Let $X=\left\{0,1,2\right\}^{\mathbb{Z}}$ and on it the following dynamics described by $T\colon X\to X$ as follows: A 1 becomes a 2, a 2 becomes a 0 and a 0 becomes a 1 if at least one of its two neighbors is 1.
Moreover, consider $$ Y=\bigcap_{n=1}^{\infty}T^n(X). $$
Statement to prove:
$\eta\in X$ is in $Y$ if and only if there exists $n\in\mathbb{Z}\cup\left\{-\infty,\infty\right\}$, such that:
Condition I: At $n$ and to the left on $n$, every 1 has a 2 to its left, every 2 has a 0 to its left, and every 0 has a 0 or a 1 to its left.
Condition II: To the the right of $n$, every 1 has a 2 to its right, every 2 has a 0 to its right and every 0 has a 0 or a 1 to its right.
Here is what I tried. Please, have a look on it and tell me, if it is okay or not.
Proof
"$\Longleftarrow$": Let $\eta\in X$ with a corresponding $n$ fulfilling the conditions of the statement. By the following procedure, one can see that $\eta\in Y$: First develop the positions n and n+1 back infinitely often, a 1 getting a 0, a 2 getting a 1 and a 0 remaining a 0.
Then for each step back of the positions left to $n$ and right to $n+1$, just make the left-shift and right-shift, respectively.
Example: Let | separate the $n$-th from the $(n+1)$-th position.
$$ 10021|2001 $$
Step 1: $$ 10021|2001\\ 0|1 \\ 0|0 \\ 0|0 \\ \text{etc.} $$
Step 2: Fill the other gaps by left-shift and right-shift, respectively: $$ \ldots 10021|2001\ldots\\ 00210|1200\\ 02100|0120\\ 21000|0012\\ \text{etc.} $$
This shows that $\eta\in Y$.
"$\Longrightarrow$": Let $\eta\in Y$. By $\eta(x)$ denote the position $x$ of $\eta$.
W.l.o.g. let $\eta(x)=1$. In particular, $\eta\in T(X)$. There are the following possible left and right neighbours of $\eta(x)=1$ (where $\eta(x)=1$ is supposed to stand in the middle):
- 012
- 212
- 210
- 112
- 211
Take Case 1: 012 exemplarily-
When we develop $012$ one step back, we have two possibilities (that is, we have at least two pre-images) to do so: $$ 012\\ 201 $$ or $$ 012\\ 001 $$ Choose the first possibility. Then, when we develop this backwards, one gets the following scheme: $$ \ldots\star 0120\star\ldots\\ \ldots\star 020120\star\ldots\\ \ldots\star 02100120\star\ldots\\ \ldots\star 0210\star\star 0120\star\ldots\\ \ldots\star 0210\star\star\star 0120\star\ldots\\ \text{etc.} $$ Choose the $\star$'s within the "pyramide" all to be $0$.
For the $\star$'s on the left, if we choose one of it to be a $1$ then all of it have to be $1$. If we choose one of it to be a $0$, then all of it have to be $0$'s. Left of the stars on the left, there $1$'s ,$2$'s and $0$'s can only appear as Condition I from the statement says.
The same holds for the $\star$'s on the right. If we choose one of it to be a $1$, all are $1$'s. And if we choose one of it to be a $0$, all are $0$'s. To the right of the $\star$'s on the right side, there $1$'s, $2$'s, and $0$'s can only appear as Condition II from the statement wants them to appear.
So in this case, we can choose the $n$ to be $x$.
The other cases work similarly. Namely, in case that we start with the neighbourhood $112$ (again $\eta(x)=1$ in the middle), we can choose $n$ to be $x-1$. For the other cases we can chose $n=x$, too, as in case 1.
(Remark: Starting with $\eta(x)=1$ as done above is w.l.o.g. because if we start with a $2$, there is a $1$ in the row before and we can choose the $n$ as in the corresponding case starting with a neighbourhood of 1 because developing forward does not change the desired order of $0$'s, $1$'s and $2$'s as Condition I and Condition II wants them to appear. If we start with a $0$ then either we always have a $0$ before each $0$ and we can choose $n$ to be any number in $\mathbb{Z}$, or at some time there is a $2$ before a $0$ and then a $1$ before this $2$ and we can again choose the $n$ as in the corresponding neighborhoud of $1$.)
Finally, if we have a $y\in Y$ consisting of $0$'s, $1$'s and $2$' such that there is a $2$ on the right (left) of every 1, a $0$ on the right (left) of every $2$ and a $0$ or a $1$ on the right (left) of every $0$, we can choose $n=+\infty$ and $n=-\infty$, respectively.
$\Box$
I really hope to get a feedback from you, because I do need this statement in a project and thus I really need a proof. I gave you my proof and hope to get an answer back from you. Thank you very much!
Best wishes
math12
I don't know whether the following helps. It is just about the workings of $T$, and there are no limit considerations. I did a few simulations with random starting sequences over the alphabet $\{0,1,2\}$. What I observed is that after only two forward $T$-steps there are just the following features left: In a soup of zeros there are little trains 12 moving to the left one unit per step and little trains 21 moving to the right one unit per step. When two such trains collide they annihilate each other within two steps. The following figure shows an example. Zeros are not printed, ones are replaced by red bullets, and twos by black bullets. I venture that staring at this figure will lead to the solution of your problem.