# A permutation group inducing a topologically transitive action without dense orbits on $\omega^*$

Let $$G$$ be a subgroup of the permutation group $$S_\omega$$ of the countable infinite set $$\omega$$. Each bijection $$g\in G$$ admits a unique extension to a homeomorphism $$\bar g$$ of the Stone-Cech compactification $$\beta\omega$$ of $$\omega$$. The homeomorphism $$\bar g$$ induces a homeomorphism of the remainder $$\omega^*=\beta\omega\setminus\omega$$ of the Stone-Cech compactification. So, we obtain a continuous action of the group $$G$$ on the compact Hausdorff space $$\omega^*$$. I am interested in properties of the obtained dynamical system $$(\omega^*,G)$$. Namely, I would like to know the answer to the following

Problem. Is there a subgroup $$G\subseteq S_\omega$$ such that the dynamical system $$(\omega^*,G)$$ is topologically transitive (=each nonempty open set has dense orbit) but does not have a dense orbit.

An example of such subgroup $$G$$ exists under the assumption $$\mathrm{non}(\mathcal M)<\mathfrak c$$. So, the question actually ask about the situation in ZFC.

Remark. If a group $$G\subseteq S_\omega$$ induces a topologically transitive action on $$\omega^*$$, then $$G$$ has large cardinality, namely, $$|G|\ge\mathsf \Sigma\ge\max\{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal M)\}$$. More information on the cardinal $$\mathsf \Sigma$$ can be found in this preprint.

• Identify $\omega$ with a dense countable order $Q$. Let $G$ be the group of piecewise monotone permutations of $Q$ (i.e., cut $Q$ into finitely many convex pieces, and rearrange them to finitely many convex pieces through order preserving or reversing partial isomorphisms). I think $G$ acts topologically transitively on $\beta^*Q$. But I don't see if there's a dense orbit. – YCor Apr 2 at 20:32
• Note: For $G$ acting on $\omega$, it's clear that if for every $I,J\subset\omega$ with $J,\omega-I$ infinite there exists $g\in G$ such that $gI\subset^* J$, then $G$ acts minimally on $\beta^*\omega$. Here $\subset^*$ means inclusion modulo finite subset. In particular $S_\omega$ acts minimally on $\beta^*\omega$ (as you told me in answer to a comment of mine in your previous question; I'm just writing the easy argument to help the reader. – YCor Apr 2 at 20:35
• You are right the group of piecewise monotone functions acts topologically transitively on $Q^*$ (because each sequence contains a monotone subsequence). Concerning dense orbit, let me think a bit. – Taras Banakh Apr 2 at 20:48
• Yes. I needed piecewise, so as to handle the case of a bounded above increasing sequence, vs an unbounded above increasing sequence. – YCor Apr 2 at 20:49
• The action of the group $G$ on $Q^*$ will have many dense orbits: just take any ultrafilter living on a monotone sequence; its orbit will be dense. – Taras Banakh Apr 2 at 21:07

It turns out that this problem is independent of ZFC because of the following simple

Theorem. Under $$\mathfrak t=\mathfrak c$$, every topologically transitive continuous action of a group $$G$$ on $$\omega^*$$ has a dense orbit.

Proof. Let $$(A_\alpha)_{\alpha\in\mathfrak c}$$ be an enumeration of all infinite subsets of $$\omega$$. By transfinite induction we shall construct a transfinite sequence of infinite subsets $$(U_\alpha)_{\alpha\in\mathfrak c}$$ of $$\omega$$ and a transfinite sequence $$(g_\alpha)_{\alpha\in\mathfrak c}$$ of elements of the group $$G$$ such that for every $$\alpha\in\mathfrak c$$ the following conditions are satisfied:

(a) $$U_\alpha\subseteq^* U_\beta$$ for all $$\beta<\alpha$$;

(b) $$g_\alpha(U_\alpha)\subseteq^* A_\alpha$$.

To start the inductive construction, put $$U_0=A_0$$ and $$g_0$$ be the identity of the group $$G$$. Assume that for some ordinal $$\alpha\in\mathfrak c$$, a transfinite sequence $$(U_\beta)_{\beta<\alpha}$$ satisfying the condition (a) has been constructed. By the definition of the tower number $$\mathfrak t$$ and the equality $$\mathfrak t=\mathfrak c>\alpha$$, there exists an infinite subset $$V_\alpha\subseteq\omega$$ such that $$V_\alpha\subseteq^* U_\beta$$ for all $$\beta<\alpha$$. The infinite sets $$V_\alpha$$ and $$A_\alpha$$ and determine clopen sets $$\overline V_\alpha=\{p\in\omega^*:V_\alpha\in p\}$$ and $$\bar A_\alpha=\{p\in\omega^*:A_\alpha\in p\}$$ in the space $$\omega^*=\beta\omega\setminus\omega$$. Since the action of the group $$G$$ on $$\omega^*$$ is topologically transitive, there exist $$g_\alpha$$ and an infinite subset $$U_\alpha\subset V_\alpha$$ such that $$g_\alpha(\overline U_\alpha)\subseteq \bar A_\alpha$$, which implies $$g_\alpha (U_\alpha)\subseteq^* A_\alpha$$. This completes the inductive step.

Adter completing the inductive construction, extend the family $$\{U_\alpha\}_{\alpha\in\mathfrak c}$$ to a free ultrafilter $$\mathcal U$$ and observe that its orbit intersetcs each clopen set $$\bar A_\alpha$$, $$\alpha\in\mathfrak c$$ and hence is dense in $$\omega^*$$. $$\qquad\square$$

• Interesting. A follow-up question would be the same question (esp. in ZFC+CH) for $G\subset\mathrm{Homeo}(\beta^*\omega)$. – YCor Apr 3 at 10:58
• What is $\beta^*\omega$? If $\beta^*\omega=\omega^*$, then the theorem answer this question under $\mathfrak t=\mathfrak c$ and hence under CH. – Taras Banakh Apr 3 at 11:09
• Yes I mean $\beta^*X=\beta X-X$ (I'm boycotting the notation $X^*$ which is incomprehensible without context, and I like to retain the letter $\beta$ to denote the Stone-Cech remainder). Ah indeed I didn't notice your answer doesn't assume $G\subset S_\omega$, so is much more general than what you initially asked. – YCor Apr 3 at 11:16