Non wandering Set

Question

Let be $T:X\to X$ a topological dynamical system, some definitions:

Omega limit set: $\omega(x)=\{y: \exists (n_j)~~\text{such that}~ T^{n_j}(y)\to x \}$

Recurrent set: $\mathcal{R}(T)=\{x\in X; x\in \omega(x)\}$

Non Wandering set: $ NW(T)=\{x; \text{for all open }~U\ni x; \text{exists} ~N~\text{such that}~T^N(U)\cap U\neq\phi\} $

It's clearly that: $\mathcal{R}(T)\subset NW(T)$

I am looking for an example showing that in general,

$\mathcal{R}(T)\neq NW(T)$ in other words, I'm looking for an example of a dynamical system and a point $x$ sutch that $x$ is non wandering, but it is not recurrent.

Answer 1

Let $X$ be the torus, thought of as ${\bf R}^2$ modulo the integer lattice, let $T(u)=Au$ where $A=\pmatrix{2&1\cr1&1\cr}$. $T$ has an eigenvalue $(3-\sqrt5)/2$ with modulus less than $1$ so any eigenvector $x$ for that eigenvalue has $T^n(x)\to0$ and is non-recurrent. But the points with rational coordinates are periodic points for $T$, and they are dense in $X$, so $x$ is non-wandering.

EDIT: A simpler example is the tent map. $X$ is the closed interval $[0,1]$, $T(u)=\min(2u,2-2u)$. Dyadic rationals $x$ (rationals with denominator a power of $2$) are non-recurrent since for any such $x$ there an $n$ such that $T^n(x)=0$. Rationals with odd denominator are periodic, and they are dense in $X$, so all $x$ are non-wandering.

Answer 2

I think you can also get an answer without calculations by considering an irrational rotatation, $x \rightarrow x+a$, with $a$ irrational. This map has no recurrent points, but every point is in the non-wandering set. To get a map on the torus$=S^1 \times S^1$, just take the product of a irrational rotation times the identity. So you get a function like $(x,y)=(x+a,y)$, where $x$ and $y$ are elements of $R\backslash Z$