Let $X$ be a compact metric space and $f:X\to X$ is a homeomorphism. A point $x$ is aid to be nonwandering if for any open set $U$ containing $x$, there is an $N>0$ such that $f^N(U)\cap U\neq \emptyset$. Donote by $NW(f)$ the set of all nonwandering points of $f$. A point in $X$ is called a $\omega$-limit point for $x\in X$ if $y$ is a limit point of the forward orbit of $x$. It is well known that the closure of the set of all $\omega$-limit points for some $x$ in $X$, denoted by $\omega(f)$, is a subset of $NW(f)$. So my question is it true that there exists some example such that this inclusion can be proper? 



Non wandering sets and limit sets
Consider the shift space $X \subset \{0,1\}^{\mathbb{Z}}$ obtained by forbidding the words $1 0^m 1^n 0$ for all $m, n \geq 1$, and denote the shift map on $X$ by $\sigma$.
Since a point of $X$ can contain at most three transitions from $0$ to $1$ or back, the only $\omega$-limit points of $X$ are the two uniform points (all-$0$ and all-$1$).
However, $x = \ldots 0 0 0 1 1 1 \ldots$ is nonwandering because arbitrarily close to it we find the point $\ldots 0 0 0 1^n 0^n 1 1 1 \ldots \in X$ for some large $n \geq 0$ that returns close to $x$ after $2 n$ shifts.
Hence $\omega(\sigma)$ is properly contained in $NW(\sigma)$.


