Let $T\colon X\to X$, with $X=\left\{0,1,2\right\}^{\mathbb{Z}}$, desribe the following dynamics:
- 1 becomes 2
- 2 becomes 0
- 0 becomes 1 if at least one of its two neighbours is 1, otherwise it remains 0
Equip $\left\{0,1,2\right\}$ with the discrete topology and $X$ with the associated product topology (with is compact by Tychonov).
Maybe a silly question, but: Why is $T$ continuous?
Is there a very simple reason I do not see?
Note that a map into a product space is continuous iff all its components are. Hence to check where $T \colon X \to X$ is continuous it suffices to look at all $T_n = \pi_n \circ T \colon X \to \{0,1,2\}$, where $\pi_n \colon X \to \{0,1,2\}$ denotes the $n$-th projection, $n \in \mathbf Z$.
Now, $$ T_n^{-1}[\{2\}] = \pi_n^{-1}[\{1\}] $$ which is open, as $\{1\}$ is open and $\pi_n$ is continuous, moreover $$ T_n^{-1}[\{1\}] = \pi_n^{-1}[\{0\}] \cap \bigl(\pi_{n-1}^{-1}[\{1\}] \cup \pi_{n+1}^{-1}[\{1\}] \bigr) $$ is open and $$ T_n^{-1}[\{0\}] = \pi_n^{-1}[\{2\}] \cup \pi_n^{-1}[\{0\}] \cap \bigl(\pi_{n-1}^{-1}[\{0,2\}] \cap \pi_{n+1}^{-1}[\{0,2\}] \bigr) $$ also is. Hence, $T_n$ is continuous, as $n$ was arbitrary, $T$ is.