Is the infinite product map $(∏_{i=1}^{∞}S_{i})×f$ topologically transitive

Question

Let $f:X→X$ be a map. We say that $f$ is topologically mixing if for every open subsets $U,V$ of $X$, there exists $N$ such that for every $n≥N$ the set $f^{n}(U)∩V$ is non-empty.

Let $S : X → X$ and $T : Y → Y$ be dynamical systems. Then the map $S × T$ is defined by:

$S × T : X × Y → X × Y$, $(S × T)(x, y) = (Sx, Ty)$.

We know that if $S$ and $T$ are topologically transitive and at least one of them is mixing then $S × T$ is topologically transitive (https://www.merry.io/dynamical-systems-lecture-notes/2016/10/3/the-shift-map).

My question is: Consider an infinite family of mixing continuous maps $S_{i}:X_{i}→X_{i}$ along with a topologically transitive map $f:X\to X$. Here $X_{i}$ and $X$ are compact sets.

Is the infinite product map $(∏_{i=1}^{∞}S_{i})×f$ topologically transitive?

Answer 1

If $\mathcal{X} = (\prod_{i=1}^\infty X_i ) \times X$ is endowed with the product topology, then to show that $F = (\prod_{i=1}^N S_i) \times f$ is topologically transitive (that is, given any pair of open subsets $U, V \subset \mathcal{X}$, there exists $n\in \mathbf{N}$ such that $F^n(U) \cap V \ne \emptyset$), it suffices to consider open subsets of the form $$U = \left(\prod_{i \le N} U_i \times \prod_{i > N} X_i\right) \times U'$$ $$V = \left(\prod_{i \le N} V_i \times \prod_{i > N} X_i\right) \times V'$$ where $U_i$ and $V_i$ (resp. $U'$ and $V'$) are open subsets of $X_i$ (resp. $X$). As $(\prod_{i=1}^N S_i) \times f$ is topologically transitive (which can be proven by induction using the result you cite in the question since every $S_i$ is topologically mixing and $f$ is topologically transitive), it follows that $F^n(U) \cap V \ne \emptyset$ for some $n \in \mathbf{N}$.