In this question there is an example of a continuous dynamical system with $NW(f) \not\subset \overline{R(f)}$. The definitions I am working with are exactly same as that question. I want to find a symbolic or a discrete dynamical system with the mentioned property. I would really appreciate a not-so-involved example, so that I can gain an intuition to distinguish non-wandering points from points in the closure of the set of all recurrent points.
Example of a symbolic or a discrete dynamical system where $NW(f) \not\subset \overline{R(f)}$
dynamical systems
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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.
In this case the set $Per(f)$ is closed because we know the set of all fixed points is closed.
Yes (when $X$ is Hausdorff).
Examples such that:
1)) $F(f) \subset \overline{Per(f)}$. Let $X$ be a set $\{1,2,3\}$ endowed with the discrete topology and $f(1)=2$, $f(2)=3$, and $f(3)=1$. Then $F(f)=\varnothing$ but $Per(f)=X$.
2)) $\Omega(f) \nsubseteq\overline{Per(f)} \subset L(f)$. Let $X$ be the unit circle $\{z\in\Bbb C: |z|=1\}$ endowed with the topology inherited from $\Bbb C$, $\varphi$ be a real number such that $\varphi/\pi$ is irrational and $f:X\to X$, $x\mapsto xe^{\varphi i}$ be a rotation at the angle $\varphi$. It is easy to check that $Per(f)=\varnothing$ and $L(f)=\Omega(f)=X$.
3)) $L(f) \subset \Omega(f)$. I assume that $(n_i)$ in the definitions of $\alpha(x)$ and $\omega(x)$ is a strictly increasing sequence of natural numbers.
Let $X={\Bbb T}^{\Bbb T}$. By Tychonov Theorem, $X$ is a compact topological group with coordinate-wise multiplication. Let $g=(g_z)_{z\in\Bbb T}\in X$ be an element such that $g_z=z$ for each $z\in\Bbb T$. For each $x\in X$ put $f(x)=gx$.
We claim that $\Omega(f)=X$. Indeed, let $x$ be any element of $X$, $U$ be any neighborhood of $x$, and $N$ be any natural number. Since $X$ is a topological group, there exists a neighborhood $V$ of the identity of $G$ such that $xVV{-1}\subset U$. A family $\{yV:y\in X\}$ is an open cover of a compact space $X$, so there exists a finite subset $F$ of $X$ such that $FV=X$. Therefore there exist an element $y\in F$ and natural numbers $n,m$ such that $g^n,g^m\in yV$ and $n>m+N$. Then $g^m\in g^nVV^{-1}$ and so $x\in x g^{n-m}VV^{-1}\subseteq g^{n-m}U$.
We claim that sets $\alpha(x)$ and $\omega(x)$ are empty for each $x=(x_z)_{z\in\Bbb T}\in X$. Since the group $G$ is Abelian, $\omega(x)=\alpha(x^{-1})^{-1}$, it suffices to show that the set $\alpha(x)$ is empty.
Suppose to the contrary that there exists an increasing sequence $\{n_i\}$ of natural numbers such that a sequence $\{g^{n_i}x\}$ converges to a point $y=(y_z)_{z\in\Bbb T}\in X$. Let $U_0=\{z\in\Bbb T: |z-1|\le 1/\sqrt{2} \}$ be a neighborhood of the identity of the group $\Bbb T$. For each natural number $i$ put $$T_i=\{z\in\Bbb T: (n_j-n_k)z\in U_0^2\mbox{ for each }j,k>i\}.$$ Since the set $ U_0^2$ is closed in $\Bbb T$, the continuity of power on the group $\Bbb T$ implies that the set $T_i$ is closed for each natural number $i$. We claim that $\Bbb T=\bigcup_{i\in\Bbb N} T_i$. Indeed, let $z\in \Bbb T $ be any element. There exists a natural number $i$ such that $n_jzx_z\in y_zU_0$ for each $j>i_z$. Then $(n_i-n_k)z\in U_0^2$ for each $j,k>i$, so $z\in T_i$.
Since $\Bbb T=\bigcup_{i\in\Bbb N} T_i$ and $\Bbb T$ is a compact metrizable space, Baire theorem implies that there exists a natural number $i$ such that a set $T_i$ has non-empty interior. Therefore there exists an open arc $U\subseteq T_i$ of the circle $\Bbb T$. Let $\ell$ be the length of $U$. Since the sequence $\{n_i\}$ is increasing, there exists a number $j>i$ such that $n_j-n_{i+1}>2\pi/\ell$. But then $U_0^2\supseteq (n_j-n_{i+1})T_i\supseteq (n_j-n_{i+1}) U=\Bbb T$, a contradiction.
4)) $Per(f)$ is not closed. Let $X$ be the unit disk $\{z\in\Bbb C: |z|\le 1\}$ and $f:X\to X$, $x\mapsto xe^{|x|i}$. Then $Per(f)=\{x\in X: |x|/\pi\in\Bbb Q\}.$
Best Answer
One example is the orbit closure of ${}^\infty 0 1 0 1 0 0 1 0 0 0 1 \cdots 1 0^n 1 0^{n+1} 1 \cdots$ with respect to the shift map. The left tail is just repeating 0s, and on the right are 1s with increasing gaps. The only recurrent point is the all-0 configuration, but ${}^\infty 0 1 0^\infty$ is nonwandering.