How to determine $\Omega(T)$?

Let $X=\left\{0,1,2\right\}^{\mathbb{Z}}$ and let $T\colon X\to X$ describe the following dynamics:

  • 1 becomes a 2,
  • 2 becomes a 0 and
  • 0 becomes a 1 if at least one of its two neighbours is 1, otherwise it remains 0.

Now define the so-called non-wandering set as follows.

For a map $f\colon X\to X$, a point $p$ is called non-wandering provided for every neighborhood $U$ of $p$ there is an integer $n>0$ such that $f^n(U)\cap U\neq\emptyset$. Thus, there is a point $q\in U$ with $f^n(q)\in U$. The set of all non-wandering points for $f$ is called the non-wandering set and is denoted by $\Omega(f)$.

Question: What is the non-wandering set $\Omega(T)$ for $T$?

Edit
Please see the definition of $L$ and $R$ below. I think it is not that difficult to show that $L\cup R\subseteq\Omega(T)$. At least, I think I managed it to show that. But obviously, there are more non-wandering points than the points that are in L or R, since

$$
c=…210210210210210210120120120120120…
$$

surely is non-wandering. Thus, there must be at least one more set $C$ with $c\in C$ so that
$$
\Omega(T)=L\cup R\cup C.
$$
Maybe the set $C$ can be determined.


Background

The background of this is the following: There is a theorem saying that in order to compute the topological entropy of $T$, it suffices to compute the topological entropy of $T$, restricted to the non-wandering set. I’ve already computed the topological entropy of $T$ (s. below), and now I would like to verify the theorem. Thus, I need to know what the non-wandering set looks like.

Consider the set $L$ consisting of those $x\in X$ that have the following form:

  • Every 1 has a 2 to its right,
  • every 2 has a 0 to its right,
  • every 0 has a 0 or a 1 to its right.

On $L$, the map $T$ acts a left-shift.

Similarly, let $R$ contain those $x\in X$ where

  • Every 1 has a 2 to its left,
  • every 2 has a 0 to its left,
  • every 0 has a 0 or a 1 to its left.

On $R$, the map $T$ acts as the right-shift.

The computation of the topological entropy gave $2\ln\rho$, where $\rho$ is the positive root of $x^3-x^3-1$.

Solutions Collecting From Web of "How to determine $\Omega(T)$?"

Here is a nearly complete answer: All points in $R$ and $L$ are non-wandering and all points with a sink (see definition below) are also. For example, an element with a sink is
$$
…21021021021021021\ 0\ 12012012012012012…
$$
We sketch the proof of the fact that these are all non-wandering points.

First let us look at some neighborhoods of points:
For $k>n$ let the projection $P_{n,k}:X=\{0,1,2\}^{\Bbb{Z}}\to \{0,1,2\}^{k-n+1}$ be given by
$$
P_{n,k}(Y)=(y_{n},y_{n+1},\dots,y_{k-1},y_k)\quad\text{for}\quad Y=(y_n)_{n\in\Bbb{Z}}\in X.
$$
Then for each $n>0$, the set
$$
V_{n,k}(Y):=\{Z\in X\ :\ P_{n,k}(Z)=P_{n,k}(Y)\}
$$
is an open neighborhood of $Y$. We also set $P_n:=P_{-n,n}$ and define correspondingly $V_n$. Moreover, any neighborhood $V(Y)$ of $Y$ contains $V_n(Y)$, for $n$ big enough.

Now we see a consequence for non-wandering sets in
$$
L=\{ \text{Every 1 has a 2 to its right,
every 2 has a 0 to its right,}
$$
$$
\text{every 0 has a 0 or a 1 to its right.}\}
$$ (or by symmetry in $R$).
Let $Y$ be a point in $L$ and let $V(Y)$ be an open neighborhood of $Y$. Then for some $n>0$ with $V_n(Y)\subset V(Y)$, we are searching for $k>0$ such that $V_n(Y)\cap f^k(V_n)\ne \emptyset$: Since $f$ is the shift to the left on $L$, it suffices to find an element $Z\in L$, such that $z_{n+k}=z_n=y_n, z_{n+k-1}=z_{n-1}=y_{n-1},\dots,z_{-n+k}=z_{-n}=y_{-n}$, or in other words, such that
$P_{-n+k,n+k}(Z)=P_n(Z)=P_n(Y)$. Then
$Z\in V_n(Y)\cap f^k(V_n)$. Clearly we can find such a $k$ and $Z$ for $k>3n+3$.

So we have determined that all points in $L$ are non-wandering (and by symmetry also the points in $R$).

Now we define a familiy of elements of $X$, the elements with a (one or two point) sink.

Definition: An element of $X$ is called an element with a one-point sink at $n_0$, if

  1. The entries at $n_0-1$, $n_0$ and $n_0+1$ are one of the following triples:
    $$
    212,\quad 020,\quad 101\quad\text{or}\quad 000.
    $$
    In the last case, there exists $k>1$ such that $y_{n_0+j}=0$ for $-k<j<k$ and $y_{n_0-k}=y_{n_0+k}=1$.

  2. To the right, we have a sequence in $L$.

  3. To the left, we have a sequence in $R$.

An element $Y$ of $X$ is called an element with a two-point sink at $n_0-1,n_0$, if

  1. The entries at $n_0-2$, $n_0-1$, $n_0$ and $n_0+1$ are one of the following cuadruples:
    $$
    2112,\quad 0220,\quad 1001\quad\text{or}\quad 0000.
    $$
    In the last case, there exists $k>1$ such that $y_{n_0+j}=0$ for $-k-1<j<k$ and $y_{n_0-k-1}=y_{n_0+k}=1$.

  2. To the right, we have a sequence in $L$.

  3. To the left, we have a sequence in $R$.

Remark: After $020$ and $0220$ the sink can change its place, and it can become two point or one-point sink regardless if it was a one or two point sink. It also can become zero or an element of $R$ or $L$.
For example, let $Y$ be an element with an sink at $0$ with
$$
P_{15}(Y)=021\ 021\ 021\ 021\ 021\ 0\ 120\ 000\ 001\ 201\ 201\ .
$$
Then $f^5(Y)$ has a sink at $2,3$ and
$$
P_{10}(f^5(Y))=0\ 210\ 210\ 210\ 2\ 100\ 120\ 120\ 1.
$$

Now let us prove that points that have a sink are nonwandering.

We shall assume that the sink is situated at
$0$ or at $0,1$(by translation we can do that). Then the positive part is $P_+(Y_+)$ of an element $Y_+$ of $L$ and the negative part
is $P_-(Y_-)$ of an element $Y_-$ of $R$, where $P_+=(y_0,y_1,\dots)$ and $P_-(Y)=(\dots,y_{-2},y_{-1})$. Let $V(Y)$ be an open neighborhood of $Y$. Then for some $n>0$ with $V_n(Y)\subset V(Y)$, we are searching for $k>0$ such that $V_n(Y)\cap f^k(V_n)\ne \emptyset.$

Take $n_1>3n+4$, and define an element $Z$ by
$$
P_{-n_1,-n_1+n-1}(Z)=P_{-n,-1}(Z)=P_{-n,-1}(Z)\quad\text{and}\quad P_{0,n}(Y)=P_{0,n}(Z)=P_{n_1-n,n_1}(Z),
$$
and complete the gaps such that $Z$ is also an element with a (one or two point) sink, and such that after $k=n_1-n$ steps the sink is again at $0$. We don’t have to assume that $Y$ is symmetric, and we need the space between $n$ and $n_1-n$ to pull the sink back to $0$, if it’s necessary.

One can prove that these are all non-wandering points.

Sketch of the proof:
Assume that $Y$ is a non-wandering point.

First step: If $Y=0$, then $Y\in L$. Else there exists $n$ such that $y_n=1$.

Second step: There must be a $2$ at the right or at the left of $y_n$.
One uses that the combinations
$$
010,\quad 011,\quad 110,\quad\text{and}\quad 111
$$
cannot appear in a non-wandering element.
Assume it is on the right (if it’s on the left the argument is symmetric).

Third step: The next element on the right must be $0$, since the combinations
$121$ and $122$ cannot appear in a non-wandering element.

Fourth step: On the right we have a string of $L$: For $k>n$, all the pairs $y_k,y_{k+1}$ must be of the permitted in $L$, i.e.,
$00$, $01$, $12$, or $20$. Else one of the pairs
$$
10,\quad 11,\quad 21,\quad 22\quad\text{or}\quad 02
$$
appear. Take the first time one of such pairs appears, then we obtain one of the prohibited
$$
010,\quad 011,\quad 121,\quad 122,
$$
or a string $20\dots 02$, which is also unrepeatable.

Fifth step: Either on the left all pairs $y_{k-1},y_{k}$ are the permitted in $L$, which means that $Y$ is in $L$, or there appears a pair
$$
10,\quad 11,\quad 21,\quad 22\quad\text{or}\quad 02.
$$
Take the first time such a pair appears, then we obtain the triples
$$
020,\quad 210,\quad 211,\quad 212\quad\text{or}\quad 220.
$$
By the same (symmetricly flipped) arguments as above, on the left must continue a sequence in $R$, and we have a sink.