Let $f: X \rightarrow X$ be a homeomorphism of a compact metric space. If the orbit of $x$ is compact, then $x$ is periodic

Question

Let $f: X \rightarrow X$ be a homeomorphism of a compact, connected metric space. If the orbit of $x$ is compact, then $x$ is periodic.

I feel like should be trivial, but I cannot seem to work out a proof.

Since $f(Orb(x)) = Orb(x)$ and $Orb(x)$ is minimal, it follows that $Orb(x)$ is nowhere dense.

Answer 1

Assuming that the orbits are two-sided orbits consider the orbit $Y$ of $x$. This is a countable compact metric space and $f$ is a homeomorphism of $Y$ onto itself. Write $Y$ as the union of its singletons and apply BCT to conclude that some point in $Y$ is open. This implies that each point of $Y$ is open because $f$ is a homeomorphism. Compactness of $Y$ implies that $Y$ is a finite set. [Look at the open cover formed by singletons].

Answer 2

Kavi Rama Murthy's proof can be modified to work when the orbit $O=\{x_0,f(x_0),f(f(x_0)),\dots\}$ is one sided. As in their proof, some point $x$ in $O$ is open in $O$. This means there is an open $U\subset X$ for which $U\cap O=\{x\}$. This implies $f(x)$ is open as well, as $$ \{f(x)\}=f(U\cap O)=f(U) \cap f(O)=(f(U)\setminus \{x_0\})\cap O $$ Note $f(A\cap B)=f(A)\cap f(B)$ since $f$ is injective.

Supposing $x$ is not the initial point $x_0$, then

\begin{align} \{f^{-1}(x)\} &=f^{-1}(U\cap O) \\&=f^{-1}(U)\cap f^{-1}(O) \\&=f^{-1}(U)\cap (O\cup \{f^{-1}(x_0)\}) \\&=(f^{-1}(U)\cap O)\cup f^{-1}(U\cap \{x_0\}) \\&=f^{-1}(U)\cap O \end{align} The last equality follows since $U\cap O=\{x\}$, so $x_0\notin U$. This shows $f^{-1}(x)$ is open.

Therefore, the fact $x$ is open implies all points in the orbit are open, and you can conclude as in the other proof.