Proof. By Kechris [Reference Kechris9, Theorem 8.38], we can fix some comeager $G_{\delta }$ set $A\subseteq X$ such that $f\upharpoonright {}A$ is continuous. Since every $\gamma \in \Gamma $ acts on X by a homeomorphism, and hence preserves comeager-ness of A, the intersection $A'=\bigcap _{\gamma \in \langle S\rangle }\gamma \cdot A\subseteq A$ over the countable subgroup $\langle S\rangle $ generated by S is also comeager $G_\delta $ . Then $f\upharpoonright {}A'$ is also continuous, and $A'$ is $E_G$ -invariant.

For each $x\in A'$ , since $A'$ is $E_{\operatorname {\mathrm {Sch}}(\Gamma ,S,X)}$ -invariant, we have $S\cdot x\subseteq A'$ and so $f\upharpoonright {}A'$ is continuous over $S\cdot x$ . The function $g:S\to \omega $ defined by $g(s)=f(s\cdot x)$ is continuous as it is a composition of the continuous functions $s\mapsto s\cdot x$ and $f\upharpoonright {}A'$ . Since S is compact, its continuous image $g[S]\subseteq \omega $ must be finite. Thus we have shown that for all x inside the comeager set $A'$ , the set $f[N_G(x)]=f[S\cdot x]=g[S]\subseteq \omega $ is finite, as desired.

Proof. Let $\{U_i:i<\omega \}$ be an open neighborhood basis at $1_\Gamma \in \Gamma $ . Fix a sequence $K_0,K_1,\dots $ of compact normal subgroups of G, such that $K_i\subseteq U_i$ and $\Gamma /K_i$ is a Lie group for all $i<\omega $ . The group is still a Lie group for $i<\omega $ , since it embeds as a closed subgroup into the Lie group $(\Gamma /K_0)\times \dots \times (\Gamma /K_i)$ . The inverse system $\langle \Gamma _i\rangle $ is as desired.

Proof. By the Katětov–Morita theorem [Reference Engelking6, Theorem 7.3.3], the three topological dimensions of a separable metrizable space agree. We will finish by showing that $\dim (\Gamma )=d(\Gamma )$ , where $d(\Gamma )$ denotes the Lebesgue covering dimension of $\Gamma $ .

First we’ll show for all finite $n<\omega $ , if $\dim (\Gamma )\le n$ then $d(\Gamma )\le n$ . If $\dim (\Gamma )\le n$ , the each $\Gamma _i$ is a manifold of dimension $\le n$ , and so each $\Gamma _i$ has Lebesgue covering dimension $d(\Gamma _i)\le n$ . By Nagami’s theorem [Reference Engelking6, Exercise 7.3.I], their inverse limit $\Gamma =\varprojlim _i\Gamma _i$ also has $d(\Gamma )\le n$ .

Next we’ll show for all finite $n<\omega $ , if $\dim (\Gamma )\ge n$ then $d(\Gamma )\ge n$ , which completes the proof. If $\dim (\Gamma )\ge n$ , then there is some $\Gamma _i$ whose manifold dimension $d_i$ satisfies $d_i\ge n$ . For each $j<\omega $ , let $\exp _j:\mathfrak {g}_j\to \Gamma _j$ be the exponential map associated with $\Gamma _j$ . One can check that taking the inverse limit as $j\to \infty $ gives a well-defined exponential map $\exp :\mathfrak {g}\to \Gamma $ , and that the projection $\pi :\Gamma \to \Gamma _i$ induces a projection $\pi _*:\mathfrak {g}\to \mathfrak {g}_i$ . Note that $\mathfrak {g}$ and $\mathfrak {g}_i$ are linearly isomorphic to $\mathfrak {g}\cong \mathbb {R}^{\dim (\Gamma )}$ and $\mathfrak {g_i}\cong \mathbb {R}^{d_i}$ , respectively.

Let $U\subseteq \mathfrak {g}_i$ be an open neighborhood of $0\in \mathfrak {g}_i$ such that the exponential map $\exp _i:\mathfrak {g}_i\to \Gamma _i$ induces a homeomorphism $U\cong \exp _i(U)$ . We lift U linearly under the projection $\pi _*:\mathfrak {g}\to \mathfrak {g_i}$ to a $d_i$ -manifold $0\in V\subseteq \mathfrak {g}$ such that $\pi _*:V\cong U$ is a homeomorphism. The composite map $\pi \circ \exp ={\exp _i}\circ {\pi _*}:\mathfrak {g}\to \Gamma _i$ thus induces a homeomorphism $V\cong \exp _i(U)$ , and so its factor map $\pi $ induces the homeomorphism $\exp (V)\cong \exp _i(U)$ .

Since $U\subseteq \mathfrak {g}_i\cong \mathbb {R}^{d_i}$ has closed subsets of Lebesgue covering dimension $d_i$ , its homeomorphic copy $\exp (V)\subseteq \Gamma $ also has closed subsets of Lebesgue covering dimension $d_i$ . Since the Lebesgue covering dimension is hereditary to closed subsets [Reference Engelking6, Theorem 7.1.8], we see that $\Gamma \supseteq \exp (V)$ has Lebesgue covering dimension $d(\Gamma )\ge d_i\ge n$ as desired.

Proof. Since $\Gamma $ is an inverse limit of Lie groups of dimension at most $\dim (\Gamma )<\infty $ , it suffices to show that every d-dimensional Lie group G has such a basis with $M_G=10^d$ .

Since G is a Lie group, it admits left-invariant Riemannian metrics, and so it also admits a complete left-invariant metric $d_G$ such that for any $\varepsilon>0$ there exists some open neighborhood $1_G\in \Omega \subseteq G$ of the identity and a $(1+\varepsilon )$ -bi-Lipschitz homeomorphism $f:D\cong \Omega $ from some open Euclidean domain $D\subseteq \mathbb {R}^d$ to $\Omega $ .

Then, for all sufficiently small $r>0$ , the open ball $U=B_{d_G}(1_G,r)\subseteq G$ is such that $U=U^{-1}$ , there is $U^9\subseteq \Omega $ and $f^{-1}[U^9]\subseteq D$ is contained in a ball of radius $9(1+\varepsilon )r$ , and every left translate $\gamma \cdot U\subseteq U^9$ is a radius-r ball $\gamma \cdot U=B_{d_G}(\gamma ,r)$ such that $f^{-1}[\gamma U]\subseteq f^{-1}[U^9]$ contains some ball of radius $(1-\varepsilon )r$ . For sufficiently small $\varepsilon>0$ , a Euclidean ball of radius $9(1+\varepsilon )r$ has volume at most $10^d$ times that of radius $(1-\varepsilon )r$ , which means at most $10^d$ many left translates of U can be packed into $U^9=(UU^{-1})^4U$ .

Proof. Fix a metric d on X, and let W be an open set in X such that $A\cup \overline {U}\subseteq W\subseteq \overline {W}\subseteq B$ . We define $V=\{x\in W:d(x,A\cup \overline {U})<d(x,C\smallsetminus U)\}$ as desired.

Proof. We proceed by induction on $r<\omega $ , noting that there is nothing to prove when $r=0$ . Assume we have proved the lemma for a fixed $r<\omega $ , and we will prove next the case with $r+1$ given closed sets $M_0,\dots ,M_r\subseteq X$ . We may assume $\dim (M_0)\le \dots \le \dim (M_r)$ .

By the inductive hypothesis, let U be an open set such that $A\subseteq U\subseteq \overline {U}\subseteq B$ , and such that $\dim (\partial U\cap M_i)<\dim (M_i)$ for $0\le i\le r-1$ . Let $M'=M_0\cup \dots \cup M_{r-1}$ . We will next work in the subspace $Y=(M'\cup M_r)\smallsetminus (\partial U\cap M')$ . Note that by the countable sum theorem [Reference Engelking6, Theorem 7.2.1] and by that the topological dimension is hereditary to subspaces [Reference Engelking6, Theorem 7.1.1], we have $\dim (Y)\le \dim (M'\cup M_r)\le \dim (M_r)$ and $\dim (\partial U\cap M')<\dim (M_r)$ .

Let $Z=U\cap M'$ , and note that Z is relatively closed in Y since $Z=\overline {U}\cap M'\cap Y$ . Then we have $(A\cap Y)\cup Z\subseteq U\cap Y$ , where $(A\cap Y)\cup Z$ is relatively closed in Y, and $U\cap Y$ is relatively open in Y. By definition of the large inductive dimension [Reference Engelking6, Section 7.1], there exists a set $V\subseteq Y$ relatively open in Y such that $(A\cap Y)\cup Z\subseteq V\subseteq U\cap Y$ , and $\dim (\partial ^Y V)<\dim (Y)\le \dim (M_r),$ where $\partial ^Y V=(\overline {V}\smallsetminus V)\cap Y$ is the relative boundary of V in Y.

Since $\partial U\cap M'$ is closed, Y is relatively open in $M'\cup M_r$ . Let $C=M'\cup M_r$ . Since V is relatively open in Y, we see that V is also relatively open in C, and we see that the relative boundary of V in C is $\partial ^{C}V\subseteq (\partial ^Y V)\cup (\partial U\cap M')$ . We have the following by the countable sum theorem [Reference Engelking6, Theorem 7.2.1], since $\partial ^Y V$ is $F_\sigma $ and $\partial U\cap M'$ is closed:

$$\begin{align*}\dim(\partial^C V\cap M_r)\le \dim(\partial^C V)\le \max\{\dim(\partial^Y V),\dim(\partial U\cap M')\}<\dim(M_r)\end{align*}$$

We also have the following:

$$\begin{align*}A\cap C=A\cap U\cap C=A\cap Y\subseteq V\subseteq\overline{V}\subseteq\overline{U\cap Y}\subseteq\overline{U}\cap C\subseteq B\cap C\end{align*}$$

Then, since $U\cap M'=Z\subseteq V\cap M'$ and $\overline {V}\cap M'\subseteq \overline {U}\cap C\cap M'=\overline {U}\cap M'$ , we have the inclusion $\partial ^{C}V\cap M'\subseteq \partial U\cap M'$ , which means by the inductive hypothesis that for all $0\le i\le r-1$ , we have $\dim (\partial ^C V\cap M_i)\le \dim (\partial U\cap M_i)<\dim (M_i)$ .

Finally, by Lemma 2.7, there is an open set $W\subseteq X$ extending $V\subseteq C$ such that $A\subseteq W\subseteq \overline {W}\subseteq B$ and $\partial W\cap C=\partial ^C V$ . This W completes the inductive step for the case $r+1$ as desired.

Proof. Since every open set is $F_\sigma $ , there is a countable closed covering $\{A_i:i<\omega \}$ refining the open covering $\{U_i:i<\omega \}$ . We can fix a function $a:\omega \to \omega $ such that $A_i\subseteq U_{a(i)}$ for all $i<\omega $ .

The first step of this proof is to construct a sequence $\langle V_i:i<\omega \rangle $ of open sets $V_i$ inductively on $i<\omega $ , such that in constructing each set $V_i$ we ensure that $A_i\subseteq V_i\subseteq \overline {V_i}\subseteq U_{a(i)}$ , and that for every $f\in F$ and $S\subseteq \{0,\dots ,i-1\}$ , if the closed set $M_{f,S}=\bigcap _{j\in S}fF^{-1}\cdot \partial V_j$ is nonempty, then $\dim (\partial V_i\cap M_{f,S})<\dim (M_{f,S})$ . (Note that when $S=\varnothing $ , the last condition just says $\dim (\partial V_i)<\dim (\Gamma )=d$ .) Since for each $i<\omega $ , the number of such pairs $(f,S)$ is finite, the construction of $\langle V_i:i<\omega \rangle $ follows from an application of Lemma 2.8.

We now prove by induction on $0\le t\le d+1$ that for every sequence $s_0<\dots <s_{t-1}<\omega $ , we have $\dim (F^{-1}\cdot \partial V_{s_0}\cap \dots \cap F^{-1}\cdot \partial V_{s_{t-1}})\le d-t$ . When $t=0$ , this says $\dim (\Gamma )\le d$ . In the inductive step going from case t to case $t+1$ , we have the following:

$$ \begin{align*} \dim\left(\bigcap_{{i\le t}}F^{-1}\partial V_{s_i}\right)&=\dim\left(\bigcup_{{f\in F}}\left(\bigcap_{{i\le t-1}}F^{-1}\partial V_{s_i}\cap f^{-1}\partial V_{s_t}\right)\right)\\ &\le \max_{f\in F}\dim\left(\bigcap_{{i\le t-1}}F^{-1}\partial V_{s_i}\cap f^{-1}\partial V_{s_t}\right)\\ &=\max_{f\in F}\dim\left(\bigcap_{i\le t-1}fF^{-1}\partial V_{s_i}\cap \partial V_{s_t}\right)\\ &\le\max_{f\in F}\max\left\{\dim\left(\bigcap_{i\le t-1}fF^{-1}\partial V_{s_i}\right)-1,\;-1\right\}\\ &=\max\left\{\dim\left(\bigcap_{i\le t-1}F^{-1}\partial V_{s_i}\right)-1,\;-1\right\}\\ &\le d-t-1. \end{align*} $$

Here we used the sum theorem for dimension [Reference Engelking6, Theorem 7.2.1], that F act on $\Gamma $ by homeomorphisms, the construction of $V_{s_t}$ , and the inductive hypothesis. Thus in particular when $t=d+1$ , we get that for every sequence $s_0<\dots <s_d<\omega $ , we have $F^{-1}\partial V_{s_0}\cap \dots \cap F^{-1}\partial V_{s_d}=\varnothing $ .

Next, we claim that any right translate $F\cdot \gamma $ of F can intersect $\bigcup _{i<\omega }\partial V_i$ at most d times. If not, then there exists a $\gamma \in \Gamma $ and distinct elements $f_0,\dots ,f_d\in F$ such that $f_i\cdot \gamma \in \bigcup _{s<\omega }\partial V_s$ for all $i\le d$ . For each i, we can pick some $s_i<\omega $ such that $f_i\cdot \gamma \in \partial V_{s_i}\subseteq U_{a(s_i)}$ . Then for $i\ne j$ , since $f_jf_i^{-1}U_{a(s_i)}\cap U_{a(s_i)}=\varnothing $ by assumption and $f_j\cdot \gamma \in f_jf_i^{-1}U_{a(s_i)}\cap U_{a(s_j)}\ne \varnothing $ , we have $a(s_i)\ne a(s_j)$ , and hence $s_0,\dots ,s_d$ are all distinct. Then $\gamma \in F^{-1}\partial V_{s_0}\cap \dots \cap F^{-1}\partial V_{s_d}\ne \varnothing $ , which contradicts our earlier arguments.

Let $W_i=V_i\smallsetminus \bigcup _{j<i}\overline {V_j}$ for $i<\omega $ . Since $\bigcup _{i<\omega }V_i\supseteq \bigcup _{i<\omega }A_i=\Gamma $ , we see that $\bigcup _{i<\omega }W_i\cup \bigcup _{i<\omega }\partial V_i=\Gamma $ , while $\{W_i:i<\omega \}$ is a family of pairwise-disjoint open sets. For $i<\omega $ , if we let $R_i=\bigcup _{a(j)=i}W_j$ , then $\mathcal {R}=\{R_i:i<\omega \}$ is a family of pairwise-disjoint open sets such that $R_i\subseteq U_i$ for all i. The previous argument shows that every right translate $F\cdot \gamma $ of F intersects the set $\bigcup _{i<\omega }R_i\supseteq \Gamma \smallsetminus \bigcup _{i<\omega }\partial V_i$ at least $n-d$ times, and a similar argument as before using our assumptions on $U_i$ shows that in each of these $n-d$ times, $F\cdot \gamma $ must intersect a distinct $R_i\in \mathcal {R}$ . The family $\mathcal {R}=\{R_i:i<\omega \}$ is as desired.

Proof. Since $\Gamma $ admits a two-sided invariant metric, $1_\Gamma $ is not a limit point of the conjugacy classes of $FF^{-1}\smallsetminus \{1_\Gamma \}$ . We can fix an open neighborhood U of the identity $1_\Gamma $ such that $(UU^{-1})^4$ is disjoint from the conjugacy classes of $FF^{-1}\smallsetminus \{1_\Gamma \}$ , and at most M left translates $\gamma \cdot U$ can be packed into $(UU^{-1})^4U$ .

Let $S\subseteq \Gamma $ be a maximal set such that the family $\{s\cdot U:s\in S\}$ is pairwise-disjoint. The family $\mathcal {B}=\{s\cdot UU^{-1}:s\in S\}$ is an open cover of $\Gamma $ , since for all $x\in \Gamma $ , if $s\in S$ is such that $sU\cap xU\ne \varnothing $ then $x\in sUU^{-1}\in \mathcal {B}$ . We’ll next check that $\mathcal {B}$ has our desired properties.

Let $V=sUU^{-1}\in \mathcal {B}$ and $g\in FF^{-1}\smallsetminus \{1_\Gamma \}$ . Then $g\cdot sUU^{-1}\cap sUU^{-1}=\varnothing $ follows from our assumption that $s^{-1}gs\notin (UU^{-1})^2\subseteq (UU^{-1})^4$ . This means $(FF^{-1}\smallsetminus \{1_\Gamma \})\cdot V\cap V=\varnothing $ .

Fix $\gamma \in \Gamma $ , and let $S_\gamma \subseteq S$ be the set of all $s\in S$ for which $\gamma UU^{-1}\cap sUU^{-1}\ne \varnothing $ . Then we see that $\gamma ^{-1}s\in (UU^{-1})^2$ for all $s\in S_\gamma $ , and since $S_\gamma \subseteq S$ , the family $\{\gamma ^{-1}sU:s\in S_\gamma \}$ is a pairwise-disjoint family of left translates of U which are subsets of $(UU^{-1})^2U\subseteq (UU^{-1})^4U$ , so by assumption we get $\mathopen |S_\gamma \mathclose |\le M$ . Since every $V\in \mathcal {B}$ is a left translate of $UU^{-1}$ , we get that every left translate of V also intersects at most M members $sUU^{-1}\in \mathcal {B}$ .

Let $V=xUU^{-1}\in \mathcal {B}$ , and let $f,g\in F$ be distinct such that there is some $A=sUU^{-1}\in \mathcal {B}$ for which both $fV\cap A\ne \varnothing $ and $gV\cap A\ne \varnothing $ . From $fxUU^{-1}\cap sUU^{-1}\ne \varnothing $ we get $x^{-1}f^{-1}s\in (UU^{-1})^2$ , and similarly $x^{-1}g^{-1}s\in (UU^{-1})^2$ . Then $s^{-1}gf^{-1}s=(x^{-1}g^{-1}s)^{-1}(x^{-1}f^{-1}s)\in (UU^{-1})^4$ , which contradicts that $gf^{-1}\in FF^{-1}\smallsetminus \{1_\Gamma \}$ and that $(UU^{-1})^4$ is disjoint from the conjugacy classes of $FF^{-1}\smallsetminus \{1_\Gamma \}$ . Thus such $f,g\in F$ cannot exist.

Finally assume $V=xUU^{-1}\in \mathcal {B}$ and $W=yUU^{-1}\in \mathcal {B}$ . By a similar argument as the above, we see that if both $FV$ and $FW$ intersect a same member $A\in \mathcal {B}$ , then we must have $y\in S\cap F^{-1}Fx(UU^{-1})^4$ . For each fixed $g\in F^{-1}F$ , the collection of all $y\in S\cap gx(UU^{-1})^4$ gives a collection of left translates $(gx)^{-1}yU$ of U which are pairwise-disjoint subsets of $(UU^{-1})^4U$ , and there are at most M such y. Since there are at most $\mathopen |F\mathclose |^2$ many such $g\in F^{-1}F$ , we see that there are at most $M\cdot \mathopen |F\mathclose |^2$ many $W=yUU^{-1}\in \mathcal {B}$ for which both $FV$ and $FW$ intersect a same member of $\mathcal {B}$ .

Proof. The proof strategy is to invoke the Lovász local lemma. Towards that goal, we will first pick $c:\mathcal {B}\to \{0,\dots ,k-1\}$ independently and uniformly randomly at each point $B\in \mathcal {B}$ . We will define a family $\mathcal {A}$ of bad events we wish c to avoid, and we will check that the probability upper-bound $p(\mathcal {A})$ and the dependency degree $d(\mathcal {A})$ satisfy $ep(\mathcal {A})(d(\mathcal {A})+1)\le 1$ . Each bad event $A\in \mathcal {A}$ will only depend on finitely many values of c, which means it is clopen in the compact product space $k^{\mathcal {B}}$ . So a compactness argument implies that the Lovász local lemma holds for $\mathcal {A}$ , and there exists a function c avoiding the bad events in $\mathcal {A}$ . Finally, we will show that such a function c is as we desired.

For a set $X\subseteq \Gamma $ , define $\mathcal {B}\upharpoonright {}X=\{B\in \mathcal {B}:B\cap X\ne \varnothing \}$ . For each $B\in \mathcal {B}$ , we define the event $A(B)$ over the domain $\mathcal {B}\upharpoonright {} F\cdot B$ of size $\le M\cdot \mathopen |F\mathclose |$ , where $c\in A(B)$ if for some $i<n$ and $j<k$ , there are only d or less $f\in F_i$ for which $c[\mathcal {B}\upharpoonright {}fB]=\{j\}$ . Let $\mathcal {A}=\{A(B):B\in \mathcal {B}\}$ .

We give an upper bound of the probability $p(\mathcal {A})$ . Note that if $c\in A(B)$ for some $B\in \mathcal {B}$ , then there exists $i<n$ , $j<k$ , and some size-d subset $D\subseteq F_i$ , such that all elements $f\in F_i$ for which $c[\mathcal {B}\upharpoonright {}fB]=\{j\}$ are contained within D. For each $f\in F_i\smallsetminus D$ , since $\mathopen |\mathcal {B}\upharpoonright {}fB\mathclose |\le M$ , the probability that c does not evaluate to j over all of $\mathcal {B}\upharpoonright {}fB$ is at most $1-k^{-M}$ . Thus as our assumption on $\mathcal {B}$ implies that $\mathcal {B}\upharpoonright {}fB$ for all $f\in F$ are pairwise-disjoint subsets of $\mathcal {B}$ , the probability for a fixed D that all of $f\in F_i\smallsetminus D$ satisfy $c[\mathcal {B}\upharpoonright {}fB]\ne \{j\}$ is at most $(1-k^{-M})^{\mathopen |F_i\smallsetminus D\mathclose |}$ . By the union bound, we have the following:

$$\begin{align*}\mathbb{P}[c\in A(B)]\le\sum_{i<n}\sum_{j<k}\binom{\mathopen|F_i\mathclose|}{d}\left(1-k^{-M}\right)^{\mathopen|F_i\mathclose|-d}\le kn\cdot N^d\left(1-k^{-M}\right)^{N-d}. \end{align*}$$

Thus we see that $p(\mathcal {A})\le knN^d(1-k^{-M})^{N-d}$ .

Next, note that two events $A(B)$ and $A(B')$ have intersecting domains exactly when $FB$ and $FB'$ intersect a same member of $\mathcal {B}$ . By our assumption on $\mathcal {B}$ , every B has at most $M\cdot \mathopen |F\mathclose |^2$ many such $B'$ , including B itself, and so we see $d(\mathcal {A})+1\le M\cdot \mathopen |F\mathclose |^2\le Mn^2N^2$ .

By our bounds on $p(\mathcal {A}),d(\mathcal {A})$ , we find that we have made N large enough specifically such that $ep(\mathcal {A})(d(\mathcal {A})+1)\le 1$ , and the Lovász local lemma applies to give a function c not lying in any bad event in $\mathcal {A}$ .

This c is as we desired, because for every $\gamma \in \Gamma $ , if $\gamma \in B\in \mathcal {B}$ , then each point $f\gamma \in F\gamma $ is only covered by sets in $\mathcal {B}\upharpoonright {}fB$ , and so $c\notin A(B)$ implies that for all $i<n$ and $j<k$ , there are $>d$ points $f\gamma $ in $F_i\gamma $ which are covered by sets in $\mathcal {B}\upharpoonright {}fB$ for which $c[\mathcal {B}\upharpoonright {}fB]=\{j\}$ .

Proof. Fix an open cover $\mathcal {B}$ as in Lemma 2.10, which applies due to Lemmas 2.5 and 2.6. Fix a function $c:\mathcal {B}\to \{0,\dots ,k-1\}$ as in Lemma 2.11 and a disjoint open family $\mathcal {R}$ as in Theorem 2.9. Inside every right translate $F_i\cdot \gamma $ , apart from the $d=\dim (\Gamma )$ or less points not covered by $\mathcal {R}$ by Theorem 2.9, Lemma 2.11 guarantees at least one more point covered by $\mathcal {R}$ and colored solely in j by c for any fixed color $j<k$ , and in particular its $\mathcal {R}$ -cover’s associated $\mathcal {B}$ -cover (in the sense of Theorem 2.9) is colored in j. Letting $D_j$ be the union of all sets $R\in \mathcal {R}$ whose associated cover in $\mathcal {B}$ is colored in j by c finishes the proof.

Proof. Since every compact Polish group admits a two-sided invariant metric [Reference Gao8, Exercise 2.1.5], Theorem 2.12 applies. Let $P_0,\dots ,P_{n-1}$ be nonempty perfect subsets of $\Gamma $ , and for each $i<n$ let $F_i\subseteq P_i$ be any finite subset of size $N=N(2,n)$ as in Theorem 2.12. Then the theorem gives a pair of disjoint open sets $D_0,D_1$ each of which dominates every $F_i$ , and we see that they also dominate every $P_i$ because $F_i\subseteq P_i$ .

Proof. First, note that D dominates P if and only if for every $\gamma \in \Gamma $ there exists $p\in P$ such that $p\cdot \gamma \in D$ , which is also equivalent to $\{p^{-1}D:p\in P\}$ being an open cover of $\Gamma $ . So since $\Gamma $ is compact, there is a finite subcover $\{p^{-1}D:p\in F\}$ for some finite $F\subseteq P$ , and we see that D dominates this finite subset F of P.

For the next part of the lemma, fix a compatible two-sided invariant metric d on $\Gamma $ . Define the function $g:\Gamma \to \mathbb {R}\cup \{\infty \}$ via $g(\gamma )=\max _{f\in F}d(f\cdot \gamma ,\Gamma \smallsetminus D)$ . Note that for each $f\in F$ , the function $\gamma \mapsto d(f\cdot \gamma ,\Gamma \smallsetminus D)$ is continuous in $\gamma $ , and so g as a finite maximum of continuous functions is also continuous. For each $\gamma $ , by assumption there exists some $f\in F$ such that $f\cdot \gamma \in D$ , which means $g(\gamma )\ge d(f\cdot \gamma ,\Gamma \smallsetminus D)>0$ for this f. Consequently g is a continuous function over a compact domain $\Gamma $ whose range is within $(0,\infty ]$ , and so compactness allows us to put a positive lower-bound $r>0$ on $g[\Gamma ]$ .

By definition of g, we see that for all $\gamma \in \Gamma $ there exists some $f\in F$ such that $B_d(f\cdot \gamma ,r)\subseteq D$ . Right-invariance of d gives us $B_d(f\cdot \gamma ,r)=B_d(f,r)\cdot \gamma \subseteq D$ .

Finally, let $U=\{x\in \Gamma :d(x,\Gamma \smallsetminus D)>r/2\}$ , so that $U\subseteq \overline {U}\subseteq D$ . For every $\gamma \in \Gamma $ there is some $f\in F$ such that $B_d(f\cdot \gamma ,r)\subseteq D$ , which means that $f\cdot \gamma \in U$ , which means that U dominates F. Since $F\subseteq P$ , we see U also dominates P.

Proof. Fix a compatible two-sided invariant metric d on $\Gamma $ . By Lemma 2.15, for each $i<n$ , there is a finite set $F_i\subseteq P_i$ and $r>0$ , such that every $\gamma \in \Gamma $ has some $f\in F_i$ for which $B_d(f,r)\cdot \gamma \subseteq U$ . For each $f\in F_i\subseteq P_i$ , we fix a nonempty perfect subset $P_{i,f}\subseteq P_i\cap B_d(f,r)$ , such that every $\gamma \in \Gamma $ has some $f\in F_i$ for which $P_{i,f}\cdot \gamma \subseteq B_d(f,r)\cdot \gamma \subseteq U$ .

Let $\mathcal {P}=\{P_{i,f}:i<n,f\in F_i\}$ , which is a finite collection of nonempty perfect sets. By the open pair property of $\Gamma $ , there are disjoint open sets $D_0,D_1\subseteq \Gamma $ each of which dominates all of $\mathcal {P}$ . Let $A_0=D_0\cap U$ and $A_1=D_1\cap U$ , which we will show are as desired.

Fix $i<n$ , $j\in \{0,1\}$ , and $\gamma \in \Gamma $ , and it remains to show that $P_i\cdot \gamma \cap A_j\ne \varnothing $ . By the above, there exists some $f\in F_i$ for which $P_{i,f}\cdot \gamma \subseteq U$ . Since $D_j$ dominates $P_{i,f}\in \mathcal {P}$ , there is some $p\in P_{i,f}$ such that $p\cdot \gamma \in D_j$ . Then $p\cdot \gamma \in D_j\cap U=A_j$ and also $p\in P_{i,f}\subseteq P_i$ , which means that $p\cdot \gamma \in P_i\cdot \gamma \cap A_j\ne \varnothing $ as desired.

Proof. By the perfect set theorem, for each $i<n$ fix a nonempty perfect set $P_i\subseteq \overline {S_i}$ .

We first construct $\langle D_j:j<\omega \rangle $ which dominates every $P_i$ . Note that $U_0=\Gamma $ is open and dominates every $P_i$ . Inductively, whenever $\{D_0,\dots ,D_{n-1},U_n\}$ is a pairwise-disjoint open family which dominates every $P_i$ , we can apply Lemmas 2.14 and 2.16 to split $U_n$ into disjoint open parts $D_n,U_{n+1}\subseteq U_n$ which still dominate every $P_i$ , and thus we get a pairwise-disjoint open family $\{D_0,\dots ,D_n,U_{n+1}\}$ which dominates every $P_i$ . Carrying on this induction through $n<\omega $ , we get a pairwise-disjoint open family $\{D_j:j<\omega \}$ that dominates every $P_i$ .

It remains to show every $D_j$ also dominates every $S_i$ . For every $\gamma \in \Gamma $ , we have $P_i\gamma \cap D_j\ne \varnothing $ , and every point in $P_i \gamma $ is a limit point of $S_i\gamma $ . So since $D_j$ is open and contains some limit point of $S_i\gamma $ , it also contains some point of $S_i\gamma $ . We get that $S_i\gamma \cap D_j\ne \varnothing $ , which means $D_j$ also dominates $S_i$ as desired.

Proof. Follows directly from Theorems 2.1 and 2.17.

Proof. We split into three cases: The case when $\overline {S}$ is countable and bounded, the case when $\overline {S}$ is uncountable, and the case when S is unbounded.

When $\overline {S}$ is countable and bounded, Theorem 2.1 implies that $\operatorname {\mathrm {Sch}}(\mathbb {R}^n,S,\mathbb {R}^n)$ does not admit a domatic $\aleph _0$ -partition with Baire measurable parts.

When $\overline {S}$ is uncountable, the set $\overline {\pi (S)}\subseteq \mathbb {R}^n/\mathbb {Z}^n$ is also uncountable, where $\pi :\mathbb {R}^n\to \mathbb {R}^n/\mathbb {Z}^n$ is the usual projection map, and this is because $\overline {\pi (S)}\supseteq \pi (\overline {S})$ and $\pi $ is countable-to-one. Then Theorem 2.17 gives a continuous domatic partial function $f:\mathbb {R}^n/\mathbb {Z}^n\rightharpoonup {}\aleph _0$ for the graph $\operatorname {\mathrm {Sch}}(\mathbb {R}^n/\mathbb {Z}^n,\pi (S),\mathbb {R}^n/\mathbb {Z}^n)$ on $\mathbb {R}^n/\mathbb {Z}^n$ , and one can check that the pullback $f\circ \pi :\mathbb {R}^n\rightharpoonup {}\aleph _0$ is a continuous domatic partial function for $\operatorname {\mathrm {Sch}}(\mathbb {R}^n,S,\mathbb {R}^n)$ .

When S is unbounded, we may use diagonalization to get a rapidly increasing sequence of positive radii $0= R_0<R_1<\dots $ such that for any $i<\omega $ and any $x\in \mathbb {R}^n$ with $\|x\|<i+1$ , the translate $S+x$ intersects the open spherical shell $B(0,R_{i+1})\smallsetminus \overline {B(0,R_i)}$ . Thus any translate of S eventually intersects with every far enough open spherical shell in this sequence. Then an $\aleph _0$ -coloring of these spherical shells where each color is used infinitely often will give a continuous domatic partial function $f:\mathbb {R}^n\rightharpoonup {}\aleph _0$ for $\operatorname {\mathrm {Sch}}(\mathbb {R}^n,S,\mathbb {R}^n)$ .

Question 2.20. Let P be a nonempty perfect subset of the Polish group $\Gamma =(\mathbb {R}/\mathbb {Z})^\omega $ . Does the graph $G=\operatorname {\mathrm {Sch}}(\Gamma ,P,\Gamma )$ admit domatic $\aleph _0$ -partitions with open, Borel, or Baire measurable parts? Does G even admit a domatic bipartition with open parts?

Proof. By Lemma 2.16, we can build a tree $\{ U_s:s\in {}^{\omega>}2\}$ of open subsets of $\Gamma $ , such that $U_{\langle \:\rangle }=\Gamma $ , for each $s\preceq t$ we have $U_s\supseteq U_t$ , for each s we have $U_{s{}^{\smallfrown }{}0}\cap U_{s{}^{\smallfrown }{}1}=\varnothing $ , and every $U_s$ dominates $P^{-1}$ . By the last part of Lemma 2.15, we may also shrink each $U_s$ along the construction in such a way that $\overline {U_{s{}^{\smallfrown }{}0}},\overline {U_{s{}^{\smallfrown }{}1}}\subseteq U_s$ for every s.

For every $x\in {}^\omega 2$ , let $D_x=\bigcap _{n<\omega }U_{x\upharpoonright {} n}=\bigcap _{n<\omega }\overline {U_{x\upharpoonright {} n}}$ , and we claim that $D_x$ dominates $P^{-1}$ . Fix a $\gamma \in \Gamma $ , and we’ll show $P^{-1}\gamma \cap D_x\ne \varnothing $ . For every $n<\omega $ , since $U_{x\upharpoonright {}n}$ dominates $P^{-1}$ , there exists some $p_n\in P^{-1}$ such that $p_n\gamma \in U_{x\upharpoonright {}n}$ . Since $P^{-1}\subseteq \Gamma $ is compact, there is a subsequence of $p_n$ ’s which converges to some $p\in P^{-1}$ . Since $\{U_{x\upharpoonright {}n}:n<\omega \}$ is a decreasing family, taking the limit of $p_n\gamma \in U_{x\upharpoonright {}n}$ along this subsequence gives $p\gamma \in \bigcap _{n<\omega }\overline {U_{x\upharpoonright {}n}}=D_x$ , which means $D_x$ dominates $P^{-1}$ . Thus $\{D_x:x\in {}^\omega 2\}$ is the family that we wanted.

Proof. We first assume that $\Gamma $ is a connected Polish group and $x\in \Gamma $ is such that $\{\gamma \in \Gamma :\gamma ^2=x\}$ has nonempty interior. We will show that such a $\Gamma $ is Boolean.

Fix a nonempty open set $U\subseteq \Gamma $ such that $\gamma ^2=x$ for all $\gamma \in U$ . Let $g\in U$ , and let $V\ni 1_\Gamma $ be a symmetric open neighborhood of the identity such that $V^2g\subseteq U$ . For every $h\in V^2$ , we have $hg,g\in U$ , which means that $ghg^{-1}=h^{-1}(hghg)(g^{-1}g^{-1})=h^{-1}xx^{-1}=h^{-1}$ .

Assume $a,b\in V$ , so then $a,b,ab\in V^2$ , and the above argument implies that $gag^{-1}=a^{-1}$ , $gbg^{-1}=b^{-1}$ , and $a^{-1}b^{-1}=(gag^{-1})(gbg^{-1})=g(ab)g^{-1}=(ab)^{-1}=b^{-1}a^{-1}$ . We find that $a^{-1}$ commutes with $b^{-1}$ for all $a,b\in V$ , and since V is symmetric, a and b also commute for every $a,b\in V$ .

Since $\Gamma $ is connected and V is an open neighborhood of $1_\Gamma $ , V is a generating set of $\Gamma $ . Since $\Gamma $ has a commutative generating set, $\Gamma $ is abelian. This then means for all $h\in V\subseteq V^2$ , we have $h=ghg^{-1}=h^{-1}$ , and so every $h\in V$ must have order at most $2$ . Finally since $\Gamma $ is an abelian group generated by a set V of elements of order at most $2$ , $\Gamma $ is Boolean as well.

For the next part of the lemma, we will show that every nontrivial connected locally compact Polish group is non-Boolean. By Corollary 2.3, every such group is an inverse limit of nontrivial connected Lie groups. We may notice that every nontrivial connected Lie group is non-Boolean, by considering group elements near the identity. Thus their inverse limit must also be non-Boolean, which finishes the proof.

Proof. For every $\gamma \in \Gamma $ , the set $P\gamma \cap D\ne \varnothing $ is nonempty and has no isolated points. So $P\gamma \cap D$ is infinite, and F being finite implies $P\gamma \cap (D\smallsetminus F)=(P\gamma \cap D)\smallsetminus F\ne \varnothing $ is nonempty, which means $D\smallsetminus F$ dominates P.

Proof. Since U dominates $U^{-1}$ , Lemma 2.15 gives a finite subset $F_0^{-1}\subseteq U^{-1}$ such that U dominates $F_0^{-1}$ . By Lemma 2.24, $U\smallsetminus F_0$ dominates $U^{-1}$ , and by another application of Lemma 2.15, there is a finite collection $B_0^{-1},\dots ,B_{n-1}^{-1}$ of nonempty open subsets of $U^{-1}$ , such that for every $\gamma \in \Gamma $ , there exists some $i<n$ for which $B_i^{-1}\gamma \subseteq U\smallsetminus F_0$ . It follows that for any finite tuple $(f_0^{-1},\dots ,f_{n-1}^{-1})\in B_0^{-1}\times \dots \times B_{n-1}^{-1}$ , the set $U\smallsetminus F_0$ dominates $\{f_0^{-1},\dots ,f_{n-1}^{-1}\}\subseteq U^{-1}$ .

We claim that for a comeager set of tuples $(f_0,\dots ,f_{n-1})\in \Gamma ^n$ , the finite set $F_1=\{f_0,\dots ,f_{n-1}\}$ satisfies $F_1F_1F^{-1}_1\cap F_0=\varnothing $ and $F_1F_1\cap F_0F_0=\varnothing $ .

To fulfill the first condition $F_1F_1F_1^{-1}\cap F_0=\varnothing $ , it suffices to guarantee for every $(i,j,k)\in \{0,\dots ,n-1\}^3$ and $g\in F_0$ , that $f_if_jf_k^{-1}\ne g$ happens comeagerly often in $\Gamma ^n$ . If one of $i,j,k$ is different from the other two, then $f_if_jf_k^{-1}\ne g$ happens comeagerly if we fix the other two f’s and move the one different f freely, and so $f_if_jf_k^{-1}\ne g$ happens comeagerly over $\Gamma ^n$ by Kuratowski–Ulam [Reference Kechris9, Theorem 8.41]. Otherwise if $i=j=k$ , then $f_i=f_if_jf_k^{-1}\ne g$ also happens comeagerly over $\Gamma ^n$ .

To fulfill the second condition $F_1F_1\cap F_0F_0=\varnothing $ , one can apply a similar argument as the previous case, using the fact that Lemma 2.23 applies.

Therefore since $B_0\times \dots \times B_{n-1}\subseteq \Gamma ^n$ is nonmeager, we can fix some tuple $(f_0,\dots ,f_{n-1})\in B_0\times \dots \times B_{n-1}$ such that the finite set $F_1=\{f_0,\dots ,f_{n-1}\}$ satisfies $F_1F_1F_1^{-1}\cap F_0=\varnothing $ and $F_1F_1\cap F_0F_0=\varnothing $ . We note here that $F_0F_1\cap F_1F_1=\varnothing $ , since otherwise there exists $(g,f)\in F_0\times F_1,$ where $gf\in F_1F_1$ , and then $g\in F_1F_1F_1^{-1}\cap F_0\ne \varnothing $ gives a contradiction. In other words, we now know that $(F_0F_0\cup F_0F_1)\cap F_1F_1=\varnothing $ .

Let $F=F_0\cup F_1\subseteq U$ , and recall from earlier that U dominates $F_0^{-1}$ and $U\smallsetminus F_0$ dominates $F_1^{-1}$ . This means that $F_0U=F_1(U\smallsetminus F_0)=\Gamma $ . We see that the following holds:

$$ \begin{align*} F\cdot(U\smallsetminus F)&=(F_0\cup F_1)\cdot\left(U\smallsetminus (F_0\cup F_1)\right)\\ &=F_0\cdot\left(U\smallsetminus (F_0\cup F_1)\right)\cup F_1\cdot\left((U\smallsetminus F_0)\smallsetminus F_1\right)\\ &\supseteq \left(\Gamma\smallsetminus (F_0F_0\cup F_0F_1)\right)\cup\left(\Gamma\smallsetminus F_1F_1\right)\\ &=\Gamma\smallsetminus \left((F_0F_0\cup F_0F_1)\cap F_1F_1\right)\\ &=\Gamma. \end{align*} $$

We conclude that $U\smallsetminus F$ additively dominates F.

Proof. By Lemma 2.15 and since domination is closed upwards, there is a finite subset $F_Q\subseteq U$ such that every $Q_j$ additively dominates $F_Q$ . By Lemma 2.24, $U\smallsetminus F_Q$ additively dominates U, and a similar argument shows that $U\smallsetminus F_{Q}$ additively dominates itself. By Lemma 2.25, there is a finite subset $F_U\subseteq U\smallsetminus F_Q$ such that $U\smallsetminus (F_Q\cup F_U)$ additively dominates $F_U$ . Letting $F=F_Q\cup F_U\subseteq U$ , we see that by upward closure of domination, the sets $Q_0,\dots ,Q_{n-1},U\smallsetminus F$ all additively dominate F, and by Lemma 2.24, the set $U\smallsetminus F$ additively dominates all of the sets $P_0,\dots ,P_{m-1}$ .

By Lemma 2.15, there is an open subset $V\subseteq \overline {V}\subseteq U\smallsetminus F$ such that V additively dominates all of the sets $F,P_0,\dots ,P_{m-1}$ , which is because we can take a finite union of all such V’s over each of the sets $F,P_0,\dots ,P_{m-1}$ . Also by Lemma 2.15, if d is a compatible two-sided invariant metric on $\Gamma $ , then for every sufficiently small $r>0$ , every set that intersects every ball $B_d(f,r)$ for $f\in F$ is additively dominated by the sets $Q_0,\dots ,Q_{n-1},V$ which additively dominated F, and the radius-r ball $B_d(F,r)$ around F satisfies $\overline {B_d(F,r)}\subseteq U\smallsetminus \overline {V}$ .

Since $\Gamma $ is perfect, we can take a pair of disjoint open subsets $W_0,W_1\subseteq B_d(F,r)$ , such that each of $W_0,W_1$ intersects every ball $B_d(f,r)$ for $f\in F$ . Thus each of $W_0,W_1$ is additively dominated by the sets $Q_0,\dots ,Q_{n-1},V$ .

The open set V additively dominates the perfect sets $P_0,\dots ,P_{m-1},\overline {W_0},\overline {W_1}$ , and so by Lemma 2.16, there are disjoint open subsets $D_0,D_1\subseteq V$ each of which additively dominates the sets $P_0,\dots ,P_{m-1},\overline {W_0},\overline {W_1}$ . By an argument in the proof of Theorem 2.17, $D_0,D_1$ each also additively dominates the sets $W_0,W_1$ .

Let $A_0=D_0\cup W_0$ and $A_1=D_1\cup W_1$ . Then $A_0,A_1$ additively dominate each other and themselves since each of $D_0,D_1$ additively dominates each of $W_0,W_1$ , also $A_0,A_1$ additively dominate $P_0,\dots ,P_{m-1}$ since $D_0,D_1$ do, $A_0,A_1$ are additively dominated by $Q_0,\dots ,Q_{n-1}$ since $W_0,W_1$ are, and $\overline {A_0},\overline {A_1}\subseteq \overline {B_d(F,r)}\cup \overline {V}\subseteq U$ .

Proof. Like Theorem 2.22, we can build a tree $\{U_s:s\in {}^{\omega>}2\}$ of open subsets of $\Gamma $ , such that $U_{\langle \:\rangle }=\Gamma $ , for each $s\prec t$ we have $U_s\supseteq \overline {U_t}$ , for each s we have $U_{s{}^{\smallfrown }{}0}\cap U_{s{}^{\smallfrown }{}1}=\varnothing $ , every $U_s$ additively dominates P, and every $U_s$ additively dominates every $U_t$ , for all $s,t\in {}^{\omega>}2$ .

The strategy is to build $\{U_s:s\in {}^{\omega>}2\}$ “one-by-one”, where we visit every node in the tree ${}^{\omega>}2$ in $\omega $ stages, while increasingly traversing along each branch. At a given stage $s\in {}^{\omega>}2$ , we split the open set $U_s$ into two disjoint open subsets $U_{s{}^{\smallfrown }{}0},U_{s{}^{\smallfrown }{}1}$ using Lemma 2.26, such that among the open sets $U_t$ we’ve constructed so far, each one additively dominates P and all of them additively dominate each other and themselves. In using Lemma 2.26, note that $U_s$ additively dominates $U_t$ if and only if $U_s$ additively dominates the perfect set $\overline {U_t}$ by an argument in the proof of Theorem 2.17.

For every $x\in {}^\omega 2$ , let $D_x=\bigcap _{n<\omega }U_{x\upharpoonright {} n}=\bigcap _{n<\omega }\overline {U_{x\upharpoonright {} n}}$ . A same argument as Theorem 2.22 implies that the family $\{D_x:x\in {}^\omega 2\}$ is as we desired.

Proof. We first prove a weak version of Lemma 2.14 without the finite dimension assumption. Namely, let $\Gamma $ be a perfect Polish group, and let $P_0,\dots ,P_{n-1}\subseteq \Gamma $ be nonempty open sets. Then we will show that there exists a pair of disjoint open sets $A_0,A_1\subseteq \Gamma $ , each of which dominates the sets $P_0,\dots ,P_{n-1}$ .

By Birkhoff–Kakutani [Reference Gao8, Theorem 2.1.1], $\Gamma $ has a compatible right-invariant metric d. Then there is some $r>0$ such that every right translate $P_i\gamma $ of a set $P_i$ contains some ball of d-radius-r. By a ball-packing argument and since $\Gamma $ is perfect, there are disjoint discrete sets $S_0,S_1\subseteq \Gamma $ which both intersect every radius-r ball in $\Gamma $ . Then let $A_0,A_1$ be disjoint open sets which separate $S_0,S_1$ , and our claimed result follows.

Then by repeating the proofs of Lemma 2.26 and Theorem 2.27 with the above notion of a weaker open pair property in place of the normal one, the full result follows.

Proof. Let $\pi :\mathbb {R}^n\to \mathbb {R}^n/\mathbb {Z}^n$ be the usual projection map, and fix a nonempty closed perfect subset $K\subseteq \pi [P]$ in $\mathbb {R}^n/\mathbb {Z}^n$ . By Theorem 2.27, there is a family $\langle D_i:i<2^{\aleph _0}\rangle $ of pairwise-disjoint closed subsets of $\mathbb {R}^n/\mathbb {Z}^n$ which additively dominate K, each other, and themselves. Similar to Corollary 2.19, the pullbacks $C_i=\pi ^{-1}[D_i]$ for $i<2^{\aleph _0}$ are as desired.

Proof. First we describe a greedy algorithm performed on a connected out-degree $\aleph _0$ -regular graph G with a fixed enumeration $V=\{v_0,v_1,v_2,\dots \}$ of its countable vertex set, which outputs a domatic $\aleph _0$ -partition for the countable graph G.

We start with the empty coloring $f_0=\varnothing $ on V, and we fix a countable enumeration $\{(v_i,c_i):i<\omega \}=\omega \times \omega $ . At stage $i<\omega $ of this algorithm, we extend a finite partial coloring $f_i$ on V to a finite partial coloring $f_{i+1}\supseteq f_i$ on V, such that $v_i\in \operatorname {\mathrm {dom}}(f_{i+1})$ , and $f_{i+1}$ colors the first $f_i$ -uncolored neighbor of $v_i$ in color $c_i$ , if $f_i$ hasn’t colored any neighbor of $v_i$ in color $c_i$ yet. We can check that this algorithm always outputs a full-domain domatic $\aleph _0$ -partition for a connected out-degree $\aleph _0$ -regular countable graph G.

Using standard descriptive graph combinatorial arguments and Lusin–Novikov uniformization [Reference Kechris9, Theorem 18.10], one sees that the above algorithm can be performed on locally countable Borel graphs when $E_G$ is smooth [Reference Kechris9, Exercise 18.20], so that it outputs Borel colorings.

Proof. Let $\kappa $ be the probability measure on $\omega $ defined by $\kappa (\{n\})=2^{-n-1}$ , and let $\lambda =\prod _{i<\omega }\kappa $ be the product Borel probability measure on $\prod _{i<\omega }\omega ={}^\omega \omega $ . Note that since $\kappa $ does not vanish on singletons, it follows that for every infinite subset $A\subseteq \omega $ , there exists a $\lambda $ -conull set of functions $r:\omega \to \omega $ such that $r[A]=\omega $ .

Let $B=\{(x,r)\in X\times {}^\omega \omega :\forall y\in [x]_{E_G}(r\circ f[N_G(y)]=\omega )\}$ . Then since $E_G$ is a countable Borel equivalence relation, B is Borel. Since for every $y\in X$ , the set $f[N_G(y)]\subseteq \omega $ is infinite, the previous arguments imply that every section $B_x\subseteq {}^\omega \omega $ of B is $\lambda $ -conull. By Fubini’s theorem, there exists an $r\in {}^\omega \omega $ such that the section $B^r\subseteq X$ is $\mu $ -conull. This means that the Borel function $r\circ f:X\to \omega $ is domatic at every $x\in B^r= [B^r]_{E_G}$ .

Since $B^r$ is $E_G$ -invariant, we can modify the function $r\circ f:X\to \omega $ over the $E_G$ -invariant $\mu $ -null set $X\smallsetminus B^r$ in the same way as Theorem 3.1, so that we get a $\mu $ -measurable function $g:X\to \omega $ domatic everywhere for G as desired.

Proof. The proof is the same as that of Lemma 3.2, with all measure-theoretic elements replaced by their Baire category counterparts.

Proof. Let $n<\omega $ be sufficiently large so that $k(1-k^{-1})^n\le \varepsilon /2$ .

By Lusin–Novikov uniformization [Reference Kechris9, Theorem 18.10], there are Borel functions $f_0,\dots ,f_{n-1}:X\to X$ , such that for every $x\in X$ , the elements $f_0(x),\dots ,f_{n-1}(x)\in N_G(x)$ are n distinct G-neighbors of x.

Let $\{A_0,A_1,\dots \}$ be a countable family of Borel subsets of X that separates points. For $i<\omega $ , let $\mathcal {P}_i$ be the finite Borel partition of X generated by $\{A_0,\dots ,A_{i-1}\}$ . Then for every $x\in X$ , the n distinct points $f_0(x),\dots ,f_{n-1}(x)$ will be completely separated by some partition $\mathcal {P}_i$ (and every partition that comes afterwards), and so we can pick a sufficiently fine partition $\mathcal {P}=\mathcal {P}_i$ such that the Borel set $B\subseteq X$ of all elements $x\in X$ for which $f_0(x),\dots ,f_{n-1}(x)$ are completely separated by $\mathcal {P}$ satisfies $\mu (B)\ge 1-\varepsilon /2$ .

Since $\mathcal {P}$ is a finite Borel partition of X, we can uniformly randomly pick a function $c:\mathcal {P}\to \{0,\dots ,k-1\}$ , and let $f_c:X\to \{0,\dots ,k-1\}$ be its corresponding Borel coloring on X that assigns the color $c(P)$ to every element inside $P\in \mathcal {P}$ . For every $x\in B$ , the probability that $f_c$ is domatic at x is lower-bounded by the probability that $f_c$ colors the $\mathcal {P}$ -separated points $f_0(x),\dots ,f_{n-1}(x)$ with all colors in $\{0,\dots ,k-1\}$ , which is at least $1-k(1-k^{-1})^n\ge 1-\varepsilon /2$ . Thus the product probability measure of the set of all pairs $(c,x)$ for which $f_c$ is domatic at x is at least $(1-\varepsilon /2)^2\ge 1-\varepsilon $ .

By Fubini’s theorem, there exists some c such that $f_c$ is domatic at a $\mu $ -measure $\ge 1-\varepsilon $ set of $x\in X$ . This Borel function $f_c:X\to \{0,\dots ,k-1\}$ is as desired.

Proof. First we explain why we may assume that G is quasi- $\mu $ -preserving, meaning that every $\mu $ -null set is contained in an $E_G$ -invariant $\mu $ -null set. By the Feldman–Moore theorem [Reference Kechris and Marks10, Section 4.1], the countable Borel equivalence relation $E_G=\bigcup _{n<\omega } T_n\subseteq X^2$ is generated by countably many Borel involutions $T_n:X\to X$ . Letting $\nu =\sum _{n<\omega }2^{-n-1}(T_n)_*(\mu )$ , we get a Borel probability measure $\nu $ such that every $\nu $ -null set is $\mu $ -null, and G is quasi- $\nu $ -preserving. Thus $\nu $ -measurable functions are $\mu $ -measurable, and we may replace $\mu $ with $\nu $ to assume without loss of generality that G is quasi- $\mu $ -preserving.

By Lemma 3.2 and since we assumed G is quasi- $\mu $ -preserving, it suffices to show the existence of a $\mu $ -measurable function $f:X\to C$ such that $\mathopen |C\mathclose |=\aleph _0$ , and for a $\mu $ -conull set of $x\in X$ , the set $f[N_G(x)]\subseteq C$ is infinite.

For every $n<\omega $ , Lemma 3.4 implies there is a Borel function $f_n:X\to \{0,\dots ,2^n-1\}$ that is domatic at a set of vertices $A_n\subseteq X$ of $\mu $ -measure $\mu (A_n)\ge 1-2^{-n}$ . The $2^n$ color classes of $f_n$ partition X, and so the color class $D_n\subseteq X$ of $f_n$ with the least $\mu $ -measure satisfies $\mu (D_n)\le 2^{-n}$ . Since $f_n$ is domatic at $A_n$ , we have $D_n\cap N_G(x)\ne \varnothing $ for all $x\in A_n$ .

Since $\mu (X\smallsetminus A_n)\le 2^{-n}$ and $\mu (D_n)\le 2^{-n}$ for all $n<\omega $ , the Borel–Cantelli lemma implies that there is a $\mu $ -conull set of $x\in X$ which lies in only finitely many of the sets $X\smallsetminus A_n$ and $D_n$ . Since we assumed G is quasi- $\mu $ -preserving, we may find an $E_G$ -invariant $\mu $ -conull subset $Y\subseteq X$ of such elements $x\in X$ that lie in finitely many of the sets $X\smallsetminus A_n$ and $D_n$ .

Since every $y\in Y$ lies in finitely many $D_n$ , we can define the $\mu $ -measurable function $f:Y\to [\omega ]^{<\omega }$ such that for every $y\in Y$ , $f(y)\subseteq \omega $ is the finite set of all $n<\omega $ for which $y\in D_n$ . We claim that for every $x\in Y$ , the set $f[N_G(x)]\subseteq [\omega ]^{<\omega }$ is infinite, which completes our proof as $C=[\omega ]^{<\omega }$ has size $\aleph _0$ as we wanted.

Let $x\in Y$ . Since x lies in finitely many $X\smallsetminus A_n$ , it lies in infinitely many $A_n$ , and so there are infinitely many $D_n$ such that $D_n\cap N_G(x)\ne \varnothing $ by the previous arguments. By definition of f, infinitely many $n<\omega $ lie in the set $\bigcup _{y\in N_G(x)}f(y)=\bigcup f[N_G(x)]\subseteq \omega $ . Since $f[N_G(x)]\subseteq [\omega ]^{<\omega }$ is a collection of finite sets whose union $\bigcup f[N_G(x)]\subseteq \omega $ is infinite, the collection $f[N_G(x)]$ itself is infinite, which completes the proof.

Proof. Follows from Theorem 3.5.

Proof. By Feldman–Moore [Reference Kechris and Marks10, Proposition 4.1], G admits a symmetric Borel edge- $\aleph _0$ -coloring $f:G\to \omega $ , meaning that for every two distinct edges $e,e'\in G$ that share a common vertex, we have $f(e)\ne f(e')$ . In particular, every vertex of G belongs to edges of infinitely many f-colors. Then the proofs of Lemmas 3.2 and 3.3 imply our desired result.