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By a dynamical system I understand a pair $(K,G)$ consisting a compact Hausdorff space and a subgroup $G$ of the homeomorphism group of $K$.

We say that a dynamical system $(K,G)$

$\bullet$ is topologically transitive if for every non-empty open set $U\subseteq K$ its orbit $GU=\{g(x):g\in G,\;x\in U\}$ is dense in $K$;

$\bullet$ has dense orbit if for some point $x\in K$ its orbit $Gx$ is dense in $K$.

It is easy to see that a dynamical system is topologically transitive if it has a dense orbit. If the space $K$ is metrizable and nonempty, then the converse also is true.

On the other hand, under $\mathrm{non}(\mathcal M)<\mathfrak c$, there exists a subgroup $G\subset S_\omega$ of cardinality $|G|\le\mathrm{non}(\mathcal M)<\mathfrak c$ that induces a topologically transitive action of the Stone-Cech remainder $\omega^*=\beta\omega\setminus\omega$. The dynamical system $(\omega^*,G)$ does not have dense orbits since the space $\omega^*$ has density $\mathfrak c>|G|$. I am interested if such an example can be constructed in ZFC.

Problem. Is there topologically transitive dynamical system without dense orbits?

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  • $\begingroup$ Does $S_\omega$ have a dense orbit on $\omega^*$? (Its action is obviously topologically transitive.) I thought I knew the answer to be negative, but I don't see right now. $\endgroup$ – YCor Apr 2 at 17:44
  • $\begingroup$ By the way the action on the empty set is trivially topologically transitive without dense orbits, so the assertion for $K$ metrizable should assume $K$ non-empty. $\endgroup$ – YCor Apr 2 at 18:27
  • $\begingroup$ @YCor The action of $S_\omega$ on $\omega^*$ is minimal: all orbits are dense. $\endgroup$ – Taras Banakh Apr 2 at 18:37
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The answer is yes:

there is a topologically transitive dynamical system without dense orbits.

Indeed,

let X be a topological space that is not separable. Let $\ K=X^{\Bbb Z},\ $ and let $\ G\ $ be the group of homeomorphism of $\ K,\ $ induced by shifts $\ s_n\ (n\in\Bbb Z)\ $ of $\ \Bbb Z:\ $

$$ \forall_{n\in\Bbb Z}\forall_{x\in\Bbb Z}\quad s_n(x):= x+n $$

Let $\ p:=(p_n)\in K\ $ be arbitrary, and let

$$ P:=\{p_n:n\in \Bbb Z\} $$

Then there exists non-empty open $\ U\ $ in $\ X,\ $ disjoint with set $\ P.\ $ Then non-empty open in $\ K\ $ set $\ W,$

$$ W\ :=\ \pi_0^{-1}(U)\ $$

is disjoint with the orbit $\ p.\ $

On the other hand, let $\ \emptyset\ne G\subseteq K, $ where $\ G\ $ is open in $\ K.\ $ Then there exists non-empty $\ H\ $ and integer $ a\ge 0\ $ such that $\ H\ $ is an open subset of $\ X^{(-a)..a}\ $ (Perl notation "s..t") and

$$ \emptyset\ \ne\ \pi_{(-a)..a}^{-1}(H)\ \subseteq G $$

Obviously, the orbit of $\ \pi_{(-a)..a}^{-1}(H)\ $, hence of $\ G,\ $ is dense in $\ K.$

Great!

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    $\begingroup$ Thank you for the great answer. And what about the case of subgroups $G\subseteq S_\omega$ acting on $\omega^*$? $\endgroup$ – Taras Banakh Apr 2 at 18:32
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    $\begingroup$ By $\omega^*$ I denote the remainder of the Stone-Cech compactification of $\omega$ and this is the principal space I am interested in. In this case the group $\mathbb Z$ (and any other countable group) does not work, unfortunately. $\endgroup$ – Taras Banakh Apr 2 at 18:43
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    $\begingroup$ @TarasBanakh at this point if you're specifically interested in $G\subset S_\omega$ acting on $\omega^*$ it would be worth a separate question. $\endgroup$ – YCor Apr 2 at 19:17
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    $\begingroup$ @YCor Thank you for the suggestion. I will then accept the answer of Wlod AA and write this specific question separately. $\endgroup$ – Taras Banakh Apr 2 at 19:23
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    $\begingroup$ Taras, thank you (and to YCor :) ) for accepting. $\endgroup$ – Wlod AA Apr 2 at 19:33

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