(x):=\underbrace{f\circ\cdots\circ f}_{\text{$k$-times}}$ denotes the composition with $k\in\mathbb{N}$.
Moreover, the orbit of $x$ under $f$ is given by $\mathcal{O}_f(x):=\{f^k(x):k\geq 0\}$.
My question is the following: Is it true that $x\in X$ is a preperiodic point of $f$ if and only if the orbit of $x$ under $f$ is finite? That is,
$$\text{$x\in X$ preperiodic point of $f$}\iff \#\mathcal{O}_f(x)<\infty.$$
The statement is true. Suppose $x$ is preperiodic, that is, there are some $n>m\geq 0$ such that $f^{n+m}(x)= f^m(x)$, then it's easy to see that $$ \mathcal{O}_f(x)= \{x, f(x), f^2(x), \ldots, f^{m}(x), \ldots, f^{n+m-1}(x)\}, $$ is finite.
Conversely, if $\mathcal{O}_f(x)$ is finite, there must be some $n\in\mathbb{N}$ such that $x, f(x), \ldots, f^{n}(x)$ are distinct, but $f^{n+1}(x)= f^m(x)$ for some $m\leq n$, otherwise it would have been infinite. This shows that $x$ is preperiodic.