# Non wandering sets and limit sets

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?

• If $x\in NW$, then its orbit crosses any its neighbourhood infinitely often; so $x$ is its own $\omega$-limit point – Ilya Bogdanov Oct 20 '19 at 5:44

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 $$NW(\sigma)$$.