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?

  • $\begingroup$ Are you assuming that all the spaces are compact Hausdorff an that the maps are continuous? $\endgroup$
    – YCor
    Feb 11, 2019 at 18:06
  • $\begingroup$ @YCor: Yes. I will add this in the question. $\endgroup$
    – Safwane
    Feb 11, 2019 at 18:12
  • 1
    $\begingroup$ Congratilations! You asked a question number 100,000. $\endgroup$ Feb 11, 2019 at 18:13
  • $\begingroup$ @PiotrHajlasz: But this number is not a prime. My fovirate numbers are primes. $\endgroup$
    – Safwane
    Feb 11, 2019 at 18:16
  • $\begingroup$ 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?) $\endgroup$
    – YCor
    Feb 11, 2019 at 18:19

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

  • $\begingroup$ Sorry for the late comment: this question seems to be a duplicate of one on MSE math.stackexchange.com/questions/3108947/… which has not been answered. Since you have a MSE account, perhaps you could flag the question on MSE for closure there as a duplicate of a question here? $\endgroup$
    – Yemon Choi
    Dec 29, 2019 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.