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?



Is the infinite product map $(∏_{i=1}^{∞}S_{i})×f$ topologically transitive
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 $n \in \mathbf{N}$.


