Non-wandering set

Remarks

So $x$ is non-wandering exactly when it is not a wandering point in the sense of wandering sets. The set $\Omega(f)$ is closed, and $f(\Omega(f)) \subseteq \Omega(f)$. If $X$ is compact and nonempty, then $\Omega(f)$ is nonempty. Every recurrent point, and so every point on a periodic orbit, is non-wandering. A non-wandering point need not be recurrent. Example: for $f(x) = x/2$ on $\R$, the point $0$ is fixed, so $0 \in f^n(U) \cap U$ for every neighbourhood $U$ of $0$. Non-example: for the same map, a point $x > 0$ has the neighbourhood $U = (2x/3, 4x/3)$, and $f^n(U) \subseteq (0, 2x/3]$ for $n \ge 1$, so $f^n(U) \cap U = \emptyset$. In fact $\Omega(f) = \set{0}$ for this map.