Proof. The implication (ii) $\implies $ (i) is trivial. Let us prove that (i) $\implies $ (ii). Assume condition (i) holds. Without loss of generality, we may further assume that the neighborhoods $U_x$ , $x\in X$ are clopen. The sets $U_x$ form an open cover of the compact space X. Hence, there are finitely many points $x_1,x_2,\ldots ,x_k\in X$ such that the sets $U_{x_i}$ , $1\le i\le k$ form a subcover. Now, let $V_1=U_{x_1}$ and $V_i=U_{x_i}\setminus (U_{x_1}\cup \cdots \cup U_{x_{i-1}})$ for $i=2,3,\ldots ,k$ . Then, the sets $V_1,V_2,\ldots ,V_k$ form a partition of X. Since each $U_{x_i}$ , $1\le i\le k$ is clopen, it follows that the sets $V_i$ , $1\le i\le k$ are clopen as well. Let $h_i=g_{x_i}$ for $1\le i\le k$ . Clearly, f coincides with $h_i$ on $V_i$ for $1\le i\le k$ . Thus, condition (ii) holds.

Proof. As the inclusion $\widehat {\mathsf {F}}(S)\subset \widehat {\mathsf {F}}(\widehat {\mathsf {F}}(S))$ is obvious, we need to prove that $\widehat {\mathsf {F}}(\widehat {\mathsf {F}}(S))\subset \widehat {\mathsf {F}}(S)$ . Suppose $f\in \widehat {\mathsf {F}}(\widehat {\mathsf {F}}(S))$ and consider an arbitrary point $x\in X$ . Since f is locally an element of $\widehat {\mathsf {F}}(S)$ , there exists an open neighborhood U of x and a map $g\in \widehat {\mathsf {F}}(S)$ such that $f|_U=g|_U$ . Since g is locally an element of S, there exists an open neighborhood W of x and a map $h\in S$ such that $g|_W=h|_W$ . Then, f coincides with h on $U\cap W$ , which is an open neighborhood of x. We conclude that f is locally an element of S.

In the case $S\subset {\mathrm {Homeo}}(X)$ , the inclusion $\mathsf {F}(S)\subset \mathsf {F}(\mathsf {F}(S))$ is obvious. Since $\mathsf {F}(S)\subset \widehat {\mathsf {F}}(S)$ , it follows that $\widehat {\mathsf {F}}(\mathsf {F}(S))\subset \widehat {\mathsf {F}}(\widehat {\mathsf {F}}(S))$ . By the above, $\widehat {\mathsf {F}}(\widehat {\mathsf {F}}(S))=\widehat {\mathsf {F}}(S)$ . Therefore, $\mathsf {F}(\mathsf {F}(S))=\widehat {\mathsf {F}}(\mathsf {F}(S))\cap {\mathrm {Homeo}}(X)\subset \widehat {\mathsf {F}}(S)\cap {\mathrm {Homeo}}(X) =\mathsf {F}(S)$ . Thus, $\mathsf {F}(\mathsf {F}(S))=\mathsf {F}(S)$ .

Proof. Suppose $S\subset C(X,X)$ is a semigroup and consider any two maps $f,g\in \widehat {\mathsf {F}}(S)$ . Given a point $x\in X$ , there exists an open neighborhood U of x and a map $\tilde g\in S$ such that $g|_U=\tilde g|_U$ . Further, there exists an open neighborhood W of the point $g(x)$ and a map $\tilde f\in S$ such that $f|_W=\tilde f|_W$ . Then, the composition $fg$ coincides with $\tilde f\tilde g$ , which belongs to S, on $g^{-1}(W)\cap U$ , which is an open neighborhood of x. Hence, $fg$ is locally an element of S. Thus, $\widehat {\mathsf {F}}(S)$ is a semigroup.

Now, assume $S\subset {\mathrm {Homeo}}(X)$ is a group. By the above, $\widehat {\mathsf {F}}(S)$ is a semigroup. Then, $\mathsf {F}(S)$ is also a semigroup as it is the intersection of two semigroups $\widehat {\mathsf {F}}(S)$ and ${\mathrm {Homeo}}(X)$ . It remains to show that for any homeomorphism $f\in \mathsf {F}(S)$ , the inverse $f^{-1}$ is also in $\mathsf {F}(S)$ . Take an arbitrary point $x\in X$ . Since f is locally an element of S, there exists an open neighborhood U of the point $f^{-1}(x)$ and a map $g\in S$ such that $f|_U=g|_U$ . Then, $f^{-1}$ coincides with $g^{-1}$ , which belongs to S, on $f(U)$ , which is an open neighborhood of x. Thus, $f^{-1}$ is locally an element of S.

Proof. Since the set stabilizer ${\mathrm {St}}_G(Y)$ is a subgroup of G, its full amplification $\mathsf {F}({\mathrm {St}}_G(Y))$ is a subgroup of $\mathsf {F}(G)=G$ . Take any map $f\in \mathsf {F}({\mathrm {St}}_G(Y))$ . Then, f is piecewise an element of ${\mathrm {St}}_G(Y)$ , which implies that $f(Y)\subset Y$ . Since ${\mathrm {St}}_G(Y)={\mathrm {St}}_G(X\setminus Y)$ , it also follows that $f(X\setminus Y)\subset X\setminus Y$ . As f is invertible, we obtain $f(Y)=Y$ so that $f\in {\mathrm {St}}_G(Y)$ . Thus, the group ${\mathrm {St}}_G(Y)$ is ample.

If Y is the empty set or X, then one of the groups ${\mathrm {St}}^\star _G(Y)$ and ${\mathrm {RiSt}}_G(Y)$ is trivial while the other coincides with G, and both are ample. Now, assume Y is different from $\emptyset $ and X. By the above, for any point $x\in X$ , the stabilizer ${\mathrm {St}}_G(x)={\mathrm {St}}_G(\{x\})$ is an ample group. Therefore, the point stabilizer ${\mathrm {St}}^\star _G(Y)$ is the intersection of ample groups ${\mathrm {St}}_G(x)$ , $x\in Y$ , while the rigid stabilizer ${\mathrm {RiSt}}_G(Y)$ is the intersection of ample groups ${\mathrm {St}}_G(x)$ , $x\notin Y$ . It is easy to observe that for any collection of ample subgroups of ${\mathrm {Homeo}}(X)$ , their intersection is ample as well.

Proof. It is enough to show that any conjugate of a generalized permutation in the group ${\mathrm {Homeo}}(X)$ is itself a generalized permutation. After that, the proof of the lemma is straightforward. Suppose a generalized permutation $g=\mu [U;f_1,f_2,\ldots ,f_n;\pi ]$ is defined. Then, $f_1(U),f_2(U),\ldots ,f_n(U)$ are disjoint clopen sets. The support of g is the union of those $f_i(U)$ for which $\pi (i)\ne i$ . For any $x\in U$ and $i\in \{1,2,\ldots ,n\}$ , we have $g(f_i(x))=f_{\pi (i)}(x)$ . Now, consider an arbitrary homeomorphism $h\in {\mathrm {Homeo}}(X)$ . It is easy to observe that ${\mathrm {supp}}(hgh^{-1})=h({\mathrm {supp}}(g))$ . Hence, the support of $hgh^{-1}$ is contained in the union of sets $hf_1(U),hf_2(U),\ldots ,hf_n(U)$ , which are disjoint clopen sets. Furthermore, for any $x\in U$ and $i\in \{1,2,\ldots ,n\}$ , we have $hgh^{-1}(hf_i(x))=hg(f_i(x))=hf_{\pi (i)}(x)$ . It follows that $hgh^{-1}=\mu [U;hf_1,hf_2,\ldots ,hf_n;\pi ]$ . Thus, $hgh^{-1}$ is a generalized permutation.

Proof. Let P be the set of all generalized permutations of the form $\mu [U;f_1,f_2,\ldots , f_n;\pi ]$ , where $f_1,f_2,\ldots ,f_n\in \mathsf {F}(G)$ . It follows from the definition that any element of P is piecewise an element of the group $\mathsf {F}(G)$ . Therefore, P is contained in the group $\mathsf {F}(\mathsf {F}(G))$ , which coincides with $\mathsf {F}(G)$ due to Lemma 2.3.

To prove the lemma, it is enough to show that any generalized permutation $g\in \mathsf {F}(G)$ is a product of elements of the set P. The map g is represented as $\mu [U;f_1,f_2,\ldots ,f_n;\pi ]$ , where maps $f_1,f_2,\ldots ,f_n$ need not be in $\mathsf {F}(G)$ . Let $\pi =\sigma _1\sigma _2\cdots \sigma _k$ be the decomposition of the permutation $\pi $ as a product of disjoint cycles. Lemma 4.2 implies that $g=g_1g_2\cdots g_k$ , where $g_j=\mu [U;f_1,f_2,\ldots ,f_n;\sigma _j]$ , $1\le j\le k$ . We are going to show that each $g_j$ belongs to P.

First, let us show that $g_j\in \mathsf {F}(G)$ . For any $i\in \{1,2,\ldots ,n\}$ , we have either $\sigma _j(i)=\pi (i)$ or $\sigma _j(i)=i$ . In the former case, $g_j$ coincides with g on the clopen set $f_i(U)$ . In the latter case, $g_j$ coincides with the identity map on $f_i(U)$ . On the complement of the union of sets $f_i(U)$ , $1\le i\le n$ , both g and $g_j$ coincide with the identity map. Hence, $g_j$ is piecewise an element of the set $\{\mathrm {id}_X,g\}$ . Since $g\in \mathsf {F}(G)$ , it follows that $g_j\in \mathsf {F}(\mathsf {F}(G))=\mathsf {F}(G)$ .

Recall that the permutation $\sigma _j$ is a cycle: $\sigma _j=(i_1\,i_2\ldots i_m)$ , where $i_1,i_2,\ldots ,i_m$ are distinct elements of $\{1,2,\ldots ,n\}$ . Let $W_s=f_{i_s}(U)$ , $1\le s\le m$ . Then, $W_1,W_2,\ldots ,W_m$ are disjoint clopen sets and $g_j$ coincides with the identity map away from them. We have $g_j(W_s)=W_{s+1}$ for $1\le s\le m-1$ and $g_j(W_m)=W_1$ . It follows by induction that $W_s=g_j^{s-1}(W_1)$ for $1\le s\le m$ . Since the cycle $\sigma _j$ has order m in the group $\mathsf {S}_n$ , Lemma 4.2 implies that $g_j^m=\mathrm {id}_X$ . As a consequence, $(g_j^{m-1})^{-1}$ coincides with $g_j$ on $W_m$ (as well as anywhere else on X). All this leads to an alternative representation of $g_j$ as a generalized permutation:

$$ \begin{align*} g_j=\mu[W_1;\mathrm{id}_X,g_j,g_j^2,\ldots,g_j^{m-1};(1\,2\ldots m)]. \end{align*} $$

Since $g_j$ and all its powers belong to the group $\mathsf {F}(G)$ , we conclude that $g_j\in P$ .

Proof. If a generalized $2$ -cycle $g=\delta _{U;f}$ is defined and $f\in \mathsf {F}(G)$ , then g is piecewise an element of the group $\mathsf {F}(G)$ . Hence, it belongs to the group $\mathsf {F}(\mathsf {F}(G))$ , which coincides with $\mathsf {F}(G)$ due to Lemma 2.3. Conversely, if $g=\delta _{U;f}$ is defined and $g\in \mathsf {F}(G)$ , we observe that the generalized $2$ -cycle $\delta _{U;g}$ is also defined and $g=\delta _{U;g}$ .

By Lemma 4.5, the group $\mathsf {S}(G)$ is generated by generalized permutations of the form $\mu [U;f_1,f_2,\ldots ,f_n;\pi ]$ , where $f_1,f_2,\ldots ,f_n\in \mathsf {F}(G)$ . Those include all maps of the form $\delta _{U;f}$ , $f\in \mathsf {F}(G)$ . Whenever $\mu [U;f_1,f_2,\ldots ,f_n;\pi ]$ is defined, all maps $\mu [U;f_1,f_2,\ldots ,f_n;\sigma ]$ , $\sigma \in \mathsf {S}_n$ are defined as well. Assuming $f_1,f_2,\ldots ,f_n\in \mathsf {F}(G)$ , they all belong to the group $\mathsf {F}(\mathsf {F}(G))=\mathsf {F}(G)$ . Moreover, $\mathsf {S}_n\ni \sigma \mapsto \mu [U;f_1,f_2,\ldots , f_n;\sigma ]$ is a homomorphism of the group $\mathsf {S}_n$ to $\mathsf {F}(G)$ due to Lemma 4.2. Since the group $\mathsf {S}_n$ is generated by $2$ -cycles $(i\,j)$ , it follows that the group $\mathsf {S}(G)$ is generated by generalized permutations of the form $\mu [U;f_1,f_2,\ldots ,f_n;(i\,j)]$ . It remains to show that each $\mu [U;f_1,f_2,\ldots ,f_n;(i\,j)]$ is a generalized $2$ -cycle of prescribed form. First, we observe that $\mu [U;f_1,f_2,\ldots ,f_n;(i\,j)]=\mu [U;f_i,f_j;(1\,2)]$ . Then, we observe that $\mu [U;f_i,f_j;(1\,2)]=\delta _{f_i(U);f_jf_i^{-1}}$ . Finally, we note that $f_jf_i^{-1}\in \mathsf {F}(G)$ .

Proof. If a point $x\in X$ is isolated, then the set $\{x\}$ is clopen. It follows that the set Y of all non-isolated points of X is closed. Any homeomorphism of X maps an isolated point to an isolated point. Hence, the set Y is invariant under the action of any group $G\subset {\mathrm {Homeo}}(X)$ . If the action is minimal, then either Y is empty or $Y=X$ . In the case $Y=X$ , there are no isolated points. Then, X is a Cantor set due to Brouwer’s characterization of Cantor sets. In the case where Y is empty, every point of X is isolated. Hence, the partition of X into points is a partition into clopen sets. Then, compactness of X implies that it is finite.

Proof. Since the group G acts minimally, the orbit of any point $x\in U$ is dense in X. In particular, the orbit has a point in the non-empty clopen set $U'$ . Hence, $g_x(x)\in U'$ for some map $g_x\in G$ . Let $U_x=U\cap g_x^{-1}(U')$ . Then, $U_x$ is a clopen subset of U containing the point x, and $g_x(U_x)\subset U'$ . The clopen sets $U_x$ , $x\in U$ cover U. Since U is compact, there are finitely many points $x_1,x_2,\ldots ,x_k\in U$ such that the sets $U_{x_i}$ , $1\le i\le k$ form a subcover. Let $V_1=U_{x_1}$ and $V_i=U_{x_i}\setminus (U_{x_1}\cup \cdots \cup U_{x_{i-1}})$ for $i=2,3,\ldots ,k$ . Then, $V_1,V_2,\ldots ,V_k$ are clopen sets that partition U. By construction, $g_{x_i}(V_i)\subset g_{x_i}(U_{x_i})\subset U'$ for all i.

Property UR. (Universal recurrence)

For any open set $U\subset X$ , any point $x\in U$ , and any element $g\in G$ , there exists an integer $k\ge 1$ such that $g^k(x)\in U$ .

Property NC. (No contraction)

If $g(U)\subset U$ for some element $g\in G$ and open set $U\subset X$ , then, in fact, $g(U)=U$ .

Proof. Assume that the group $G\subset {\mathrm {Homeo}}(X)$ has Property NC. Take any open set $U\subset X$ and element $g\in G$ . Since g is a homeomorphism, the sets $g^{-1}(U),g^{-2}(U),\ldots $ are open as well and so is the union $W=U\cup g^{-1}(U)\cup g^{-2}(U)\cup \cdots $ . Note that $g^{-1}(W)=g^{-1}(U)\cup g^{-2}(U)\cup g^{-3}(U)\cup \cdots $ , which implies that $g^{-1}(W)\subset W$ . Since $g^{-1}\in G$ and the group G has Property NC, we have $g^{-1}(W)=W$ . As a consequence, any point $x\in U$ belongs to $g^{-k}(U)$ for some $k\ge 1$ . Then, $g^k(x)\in U$ . Thus, G has Property UR.

Conversely, assume that the group G does not have Property NC. Then, there exist an open set $U\subset X$ and an element $g\in G$ such that $g(U)$ is a proper subset of U. The inverse map $h=g^{-1}$ is also an element of G. Take any point $x\in U$ not in $g(U)$ . The condition $g(U)\subset U$ implies that $h(X\setminus g(U))\subset X\setminus U$ and $h(X\setminus U)\subset X\setminus U$ . It follows that the sequence $h(x),h^2(x),h^3(x),\ldots $ never visits U. We conclude that the group G does not have Property UR.

Proof. Lemma 5.4 implies that the group G does not have Property UR. Hence, there exist an open set $W\subset X$ , a point $x\in W$ , and a map $h\in G$ such that the sequence $h(x),h^2(x),h^3(x),\ldots $ never visits W. The inverse map $g=h^{-1}$ is also an element of G. Let $S=\{h^k(x)\mid k\ge 1\}$ and $U=X\setminus \overline {S}$ , the complement of the closure of S. Since g is a homeomorphism, it follows that $g(\overline {S})=\overline {g(S)}$ . It is easy to see that $g(S)=S\cup \{x\}$ . Hence, $\overline {g(S)}=\overline {S}\cup \{x\}$ . Consequently, $g(U)=X\setminus g(\overline {S})=X\setminus (\overline {S}\cup \{x\}) =U\setminus \{x\}$ . By construction, U is an open set. In addition, $W\subset U$ since W is an open set disjoint from S. In particular, $x\in U$ . Thus, $g(U)\subset U$ and $U\setminus g(U)$ consists of a single point.

Property SC. (Strong contraction)

For any clopen set $U\subset X$ different from the empty set and X, there exist maps $f_1,f_2\in G$ such that the images $f_1(U)$ and $f_2(U)$ are disjoint subsets of U.

Property UC. (Universal contraction)

For any clopen sets $A,B\subset X$ different from the empty set and X, there exists a map $f\in G$ such that $f(A)\subset B$ .

Proof. Assume a group $G\subset {\mathrm {Homeo}}(X)$ has Property UC. Since the topology on X is generated by clopen sets, Property UC clearly implies that every orbit of the group G is dense. Hence, G acts minimally on X. Since X is a Cantor set, any clopen set $U\subset X$ different from $\emptyset $ and X consists of more than one point. Therefore, it can be represented as a disjoint union of two non-empty clopen sets $U_1$ and $U_2$ . Property UC implies that $f_1(U)\subset U_1$ and $f_2(U)\subset U_2$ for some maps $f_1,f_2\in G$ . Then, $f_1(U)$ and $f_2(U)$ are disjoint subsets of U. Thus, the group G has Property SC.

Conversely, assume the group G is ample, acts minimally on X, and has Property SC. For any clopen set $W\subset X$ different from $\emptyset $ and X, there exist maps $f_1,f_2\in G$ such that the images $f_1(W)$ and $f_2(W)$ are disjoint subsets of W. Then, $f_1(W)$ is a proper subset of W. It follows that $f_1^{j+1}(W)$ is a proper subset of $f_1^j(W)$ for all $j\ge 0$ . Let $D=W\setminus f_1(W)$ . Then, $f_1^j(D)=f_1^j(W)\setminus f_1^{j+1}(W)$ for all $j\ge 0$ , which implies that $D,f_1(D),f_1^2(D),\ldots $ are disjoint subsets of W.

Let us show that any clopen set $U\subset X$ different from $\emptyset $ and X can be mapped to its complement by an element of G. By the above, there exist a non-empty clopen set D and a map $f\in G$ such that $D,f(D),f^2(D),\ldots $ are disjoint subsets of $X\setminus U$ . By Lemma 5.3, for some $k\ge 1$ , there exist disjoint clopen sets $V_1,V_2,\ldots ,V_k$ and maps $g_1,g_2,\ldots ,g_k\in G$ such that $U=V_1\sqcup V_2\sqcup \cdots \sqcup V_k$ and $g_i(V_i)\subset D$ for all i. For any i, $1\le i\le k$ , let $h_i=f^{i-1}g_i$ . Then, $h_i\in G$ and $h_i(V_i)\subset f^{i-1}(D)$ . Therefore, $h_1(V_1),h_2(V_2),\ldots ,h_k(V_k)$ are disjoint subsets of $X\setminus U$ . Note that the generalized $2$ -cycles $\delta _{V_i;h_i}$ , $1\le i\le k$ are well defined. Since the group G is ample, it contains all of them as well as their product $h=\delta _{V_1;h_1}\delta _{V_2;h_2}\cdots \delta _{V_k;h_k}$ . As the sets $V_1,V_2,\ldots ,V_k$ and $h_1(V_1),h_2(V_2),\ldots ,h_k(V_k)$ are disjoint from one another, $h(V_i)=\delta _{V_i;h_i}(V_i)=h_i(V_i)$ for all i so that $h(V_i)\subset X\setminus U$ for all i. Then, $h(U)\subset X\setminus U$ as well.

Now, we can establish Property UC. Take any clopen sets $A,B\subset X$ different from the empty set and X. We need to show that A can be mapped to B by an element of the group G. First, consider the case where $A\cup B\ne X$ . Then, $C=X\setminus (A\cup B)$ is a non-empty clopen set. By the above, $h_1(A\cup B)\subset C$ and $h_2(X\setminus B)\subset B$ for some maps $h_1,h_2\in G$ . Since $C\subset X\setminus B$ , we obtain that $h_2h_1(A)\subset h_2h_1(A\cup B)\subset h_2(C)\subset h_2(X\setminus B)\subset B$ . As for the case where $A\cup B=X$ , we have $h(A)\subset X\setminus A$ for some $h\in G$ . If $A\cup B=X$ , then the complement $X\setminus A$ is contained in B.

Property CI. (Cancellation of intersection)

For any non-empty clopen sets $A,B,C\subset X$ such that C is disjoint from $A\cup B$ , and the unions $A\cup C$ and $B\cup C$ are different from X, we have $f(A\cup C)=B\cup C$ for some $f\in G$ if and only if $h(A)=B$ for some $h\in G$ .

Proof. Since the set D is absorbable under the action of the group G, we have ${f(D\cup E_0)=E_0}$ for some $f\in G$ and clopen set $E_0$ disjoint from D. Note that $E_0$ is different from $\emptyset $ and $X\setminus D$ . Consider an arbitrary clopen set $E\subset X\setminus D$ different from $\emptyset $ and $X\setminus D$ . In the case where $E\cap E_0\ne \emptyset $ , Property CI first implies that ${g(D\cup (E\cap E_0))=E\cap E_0}$ for some $g\in G$ , then implies that $h(D\cup E)=E$ for some $h\in G$ . Similarly, in the case where $E\cup E_0\ne X\setminus D$ , Property CI first implies that $g(D\cup (E\cup E_0))=E\cup E_0$ for some ${g\in G}$ , then implies that $h(D\cup E)=E$ for some $h\in G$ . Finally, in the case where $E\cap E_0=\emptyset $ and $E\cup E_0=X\setminus D$ , we have $f(E)=D\cup E$ so that ${f^{-1}(D\cup E)=E}$ . Thus, D can be absorbed by E.

Proof. Suppose a group $G\subset {\mathrm {Homeo}}(X)$ has Property CI. For any disjoint non-empty sets $A,B,C\subset X$ , the unions $A\cup C$ and $B\cup C$ are both different from X. Hence, condition (CI-1) follows directly from Property CI. As for condition (CI-2), it easily follows from Lemma 5.8.

Now, let us prove that any ample group $G\subset {\mathrm {Homeo}}(X)$ satisfying conditions (CI-1) and (CI-2) has Property CI. Given non-empty clopen sets $A,B,C\subset X$ , where C is disjoint from both A and B, $A\cup C\ne X$ and $B\cup C\ne X$ , we need to show that the set $A\cup C$ can be mapped onto $B\cup C$ under the action of G if and only if the set A can be mapped onto B. This is trivial when $A=B$ . Consider the case where both $A\setminus B$ and $B\setminus A$ are non-empty sets. If $f(A\cup C)=B\cup C$ or $f(A)=B$ for some $f\in G$ , condition (CI-1) implies that $g(A\setminus B)=B\setminus A$ for some $g\in G$ . Since the clopen sets $A\setminus B$ and $B\setminus A$ are disjoint, a generalized $2$ -cycle $h=\delta _{A\setminus B;g}$ is well defined. It belongs to the group G as G is ample. Since h coincides with the identity map on $A\cap B$ and on C, we obtain that $h(A)=B$ and $h(A\cup C)=B\cup C$ .

It remains to consider the case where one of the sets A and B is a proper subset of the other. Without loss of generality, we may assume that B is a proper subset of A. Note that the set $B\cup C$ is disjoint from $A\setminus B$ and $(A\setminus B)\cup (B\cup C)=A\cup C\ne X$ . Hence, X is a disjoint union of four non-empty clopen sets $A\setminus B$ , B, C, and $X\setminus (A\cup C)$ . Condition (CI-2) first implies that the set $A\setminus B$ can be absorbed by $B\cup C$ if and only if it can be absorbed by $X\setminus (A\cup C)$ , then implies that $A\setminus B$ can be absorbed by $X\setminus (A\cup C)$ if and only if it can be absorbed by B. Consequently, $f(A\cup C)=B\cup C$ for some $f\in G$ if and only if $h(A)=B$ for some $h\in G$ .

Proof. Suppose a group $G\subset {\mathrm {Homeo}}(X)$ is ample and has Property NC. To establish Property CI, we are going to show that G satisfies conditions (CI-1) and (CI-2) in Lemma 5.9. Since the group G allows no contraction of open sets, it also allows no absorbable sets. Hence, condition (CI-2) holds vacuously. Let us verify condition (CI-1). Consider any disjoint non-empty clopen sets $A,B,C\subset X$ such that $f(A\cup C)=B\cup C$ for some $f\in G$ . We need to find an element of G that maps A onto B. By Lemma 5.4, Property NC is equivalent to Property UR. Therefore, the group G has Property UR. As a consequence, for any point $x\in A$ , the sequence $f(x),f^2(x),f^3(x),\ldots $ visits the set A. In particular, this sequence is not contained in C. Let $p(x)$ denote the least positive integer k such that $f^k(x)\notin C$ . Since $f(A\cup C)=B\cup C$ , it follows that $f^{p(x)}(x)\in B$ . Likewise, Property UR implies that for any point $y\in B$ , the sequence $f^{-1}(y),f^{-2}(y),f^{-3}(y),\ldots $ visits B and, hence, is not confined to C. Let $m(y)$ denote the least positive integer k such that $f^{-k}(y)\notin C$ . Then, $f^{-m(y)}(y)\in A$ . For any integer $k\ge 1$ , let $A_k=\{x\in A\mid p(x)=k\}$ and $B_k=\{y\in B\mid m(y)=k\}$ . The sets $A_1,A_2,A_3,\ldots $ form a partition of A, while the sets $B_1,B_2,B_3,\ldots $ form a partition of B. By construction, $f^k(A_k)\subset B_k$ and $f^{-k}(B_k)\subset A_k$ for all k, which implies that $f^k(A_k)=B_k$ for all k.

Now, we define a map $h:X\to X$ by $h(x)=f^{p(x)}(x)$ if $x\in A$ , $h(x)=f^{-m(x)}(x)$ if $x\in B$ , and $h(x)=x$ otherwise. For any $k\ge 1$ , the map h coincides with $f^k$ on $A_k$ and with $f^{-k}$ on $B_k$ . Since $f^k(A_k)=B_k$ for all k, it follows that h is an involution. In addition, $h(A)=B$ .

Notice that all sets of the form $A_k$ and $B_k$ are clopen. Indeed, $A_k$ is the intersection of clopen sets A, $f^{-k}(B)$ and $f^{-i}(C)$ , $1\le i\le k-1$ . Likewise, $B_k$ is the intersection of clopen sets B, $f^k(A)$ and $f^i(C)$ , $1\le i\le k-1$ . Since the set A, which is compact, is the disjoint union of clopen sets $A_1,A_2,A_3,\ldots, $ all but finitely many of those sets have to be empty. Likewise, all but finitely many of the sets $B_1,B_2,B_3,\ldots $ are empty. As a consequence, the map h is piecewise an element of the group G. Since h is invertible, we conclude that h belongs to the group $\mathsf {F}(G)$ , which coincides with G as G is ample.

Proof. Suppose a group $G\subset {\mathrm {Homeo}}(X)$ is ample and has Property UC. To establish Property CI, we are going to show that the group G satisfies conditions (CI-1) and (CI-2) in Lemma 5.9. We begin with condition (CI-1). Take any disjoint non-empty clopen sets $A,B,C\subset X$ such that $f(A\cup C)=B\cup C$ for some $f\in G$ . First, consider the case where the image $f(A)$ intersects B. Then, $A\cap f^{-1}(B)$ is a non-empty clopen set. Property UC implies that $g(C)\subset A\cap f^{-1}(B)$ for some $g\in G$ . Let $O_1=g(C)$ and $O_2=f(O_1)$ . Then, $O_1$ and $O_2$ are non-empty clopen subsets of respectively A and B. In particular, $O_1$ and $O_2$ are both disjoint from C. Therefore, the generalized $2$ -cycles $\delta _{C;g}$ and $\delta _{C;fg}$ are well defined. Since the group G is ample, it contains $\delta _{C;g}$ and $\delta _{C;fg}$ as well as the map $h=\delta _{C;fg}f\delta _{C;g}$ . Let us find $h(A)$ . The map $\delta _{C;g}$ fixes all points in $A\setminus O_1$ while $O_1$ is mapped onto C. Hence, $\delta _{C;g}$ maps A onto $(A\setminus O_1)\cup C=(A\cup C)\setminus O_1$ . Further, f maps $(A\cup C)\setminus O_1$ onto $(B\cup C)\setminus O_2$ . Finally, $\delta _{C;fg}$ coincides with the identity map on $B\setminus O_2$ while mapping C onto $O_2$ . We conclude that $h(A)=B$ .

Now, consider the case where $f(A)$ is disjoint from B. In this case, $f(A)\subset C$ and also $f^{-1}(B)\subset C$ . Property UC implies that $g_1(C)\subset A$ and $g_2(A)\subset B$ for some maps $g_1,g_2\in G$ . Let $O_1=g_1(C)$ , $O_2=g_2(O_1)$ , and $O_3=f^{-1}(O_2)$ . Then, $O_1$ , $O_2$ , and $O_3$ are non-empty clopen subsets of A, B, and C, respectively. In particular, $O_1$ and $O_2$ are both disjoint from C and $O_3$ . Therefore, the generalized $2$ -cycles $\delta _{C;g_1}$ , $\delta _{C;g_2g_1}$ , and $\delta _{O_1;f^{-1}g_2}$ are well defined. The ample group G contains all these generalized $2$ -cycles as well as the map $h=\delta _{C;g_2g_1}f\delta _{O_1;f^{-1}g_2}\delta _{C;g_1}$ . Let us find $h(A)$ . The map $\delta _{C;g_1}$ fixes all points in $A\setminus O_1$ while $O_1$ is mapped onto C. Hence, $\delta _{C;g_1}$ maps A onto $(A\cup C)\setminus O_1$ . Then, $\delta _{O_1;f^{-1}g_2}$ maps $(A\cup C)\setminus O_1$ onto $(A\cup C)\setminus O_3$ . Further, f maps $(A\cup C)\setminus O_3$ onto $(B\cup C)\setminus O_2$ . Finally, $\delta _{C;g_2g_1}$ coincides with the identity map on $B\setminus O_2$ while mapping C onto $O_2$ . Thus, $h(A)=B$ .

We proceed to verification of condition (CI-2). Consider any disjoint non-empty clopen sets $D,E_0,E\subset X$ such that $f(D\cup E_0)=E_0$ for some $f\in G$ . Let $E_1=f(D)$ and $E_2=f(E_0)$ . Then, $E_1$ and $E_2$ are non-empty clopen sets that partition $E_0$ . Property UC implies that $g(E_0)\subset E$ for some $g\in G$ . Let $U=g(E_0)$ , $U_1=g(E_1)$ , and $U_2=g(E_2)$ . Then, $U_1$ and $U_2$ are non-empty clopen sets that partition U. Now, we define a continuous map $h:X\to X$ that is piecewise an element of the group G. The map h coincides with $gf$ on D, with $f^{-1}$ on $E_0$ , with $gfg^{-1}$ on U, and with the identity map everywhere else. We have $h(D)=g(E_1)=U_1$ , $h(E_0)=D\cup E_0$ and $h(U)=gf(E_0)=g(E_2)=U_2$ . Since the sets $h(D)$ , $h(E_0)$ , and $h(U)$ form a partition of $D\cup E_0\cup U$ , it follows that the map h is invertible and, hence, a homeomorphism. By construction, h belongs to the group $\mathsf {F}(G)$ , which coincides with G as G is ample. Note that $h(D\cup U)=U_1\cup U_2=U$ . In addition, $U\subset E$ and h fixes every point in $E\setminus U$ . Therefore, $h(D\cup E)=E$ . Thus, the set D can be absorbed by E.

Proof. If a clopen set U can be absorbed by a clopen set E under the action of a group $G\subset {\mathrm {Homeo}}(X)$ , then for any $g\in G$ , the set $g(U)$ can be absorbed by $g(E)$ . Indeed, E is disjoint from U and $f(U\cup E)=E$ for some $f\in G$ . Then, for any $g\in G$ , the clopen set $g(E)$ is disjoint from $g(U)$ , the map $h=gfg^{-1}$ belongs to G, and $h(g(U)\cup g(E))=hg(U\cup E)=gf(U\cup E)=g(E)$ .

Now, suppose the group G has Property CI and $U_1,U_2\subset X$ are two absorbable sets. We need to show that $h(U_1)=U_2$ for some $h\in G$ . First, consider the case where both $U_1$ and $U_2$ can be absorbed by the same clopen set E. In this case, $f_1(U_1\cup E)=E$ and $f_2(U_2\cup E)=E$ for some $f_1,f_2\in G$ . Then, $f_2^{-1}f_1\in G$ and $f_2^{-1}f_1(U_1\cup E)=U_2\cup E$ . Since E is disjoint from both $U_1$ and $U_2$ , Property CI implies that $h(U_1)=U_2$ for some $h\in G$ .

If one of the sets $U_1$ and $U_2$ contains the other, it follows from Lemma 5.8 that the smaller set can be absorbed by any clopen set that can absorb the larger one. Next, consider the case where neither of the sets $U_1$ and $U_2$ contains the other, and also $U_1\cup U_2\ne X$ . Then, the clopen set $E=X\setminus (U_1\cup U_2)$ is non-empty, and the unions $U_1\cup E$ and $U_2\cup E$ are both different from X. By Lemma 5.8, both $U_1$ and $U_2$ can be absorbed by E.

It remains to consider the case where $U_1\cup U_2=X$ . Take any clopen set E that can absorb $U_1$ . Then, E is disjoint from $U_1$ and $f(U_1\cup E)=E$ for some $f\in G$ . The set $U=f(U_1)$ is also absorbable. Indeed, $f(U\cup (E\setminus U))=f(E)=E\setminus U$ so that U can be absorbed by $E\setminus U$ . Since U is disjoint from $U_1$ and $U_1\cup U_2=X$ , the set $U_2$ contains U. By the above, $h(U)=U_2$ for some $h\in G$ . Then, $hf\in G$ and $hf(U_1)=U_2$ .

Proof. Suppose $G\in {\mathrm {Homeo}}(X)$ is an ample group that acts minimally on X, admits an absorbable set, and has Property CI. We are going to show that the group G has the following property, which is a weakened form of Property SC: for any non-empty clopen set $W\subset X$ , there exist a clopen set $W_0\subset W$ and a map $f\in G$ such that $f(W_0)$ is a proper subset of $W_0$ . After that, Property UC can be established in the same way as it was done in the proof of Lemma 5.6.

Take any non-empty clopen set $D\subset X$ absorbable under the action of the group G. Let E be any clopen set that can absorb D. Then, E is disjoint from D and $g(D\cup E)=E$ for some $g\in G$ . The set $D_1=g(D)$ , which is a proper subset of E, is absorbable due to Lemma 5.12. Since the group G has Property CI, it follows from Lemma 5.8 that every non-empty clopen subset of D can absorb $D_1$ , every non-empty clopen subset of $D_1$ can absorb D, and every non-empty clopen subset of $X\setminus (D\cup D_1)$ can absorb both D and $D_1$ . Any non-empty clopen set $W\subset X$ intersects at least one of the sets D, $D_1$ , and $X\setminus (D\cup D_1)$ . Hence, W has a clopen subset $W_0$ that can absorb an absorbable set $D_0$ , where $D_0=D$ or $D_1$ . Then, $W_0$ is disjoint from $D_0$ and $f(D_0\cup W_0)=W_0$ for some $f\in G$ . Since $W_0$ is a proper subset of $D_0\cup W_0$ , it follows that $f(W_0)$ is a proper subset of $W_0$ .

Question 5.14. Let G be an ample group that acts minimally on a Cantor set.

• • Suppose the group G admits contraction of a clopen set. Is it possible that G does not have Property UC?

• • Suppose the group G does not allow contraction of any clopen set. Is it possible that G does not have Property NC?

Property HS. (Homologous sets can be mapped onto each other)

Whenever the characteristic functions of two clopen sets $A,B\subset X$ different from the empty set and X represent the same homology class in $H_0(G)$ , we have $g(A)=B$ for some $g\in G$ .

Proof. If A, B, and C are clopen subsets of X such that C is disjoint from both A and B, then $\chi _{A\cup C}-\chi _{B\cup C}=\chi _A-\chi _B$ . Hence, for any group $G\subset {\mathrm {Homeo}}(X)$ , the characteristic functions $\chi _{A\cup C}$ and $\chi _{B\cup C}$ represent the same homology class in $H_0(G)$ if and only if the characteristic functions $\chi _A$ and $\chi _B$ represent the same homology class in $H_0(G)$ . It follows that Property HS implies Property CI.

Property E. (Entanglement)

Whenever clopen sets $U_1$ and $U_2$ intersect, the local subgroup $\mathsf {F}_{U_1\cup U_2}(G)$ is generated by the union of local subgroups $\mathsf {F}_{U_1}(G)\cup \mathsf {F}_{U_2}(G)$ .

Proof. Clearly, the entire group $\mathcal {G}$ is the stabilizer of X and of the empty set. If Y is not empty, we can find a point $x\in Y$ . Since the group $\mathcal {G}$ acts minimally on X, the orbit ${\mathrm {Orb}}_{\mathcal {G}}(x)$ is dense in X. If $Y\ne X$ , then the open set $X\setminus Y$ is not empty. Hence, ${\mathrm {Orb}}_{\mathcal {G}}(x)$ has a point in $X\setminus Y$ . This means that $g(x)\notin Y$ for some $g\in \mathcal {G}$ . Then, $g\notin {\mathrm {St}}_{\mathcal {G}}(Y)$ so that ${\mathrm {St}}_{\mathcal {G}}(Y)\ne \mathcal {G}$ .

Proof. For any point $z\in Z$ , we can find a clopen neighborhood $U_z$ that contains neither x nor y. Since the set Z is closed and hence compact, it can be covered by finitely many clopen sets of the form $U_z$ , $z\in Z$ . Taking the union of all sets in that finite cover, we obtain a clopen set U containing Z and disjoint from $\{x,y\}$ . Next, we can find disjoint clopen sets $V_x,V_y\subset X\setminus U$ such that $x\in V_x$ and $y\in V_y$ . Since the points x and y are in the same orbit of the group $\mathcal {G}$ , we have $y=g(x)$ for some $g\in \mathcal {G}$ . Then, $W=V_x\cap g^{-1}(V_y)$ is a clopen neighborhood of x. The set W is disjoint from $g(W)$ as $W\subset V_x$ , while $g(W)\subset V_y$ . Therefore, the generalized $2$ -cycle $f=\delta _{W;g}$ is defined. Since $g\in \mathcal {G}$ , the map f belongs to the group $\mathsf {F}(\mathcal {G})=\mathcal {G}$ . By construction, $f(x)=g(x)=y$ and $f(y)=g^{-1}(y)=x$ . All points in the complement of the set $W\cup g(W)$ are fixed by f. Since $W\cup g(W)\subset V_x\cup V_y\subset X\setminus U$ and $Z\subset U$ , it follows that $f(z)=z$ for all $z\in Z$ .

Proof. Recall that ${\mathrm {St}}_{\mathcal {G}}(X\setminus E)={\mathrm {St}}_{\mathcal {G}}(E)$ for any set $E\subset X$ . If $f:X\to X$ is a homeomorphism, then $f(\overline {E})=\overline {f(E)}$ . Hence, $f(\overline {E})=\overline {E}$ whenever $f(E)=E$ . As a consequence, ${\mathrm {St}}_{\mathcal {G}}(E)\subset {\mathrm {St}}_{\mathcal {G}}(\overline {E})$ .

Assume that sets $Y_1,Y_2\subset X$ satisfy conditions $Y_1=\overline {X\setminus Y_2}$ and $Y_2=\overline {X\setminus Y_1}$ . By the above, ${\mathrm {St}}_{\mathcal {G}}(Y_2)={\mathrm {St}}_{\mathcal {G}}(X\setminus Y_2)\subset {\mathrm {St}}_{\mathcal {G}}(\overline {X\setminus Y_2})={\mathrm {St}}_{\mathcal {G}}(Y_1)$ . Likewise, ${\mathrm {St}}_{\mathcal {G}}(Y_1)\subset {\mathrm {St}}_{\mathcal {G}}(Y_2)$ . Hence, ${\mathrm {St}}_{\mathcal {G}}(Y_1)={\mathrm {St}}_{\mathcal {G}}(Y_2)$ . Further, either of the two assumed conditions implies that $Y_1\cup Y_2=X$ . The first condition also implies that no interior point of $Y_2$ belongs to $Y_1$ . Likewise, the second of the two implies that no interior point of $Y_1$ belongs to $Y_2$ . Therefore, any common point of $Y_1$ and $Y_2$ is their common boundary point: $Y_1\cap Y_2\subset \partial Y_1\cap \partial Y_2$ . Since the sets $Y_1$ and $Y_2$ are clearly closed, they contain their own boundaries so that $Y_1\cap Y_2=\partial Y_1\cap \partial Y_2$ . Conversely, assume $Y_1,Y_2\subset X$ are closed sets such that $Y_1\cup Y_2=X$ and $Y_1\cap Y_2=\partial Y_1\cap \partial Y_2$ . Then, X is a disjoint union of three sets $Y_1\setminus Y_2$ , $Y_2\setminus Y_1$ , and $Y_1\cap Y_2$ . Moreover, the sets $Y_1\setminus Y_2=X\setminus Y_2$ and $Y_2\setminus Y_1= X\setminus Y_1$ are open. It follows that $Y_1\setminus Y_2$ and $Y_2\setminus Y_1$ are disjoint from both $\partial Y_1$ and $\partial Y_2$ . Hence, $\partial Y_1\cup \partial Y_2\subset Y_1\cap Y_2$ , which implies that $\partial Y_1=\partial Y_2=Y_1\cap Y_2$ . Since $\partial Y_1=\partial (X\setminus Y_1)=\partial (Y_2\setminus Y_1)$ , we obtain that $\overline {X\setminus Y_1}=\overline {Y_2\setminus Y_1}=(Y_2\setminus Y_1)\cup \partial (Y_2\setminus Y_1)=(Y_2\setminus Y_1)\cup (Y_1\cap Y_2)=Y_2$ . Likewise, $\overline {X\setminus Y_2}=Y_1$ .

It remains to show that ${\mathrm {St}}_{\mathcal {G}}(Y_1)\ne {\mathrm {St}}_{\mathcal {G}}(Y_2)$ for any distinct closed sets $Y_1,Y_2\subset X$ that do not satisfy at least one of the conditions $Y_1\cup Y_2=X$ and $Y_1\cap Y_2=\partial Y_1\cap \partial Y_2$ . First, we consider the case where $Y_1\cup Y_2\ne X$ . Then, $U=X\setminus (Y_1\cup Y_2)$ is a non-empty open set. Since $Y_1\ne Y_2$ , there is a point $x\in X$ that belongs to one of these sets but not to the other. We may assume without loss of generality that $x\in Y_1$ and $x\notin Y_2$ . Since the group $\mathcal {G}$ acts minimally on X, the orbit ${\mathrm {Orb}}_{\mathcal {G}}(x)$ is dense in X. In particular, it has a point y in U. Clearly, $y\notin Y_1$ and $y\notin Y_2$ . By Lemma 6.3, there exists a map $f\in \mathcal {G}$ such that $f(x)=y$ and $f(z)=z$ for all $z\in Y_2$ . Then, $f\in {\mathrm {St}}^\star _{\mathcal {G}}(Y_2)\subset {\mathrm {St}}_{\mathcal {G}}(Y_2)$ . However, $f\notin {\mathrm {St}}_{\mathcal {G}}(Y_1)$ as $x\in Y_1$ , while $y\notin Y_1$ . Therefore, ${\mathrm {St}}_{\mathcal {G}}(Y_1)\ne {\mathrm {St}}_{\mathcal {G}}(Y_2)$ .

Now, consider the case where $Y_1\cup Y_2=X$ , but there exists a point $x\in Y_1\cap Y_2$ that is not a boundary point for at least one of the sets $Y_1$ and $Y_2$ . We may assume without loss of generality that x is an interior point of $Y_2$ . Furthermore, we may assume that $Y_1\ne X$ as otherwise, ${\mathrm {St}}_{\mathcal {G}}(Y_2)\ne \mathcal {G}={\mathrm {St}}_{\mathcal {G}}(Y_1)$ due to Lemma 6.2. Then, $X\setminus Y_1$ is a non-empty open set. Just as above, minimality of the action of the group $\mathcal {G}$ on X implies that the orbit ${\mathrm {Orb}}_{\mathcal {G}}(x)$ has a point y in $X\setminus Y_1$ . Note that $X\setminus Y_1\subset Y_2$ since $Y_1\cup Y_2=X$ . Therefore, every point of $X\setminus Y_1$ (including y) is an interior point of $Y_2$ . Hence, the closed set $\overline {X\setminus Y_2}$ contains neither x nor y. By Lemma 6.3, there exists a map $f\in \mathcal {G}$ such that $f(x)=y$ and $f(z)=z$ for all $z\in \overline {X\setminus Y_2}$ . Then, $f\in {\mathrm {St}}_{\mathcal {G}}(X\setminus Y_2)={\mathrm {St}}_{\mathcal {G}}(Y_2)$ . However, $f\notin {\mathrm {St}}_{\mathcal {G}}(Y_1)$ as $x\in Y_1$ , while $y\notin Y_1$ . Thus, ${\mathrm {St}}_{\mathcal {G}}(Y_1)\ne {\mathrm {St}}_{\mathcal {G}}(Y_2)$ .

Proof. If $Y_2=X$ , then ${\mathrm {St}}_{\mathcal {G}}(Y_1)\ne \mathcal {G}={\mathrm {St}}_{\mathcal {G}}(Y_2)$ due to Lemma 6.2. Otherwise, $Y_1\cup Y_2=Y_2\ne X$ . Then, ${\mathrm {St}}_{\mathcal {G}}(Y_1)\ne {\mathrm {St}}_{\mathcal {G}}(Y_2)$ due to Lemma 6.4.

Proof. Since the group H does not act minimally on X, it admits a closed invariant set Y different from the empty set and X. We have $h(Y)=Y$ for any $h\in H$ . Hence, $H\subset {\mathrm {St}}_{\mathcal {G}}(Y)$ . Lemma 6.2 implies that ${\mathrm {St}}_{\mathcal {G}}(Y)\ne \mathcal {G}$ . As H is a maximal subgroup of $\mathcal {G}$ , it follows that $H={\mathrm {St}}_{\mathcal {G}}(Y)$ .

Now, assume that a group $H={\mathrm {St}}_{\mathcal {G}}(Y)$ , where Y is a closed subset of X, is indeed a maximal subgroup of $\mathcal {G}$ . To prove that the induced action of H on Y is minimal, we need to show that any closed set $Y_0\subset Y$ invariant under it coincides with the empty set or Y. We have $h(Y_0)=Y_0$ for any $h\in H$ . Hence, $H\subset {\mathrm {St}}_{\mathcal {G}}(Y_0)$ . Note that $Y\ne X$ since $H\ne \mathcal {G}$ . Then, $Y_0\ne X$ as well. Assume that $Y_0\ne \emptyset $ . Just as above, Lemma 6.2 implies that ${\mathrm {St}}_{\mathcal {G}}(Y_0)\ne \mathcal {G}$ and then maximality of H implies that $H={\mathrm {St}}_{\mathcal {G}}(Y_0)$ . Since ${\mathrm {St}}_{\mathcal {G}}(Y_0)={\mathrm {St}}_{\mathcal {G}}(Y)$ and $Y_0\subset Y$ , it follows from Lemma 6.5 that $Y_0=Y$ .