This is related to my previous question here:


[Antichains and measure-preserving actions on Boolean algebras](https://mathoverflow.net/questions/76422/antichains-and-measure-preserving-actions-on-boolean-algebras)


This time I will ask something more precise.


Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is *minorising* (for $G$) if the following holds:


For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.


Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.


The condition $n=2$ is equivalent to the condition that *every* proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)


Questions: Can $n$ be greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?


If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?


NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Can $G$ act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$.



Vigorous actions on the Cantor set
The answer to your last question is YES. Maybe there is a simpler way to see this, but here is *a* way:


If $G$ acts freely on $X$ and $\alpha$ is minorising (with $n=2$ if you wish), then there is a zero-dimensional compact space $Z$ of uncountable weight, a suryective map $\pi:Z \to X$ and an action of $G$ on $Z$ commuting with $\pi$ such that $\pi^{-1}\alpha$ is minorising. If $G$ was countable then necessarily $G$ has infinitely many (in fact uncountably many) orbits on the clopen subsets of $Z$. Now you can just consider a suitable countable subalgebra of clopens of $Z$ for which you still have infinitely many orbits and a minorising element (still with $n=2$ if you wish).


**Edit:** The answer to the first question is also YES. If $G$ acts on $X$ then $G \times \mathbb{Z}\_2$ acts in a natural way on $Z$ the disjoint union of two Cantor spaces (which is again a Cantor space), so that the $n$ of a minorising subset $\alpha$ of $X$ gets doubled when $\alpha$ is viewed as a subset of one of the copies in $Z$.
Unfortunately $G \times \mathbb{Z}\_2$ is not simple. Also, I don´t know if one can get $n=3$ (or any odd number).


