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.

  • $\begingroup$ 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. $\endgroup$ – YCor Apr 2 '20 at 20:32
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    $\begingroup$ 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. $\endgroup$ – YCor Apr 2 '20 at 20:35
  • $\begingroup$ 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. $\endgroup$ – Taras Banakh Apr 2 '20 at 20:48
  • $\begingroup$ Yes. I needed piecewise, so as to handle the case of a bounded above increasing sequence, vs an unbounded above increasing sequence. $\endgroup$ – YCor Apr 2 '20 at 20:49
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    $\begingroup$ 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. $\endgroup$ – Taras Banakh Apr 2 '20 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 $\omega^*$. $\qquad\square$

  • $\begingroup$ Interesting. A follow-up question would be the same question (esp. in ZFC+CH) for $G\subset\mathrm{Homeo}(\beta^*\omega)$. $\endgroup$ – YCor Apr 3 '20 at 10:58
  • $\begingroup$ What is $\beta^*\omega$? If $\beta^*\omega=\omega^*$, then the theorem answer this question under $\mathfrak t=\mathfrak c$ and hence under CH. $\endgroup$ – Taras Banakh Apr 3 '20 at 11:09
  • $\begingroup$ 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. $\endgroup$ – YCor Apr 3 '20 at 11:16

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