Let $\mathcal{B}$ be the (unique up to isomorphism) countable atomless boolean algebra, and $\mathrm{Aut}(\mathcal{B})$ its automorphism group, with pointwise convergence topology.


My question: Does $\mathrm{Aut}(\mathcal{B})$ contain a **closed** (non-abelian) free subgroup?


This question arises from reading about the following two results (whose proofs I am not overly familiar with):


$\mathrm{Aut}(\mathcal{B})$ contains many free subgroups; by a result of Machpherson (*Groups of Automorphisms of $\aleph\_0$-Categorical Structures*, QJM, 1986), it has a dense subgroup freely generated by countably many generators.


$\mathrm{Aut}(\mathbb{Q},<)$ contains a closed non-abelian free group; this is a result of Pestov (*On free actions, minimal flows, and a problem by Ellis*, TAMS, 1998). In particular, closed subgroups of extremely amenable topological groups need not be amenable.



Closed free subgroups of the automorphism group of the countable atomless boolean algebra
Yes.


This group is widely documented as "(self-)homeomorphism group of the Cantor set" (by Stone duality). 


It admits, for every prime $p$ and $p$-adic field $K$, and $d\ge 2$, the group $\mathrm{PGL}\_d(K)$ as a closed subgroup (viewed as acting on the projective space $\mathbb{P}^{d-1}(K)$ which is homeomorphic to a Cantor space), and this group contains a non-abelian free discrete subgroup. 




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Added: here's a direct "Schottky" construction. Suppose that we have a partition of the Cantor set $X=A\_+\sqcup A\_-\sqcup B\_+\sqcup B\_-\sqcup Y$ into nonempty clopen subsets, and self-homeomorphisms $a,b$ of $X$ such that $a^{\pm 1}A\_{\mp}^c\subset A\_\pm$ and $b^{\pm 1}B\_{\mp}^c\subset B\_\pm$ (ping-pong condition). Then $(a,b)$ freely generates a discrete free subgroup (discrete because the stabilizer of the clopen subset $Y$ is reduced to $\{1\}$). There are plenty of such pairs, for instance coming from suitable loxodromic isometries of trees acting on the boundary.


