# A topologically transitive dynamical system without dense orbits

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?

• 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.
– YCor
Apr 2, 2020 at 17:44
• 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.
– YCor
Apr 2, 2020 at 18:27
• @YCor The action of $S_\omega$ on $\omega^*$ is minimal: all orbits are dense. Apr 2, 2020 at 18:37

## 1 Answer

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!

• Thank you for the great answer. And what about the case of subgroups $G\subseteq S_\omega$ acting on $\omega^*$? Apr 2, 2020 at 18:32
• 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. Apr 2, 2020 at 18:43
• @TarasBanakh at this point if you're specifically interested in $G\subset S_\omega$ acting on $\omega^*$ it would be worth a separate question.
– YCor
Apr 2, 2020 at 19:17
• @YCor Thank you for the suggestion. I will then accept the answer of Wlod AA and write this specific question separately. Apr 2, 2020 at 19:23
• Taras, thank you (and to YCor :) ) for accepting. Apr 2, 2020 at 19:33