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

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?

• Are you assuming that all the spaces are compact Hausdorff an that the maps are continuous? – YCor Feb 11 at 18:06
• @YCor: Yes. I will add this in the question. – Germany Feb 11 at 18:12
• Congratilations! You asked a question number 100,000. – Piotr Hajlasz Feb 11 at 18:13
• @PiotrHajlasz: But this number is not a prime. My fovirate numbers are primes. – Germany Feb 11 at 18:16
• You say that it's true for a family of 1 element, and asking about infinite families. But what about a family of two mixing maps (i.e., if $f$ is topologically transitive and $S_1,S_2$ are topologically mixing, is $S_1\times S_2\times f$ topologically transitive?) – YCor Feb 11 at 18:19

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 $$n \in \mathbf{N}$$.