Dynamical systems with disjoint $\omega$-limits of single points For $X$ compact metric spaces and $f:X\to X$ continuous, is there a nice characterization of the systems $(X,f)$ for which, for every pair of points $x,y\in X$ with disjoint orbits, we have $\omega(x)\cap\omega(y)=\emptyset$?
 A: This seems to be true if and only if every $x \in X$ is eventually periodic.
The reverse direction is obvious; if all points of $X$ are eventually periodic and $x, y$ have disjoint orbits, then each has $\omega$-limit set equal to a different periodic orbit, which must be disjoint.
For the forward direction, assume your condition. For every $x \in X$, $\omega(x)$ is a closed $f$-invariant set, and therefore contains a minimal subsystem $M$. If the orbit of $x$ does not eventually enter $M$, you already contradict your condition; for arbitrary $m \in M$, $\omega(m) = M$. Then $x$ and $m$ are in disjoint orbits, but $\omega(x) \cap \omega(m) = M \neq \varnothing$.
If $M$ does not consist of a single orbit, then it contains $x \neq y$ in different orbits, but by minimality, $\omega(x) = \omega(y) = M$, again contradicting your condition. So $M$ consists of a single orbit, say $O(m)$. If $m$ is not periodic, then $M$ is a countable closed set in a complete metric space, therefore it has an isolated point $f^n m$. But then $O(f^{n+1} m)$ is not dense in $M$, contradiction to minimality.
So, every minimal subsystem $M$ is a periodic orbit, and every $x \in X$ has orbit eventually in one of these periodic orbits, therefore every $x$ is eventually periodic.
Throughout I've not assumed invertibility; if $(X, f)$ were invertible, then you remove the word "eventually."
