1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are you assuming that all the spaces are compact Hausdorff an that the maps are continuous? $\endgroup$ – YCor Feb 11 at 18:06
  • $\begingroup$ @YCor: Yes. I will add this in the question. $\endgroup$ – China Feb 11 at 18:12
  • 1
    $\begingroup$ Congratilations! You asked a question number 100,000. $\endgroup$ – Piotr Hajlasz Feb 11 at 18:13
  • $\begingroup$ @PiotrHajlasz: But this number is not a prime. My fovirate numbers are primes. $\endgroup$ – China Feb 11 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 at 18:19
2
$\begingroup$

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 $n \in \mathbf{N}$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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