Eventual Image $Y\equiv\bigcap_{n\geq 1}T^n(X)$

Question

Consider $X=\left\{0,1,2\right\}^{\mathbb{Z}}$ and $T\colon X\to X$ is defined 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.

Now there is the following:

One notes that if $\sigma\in\left\{0,1,2\right\}^{\mathbb{Z}^2}$ is a possible evolution of the system, then each row of $\sigma$ is in the eventual image of $T$, $\bigcap_{n=1}^{\infty}T^n(X)\equiv Y$.

This is the first thing I do not understand.

If $r$ is a row of $\sigma$, why is it in $T^n(X)~\forall n\geq 1$?

Answer 1

The row $\sigma(i,\cdot)$ is in $T^n(X)$ because it is $T^n(\sigma(i-n,\cdot))$.

EDIT: To give you a start on the lemma: note that if $\sigma(i,j) = 1$, we must have $\sigma(i-1,j) = 0$ and at least one of $\sigma(i-1,j-1)$ and $\sigma(i-1,j+1)$ is $1$. Let's say $\sigma(i-1,j-1) = 1$. Now $\sigma(i-2,j)$ can't have been $1$ so it must have been $\sigma(i-2,j-2) = 1$ and $\sigma(i-2,j-1)=0$. By induction, $\sigma(i-k,j-k) = 1$ and $\sigma(i-k,j-k+1) = 0$ for all $k \ge 1$.