Proof. Suppose $x\stackrel {l}{\to }y$ . Then there exists a left Cauchy sequence $(g_n)$ in G such that $g_nx\to y$ . Let d be a compatible metric on G and $\rho $ a compatible metric on X.

By the continuity of the action of G on X, for any $k\in \omega $ , there exists a $\delta>0$ such that, for any $g\in G$ , if $d(g,1_G)<\delta $ , then $\rho (gx,x)<2^{-(k+1)}$ . Since $(g_n)$ is left Cauchy, for this $\delta>0$ , there exists an $m_k\in \omega $ such that, for any $n>m_k$ , we have $d(g_{m_k}^{-1}g_n,1_G)<\delta $ , and hence $\rho (g_{m_k}^{-1}g_nx,x)<2^{-(k+1)}$ . By $g_nx\to y$ , we have $g_{m_k}^{-1}g_nx\to g_{m_k}^{-1}y$ . There exists an $n>m_k$ such that $\rho (g_{m_k}^{-1}g_nx,g_{m_k}^{-1}y)<2^{-(k+1)}$ . So we have $\rho (g_{m_k}^{-1}y,x)<2^{-k}$ . Without loss of generality, we can assume that $(m_k)$ is strictly increasing. Let $h_k=g_{m_k}^{-1}$ for each k. Then $(h_k)$ witnesses that $x\stackrel {r}{\to }y$ .

Proof. Let d be a left-invariant compatible metric on G, and let $d_u$ be the supremum metric on $c(G)$ . Since G is not CLI, d is not complete. So there exists a d-Cauchy sequence $(g_n)$ in G such that $(g_n)$ diverges. For $m\in \omega $ , we define $\gamma _m\in c(G)$ as

$$ \begin{align*}\gamma_m(n)=\left\{\begin{array}{ll}g_n, & n<m,\\ g_m, & n\ge m.\end{array}\right.\end{align*} $$

For $m<k$ , we have $d_u(\gamma _m^{-1}\gamma _k,1_{c(G)})=\max \{d(g_m,g_n):m\le n\le k\}$ . So $(\gamma _m)$ is left Cauchy, and hence $(\gamma _m^{-1})$ is right Cauchy.

We define $z\in G^\omega $ as $z(n)=g_n$ for each $n\in \omega $ . Then $z\notin c(G)$ . We claim that for any $c(G)$ -invariant dense $G_\delta $ subset C of $G^\omega $ there exist $x,y\in C$ with different orbits such that $x\preceq _B y$ . Then by [Reference Lupini and Panagiotopoulos25, Theorem 2.9] we have $E(G)\not \le _BE_H^Y$ .

Fix a $c(G)$ -invariant dense $G_\delta $ subset C of $G^\omega $ . Since $z^{-1}C$ is also dense $G_\delta $ , $C\cap z^{-1}C\ne \emptyset $ . Let $x\in C\cap z^{-1}C$ and $y=zx$ , then $x,y\in C$ and $[x]\ne [y]$ . Note that $\gamma _m^{-1}y\to z^{-1}y=x$ , so $x\stackrel {r}{\to } y$ . Thus $x\preceq _B y$ .

Proof. It well known that $S_\infty $ is not CLI (cf. [Reference Gao14, Example 2.2.7]). And by Proposition 3.3, $c(H)$ is CLI.

Proof. Let d be a left-invariant compatible metric on G. Since G is not non-Archimedean, there exists an $\varepsilon _0>0$ such that no open subgroup of G is contained in $V=\{g\in G:d(1_G,g)<\varepsilon _0\}$ . For each $n\in \omega $ , we define $V_n=\{g\in G:d(1_G,g)<2^{-n}\}$ . Then $V_n^{-1}=V_n$ and $G_n=\bigcup _mV_n^m$ is an open subgroup of G. Thus $G_n\not \subseteq V$ for each $n\in \omega $ , and hence there is an integer $m_n>0$ such that $V_n^{m_n}\not \subseteq V$ . So we can find $1_G=g_{n,0},g_{n,1},\dots ,g_{n,m_n}\in G$ such that $g_{n,i}^{-1}g_{n,i+1}\in V_n$ for each $i<m_n$ and $g_{n,m_n}\notin V$ .

Let $X_n=\{g_{n,i}:0\le i\le m_n\}$ for each $n\in \omega $ . Then $(X_n,d)$ is a finite metric space. We define $\theta :\prod _nX_n\to G^\omega $ as, for $x\in \prod _nX_n$ and each $k\in \omega $ ,

$$ \begin{align*}\theta(x)(k)=\left\{\begin{array}{ll}x(n)^{-1}, & k=2n,\\ 1_G, & k=2n+1.\end{array}\right.\end{align*} $$

It is trivial to check that $\theta $ is a Borel reduction of $D(X_n,d)$ to $E(G)$ .

Note that $\delta (\varepsilon _0,X_n)\le 2^{-n}\to 0$ as $n\to \infty $ . By [Reference Farah13, Theorem 3.7(a)], there exists a turbulent $c_0$ -equality D such that $D\le _BD(X_n,d)\le _BE(G)$ .

Proof. Since H is topologically isomorphic to a closed subgroup of $S_\infty $ , by [Reference Gao14, Theorem 3.5.2], we only need to consider the case of $H=S_\infty $ .

Then the result follows from the preceding lemma.

By Proposition 3.3, $c(S_\infty )$ is non-Archimedean, so $E(G)\not \le _BE(S_\infty )$ .

Proof. It follows from Theorem 3.5(4) and the preceding theorem.

Proof. Let d be a compatible metric on G and $d_u$ the supremum metric on $G^\omega $ . Note that $1_{c(G)}(n)=1_{G^\omega }(n)=1_G$ for each $n\in \omega $ . Let $V_m=\{\gamma \in c(G):d_u(\gamma ,1_{c(G)})<2^{-m}\}$ for $m\in \omega $ . Then $(V_m)$ is a neighborhood basis of $1_{c(G)}$ . Note that $V_m$ is dense in $\{x\in G^\omega :d_u(x,1_{G^\omega })<2^{-m}\}$ under the topology of $G^\omega $ . For $x,y\in G^\omega $ , we have $x\in \overline {V_my}\iff xy^{-1}\in \overline {V_m}$ , so

$$ \begin{align*}d_u(xy^{-1},1_{G^\omega})<2^{-m}\Rightarrow x\in\overline{V_my}\Rightarrow d_u(xy^{-1},1_{G^\omega})\le 2^{-m}.\end{align*} $$

( $\Rightarrow $ ). Let $x\leftrightsquigarrow y$ . Then for each $m\in \omega $ , there exist $\gamma ^x_m,\gamma ^y_m\in c(G)$ such that $\gamma ^x_mx\in \overline {V_m(\gamma ^y_my)}$ . So $d_u(\gamma ^x_mxy^{-1}(\gamma ^y_m)^{-1},1_{G^\omega })\le 2^{-m}$ . Let $\gamma _m=\gamma ^x_m$ and $\eta _m=(\gamma ^y_m)^{-1}$ , then $\gamma _mxy^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ .

( $\Leftarrow $ ). Let $(\gamma _m),(\eta _m)\in c(G)^\omega $ with $\gamma _mxy^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ . It is easy to check that $\eta _m^{-1}yx^{-1}\gamma _m^{-1}\rightrightarrows 1_{G^\omega }$ . For any open $V\ni 1_{c(G)}$ and open set U of $G^\omega $ with $U\cap ([x]\cup [y])\ne \emptyset $ , there exists $k\in \omega $ with $V_k\subseteq V$ . Without loss of generality, we may assume that $U\cap [x]\ne \emptyset $ and $U=U_0\times \cdots \times U_l\times G^\omega $ with $U_0,\dots ,U_l$ open in G. So there exists $\gamma \in c(G)$ with $\gamma x\in U$ , i.e., $\gamma (n)x(n)\in U_n$ for each $n\le l$ .

Fix a large enough m such that

$$ \begin{align*}d_u(\gamma_mxy^{-1}\eta_m,1_{G^\omega})<2^{-k},\quad d_u(\eta_m^{-1}yx^{-1}\gamma_m^{-1},1_{G^\omega})<2^{-k}.\end{align*} $$

Now we define

$$ \begin{align*}\gamma^x(n)=\left\{\begin{array}{ll}\gamma(n), & n\le l,\\ \gamma_m(n), & n>l,\end{array}\right. \quad\gamma^y(n)=\left\{\begin{array}{ll}\gamma(n)x(n)y(n)^{-1}, & n\le l,\\ \eta_m(n)^{-1}, & n>l.\end{array}\right.\end{align*} $$

Then we have $\gamma ^x,\gamma ^y\in c(G)$ and $\gamma ^xx,\gamma ^yy\in U$ . Note that

$$ \begin{align*}\gamma^x(n)x(n)y(n)^{-1}(\gamma^y(n))^{-1}=\left\{\begin{array}{ll}1_G, & n\le l,\\ \gamma_m(n)x(n)y(n)^{-1}\eta_m(n), & n>l.\end{array}\right.\end{align*} $$

It is easy to see that

$$ \begin{align*}d_u(\gamma^xxy^{-1}(\gamma^y)^{-1},1_{G^\omega})\le d_u(\gamma_mxy^{-1}\eta_m,1_{G^\omega})<2^{-k},\end{align*} $$

$$ \begin{align*}d_u(\gamma^yyx^{-1}(\gamma^x)^{-1},1_{G^\omega})\le d_u(\eta_m^{-1}yx^{-1}\gamma_m^{-1},1_{G^\omega})<2^{-k}.\end{align*} $$

It gives that $\gamma ^xx\in \overline {V_k(\gamma ^yy)}\subseteq \overline {V(\gamma ^yy)}$ and $\gamma ^yy\in \overline {V_k(\gamma ^xx)}\subseteq \overline {V(\gamma ^xx)}$ . Thus we have $x\leftrightsquigarrow y$ .

Proof. Let G be a Polish group and $c(G)\curvearrowright G^\omega $ the action of left multiplication. Assume that $x\leftrightsquigarrow y$ for some $x,y\in G^\omega $ . From Lemma 5.2, there exist $(\gamma _m),(\eta _m)\in c(G)^\omega $ with $\gamma _mxy^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ . Let d be a two-sided invariant compatible metric on G and $d_u$ the supremum metric on $G^\omega $ . Then $d_u$ is also two-sided invariant, so

$$ \begin{align*}d_u(\eta_m\gamma_mxy^{-1},1_{G^\omega})=d_u(\gamma_mxy^{-1}\eta_m,1_{G^\omega})\to 0,\end{align*} $$

thus we have $\eta _m\gamma _mxy^{-1}\rightrightarrows 1_{G^\omega }$ . It follows that $\eta _m\gamma _m\rightrightarrows yx^{-1}$ . Since $c(G)$ is closed under uniform convergence, $yx^{-1}\in c(G)$ . So $[x]=[y]$ .

Proof. It follows from Theorems 3.5(2) and 5.5.

Proof. We prove by induction on $\alpha $ . If $\alpha =0$ , since $x\leftrightsquigarrow _V^0y$ means $x\in \overline {Vy}$ and $y\in \overline {Vx}$ , we have $gx\in \overline {gVy}=\overline {(gVh^{-1})(hy)}\subseteq \overline {V'(hy)}$ and $hy\in \overline {hVx}=\overline {(hVg^{-1})(gx)}\subseteq \overline {V'(gx)}$ , so $gx\leftrightsquigarrow _{V'}^0hy$ .

Assume that the result holds for all $\beta <\alpha $ . If $x\leftrightsquigarrow _V^\alpha y$ , then for any open $W\ni 1_G$ and any open set $U\subseteq X$ such that U contains $gx$ or $hy$ . Without loss of generality, suppose $gx\in U$ , i.e., $x\in g^{-1}U$ . Then there exist $g^x,g^y\in V$ with $g^xx,g^yy\in g^{-1}U$ , so that $g^xx\leftrightsquigarrow _{g^{-1}Wg}^\beta g^yy$ for all $\beta <\alpha $ . By the inductive hypothesis, $(gg^xg^{-1})(gx)=gg^xx\leftrightsquigarrow _W^\beta gg^yy=(gg^yh^{-1})(hy)$ for all $\beta <\alpha $ . It is trivial to check that $gg^xg^{-1}$ and $gg^yh^{-1}$ witnesses that $gx\leftrightsquigarrow _{V'}^\alpha hy$ .

Proof. For any $x,y\in X$ and $g,h\in G$ , by Lemma 5.9, $x\leftrightsquigarrow _G^\alpha y$ implies $gx\leftrightsquigarrow _G^\alpha hy$ . So $\leftrightsquigarrow _G^\alpha $ is G-invariant.

By their definitions, it is trivial that $x\leftrightsquigarrow y$ implies $x\leftrightsquigarrow _G^1y$ .

On the other hand, if $x\leftrightsquigarrow _G^1y$ , we will show $x\leftrightsquigarrow y$ . For any open $V\ni 1_G$ and any open set $U\subseteq X$ with $U\cap ([x]\cup [y])\ne \emptyset $ , without loss of generality, we can assume that $U\cap [x]\ne \emptyset $ . There exists $g\in G$ with $gx\in U$ , i.e., $x\in g^{-1}U$ . So there are $g^x,g^y\in G$ such that $g^xx,g^yy\in g^{-1}U$ , $g^xx\in \overline {(g^{-1}Vg)g^yy}$ , and $g^yy\in \overline {(g^{-1}Vg)g^xx}$ . Then $gg^x$ and $gg^y$ witness that $x\leftrightsquigarrow y$ .

Proof. Given an open set $W\ni 1_G$ and an open set U of X with $x\in U$ or $y\in U$ . We can find an open $W_0\ni 1_G$ with $W_0^{-1}=W_0$ and $W_0^3\subseteq W$ . Then there exist $h^x,h^y\in V$ such that $h^xx,h^yy\in U$ and $h^xx\leftrightsquigarrow _{W_0}^\beta h^yy$ for all $\beta <\alpha $ . We can find an open $V_0\ni 1_G$ such that $(V_0h^xx\cup V_0h^yy)\subseteq U$ . By Lemma 5.9, $w^xh^xx\leftrightsquigarrow _W^\beta w^yh^yy$ holds for all $w^x,w^y\in W_0$ . Then we can put $V^x=W_0h^x\cap V_0h^x\cap V$ and $V^y=W_0h^y\cap V_0h^y\cap V$ as desired.

Proof. Let C be the dense $G_\delta $ subset of X in [Reference Lupini and Panagiotopoulos25, Lemma 2.5].

Proof of Claim.

By [Reference Lupini and Panagiotopoulos25, Lemma 2.5], the following conditions hold:

1. (i) $f\upharpoonright C$ is continuous;

2. (ii) $\forall x\in C\,\forall ^*g\in G\,(gx\in C)$ ;

3. (iii) for any open $W'\ni 1_H$ , there exist an open set $V'\ni 1_G$ and an open neighborhood $U'$ of $x_0$ such that

   $$ \begin{align*}\forall x\in(U'\cap C)\,\forall^*g\in V'\,(f(gx)\in W'f(x)).\end{align*} $$

Let $W'\ni 1_H$ be any open subset of H. Then we find two open sets $V'\ni 1_G$ and $U'\ni x_0$ satisfying (iii). Given any open set $O\subseteq Y$ with $f(x_0)\in O$ or $f(y_0)\in O$ , without loss of generality, we can assume that $f(x_0)\in O$ . Since f is continuous on C and $x_0\in U'\cap C$ , by shrinking, we can assume that $f(U'\cap C)\subseteq O$ . Since $x_0\leftrightsquigarrow _V^\alpha y_0$ , by Lemma 5.11, we have

$$ \begin{align*}\exists^*g^x\in V\,\exists^*g^y\in V\,(g^xx_0,g^yy_0\in U'\wedge\forall\beta<\alpha\,(g^xx_0\leftrightsquigarrow_{V'}^\beta g^yy_0)).\end{align*} $$

Moveover, since $x_0,y_0\in U\cap C$ , by (1) and (ii), we can find $g^x,g^y\in V$ such that $g^xx_0,g^yy_0\in (U'\cap C)$ , $g^xx_0\leftrightsquigarrow _{V'}^\beta g^yy_0~(\forall \beta <\alpha )$ , $f(g^xx_0)\in Wf(x_0)$ and $f(g^yy_0)\in Wf(y_0)$ . Thus there exist $h^x,h^y\in W$ so that $f(g^xx_0)=h^xf(x_0)$ and $f(g^yy_0)=h^yf(y_0)$ .

Now we prove $f(x_0)\leftrightsquigarrow _W^\alpha f(y_0)$ by induction on $\alpha $ . Since $g^xx_0,g^yy_0\in (U'\cap C)$ , we have $h^xf(x_0),h^yf(y_0)\in O$ . Thus we only need to show that $h^xf(x_0)\leftrightsquigarrow _{W'}^\beta h^yf(y_0)$ for $\beta <\alpha $ .

For any open neighborhood N of $h^xf(x_0)=f(g^xx_0)$ , we have $g^xx_0\in f^{-1}(N)$ . By the continuity of $f\upharpoonright C$ , there is an open set $V_0\ni 1_G$ such that $(V_0g^xx_0\cap C)\subseteq f^{-1}(N)$ . Since $g^yy_0\in (U'\cap C)$ , we can find $v\in V_0\cap V'$ such that $vg^yy_0\in C$ and $f(vg^yy_0)\in W'f(g^yy_0)=W'h^yf(y_0)$ . So $N\cap (W'h^yf(y_0))\ne \emptyset $ . It follows that $h^xf(x_0)\in \overline {W'h^yf(y_0)}$ . Similarly, we have $h^yf(y_0)\in \overline {W'h^xf(x_0)}$ . Therefore, $h^xf(x_0)\leftrightsquigarrow _{W'}^0 h^yf(y_0)$ .

If $\alpha =1$ , we have already proved $f(x_0)\leftrightsquigarrow _W^1 f(y_0)$ .

If $\alpha>1$ , then for all $0<\beta <\alpha $ , by inductive hypothesis, $g^xx_0\leftrightsquigarrow _{V'}^\beta g^yy_0$ implies $h^xf(x_0)=f(g^xx_0)\leftrightsquigarrow _{W'}^\beta f(g^yy_0)=h^yf(y_0)$ . This completes the proof.

Put $V=G,W=H$ , and $U=X$ , then the proceeding Claim gives that, $x\leftrightsquigarrow _G^\alpha y$ implies $f(x)\leftrightsquigarrow _H^\alpha f(y)$ for $x,y\in C$ and $0<\alpha <\omega _1$ .

Proof. We prove by induction on $\alpha $ .

If $\alpha =0$ , then we have $x\in \overline {Vy}$ and $y\in \overline {Vx}$ . So $xz\in \overline {Vyz}$ and $yz\in \overline {Vxz}$ , i.e., $xz\leftrightsquigarrow _V^0 yz$ .

If $\alpha>0$ , then for any open $W\ni 1_{c(G)}$ and any open $U\subseteq G^\omega $ with $xz\in U$ or $yz\in U$ , we have $x\in Uz^{-1}$ or $y\in Uz^{-1}$ , so there exist $g^x,g^y\in V$ such that $g^xx,g^yy\in Uz^{-1}$ and $g^xx\leftrightsquigarrow _W^\beta g^yy$ for $\beta <\alpha $ . Then we have $g^xxz,g^yyz\in U$ , and by induction hypothesis, $g^xxz\leftrightsquigarrow _W^\beta g^yyz$ for $\beta <\alpha $ . Therefore, $xz\leftrightsquigarrow _V^\alpha yz$ .

Proof. Assume for contradiction that there is a Borel reduction $\theta $ of $E(G)$ to $E(H)$ . From Theorem 5.12, there is a comeager set $C\subseteq G^\omega $ such that for any $x,y\in C$ , if $x\leftrightsquigarrow _{c(G)}^\alpha y$ , then $\theta (x)\leftrightsquigarrow _{c(H)}^\alpha \theta (y)$ . Since G is $\alpha $ -l.m.-unbalanced, there exists $z\in G^\omega \setminus c(G)$ such that $z\leftrightsquigarrow _{c(G)}^\alpha 1_{G^\omega }$ . Since $C\cap zC\ne \emptyset $ , we can find an $x\in (C\cap zC)$ and put $x=zy$ . Then $x,y\in C$ and $xy^{-1}=z\not \in c(G)$ , so $[x]\ne [y]$ . By Lemma 5.13, $x\leftrightsquigarrow _{c(G)}^\alpha y$ . It follows that $\theta (x)\leftrightsquigarrow _{c(H)}^\alpha \theta (y)$ and $[\theta (x)]\ne [\theta (y)]$ , contradicting that H is not $\alpha $ -l.m.-unbalanced.

Proof. Since G is not distal, there exist $h\ne 1_G$ in G and a sequence $(g_m)$ in G with $g_mhg_m^{-1}\to 1_G$ . For each $m,n\in \omega $ , let $\gamma _m(n)=g_m$ and $\eta _m(n)=g_m^{-1}$ . Then $\gamma _m,\eta _m\in c(G)$ . We define $z\in G^\omega $ as

$$ \begin{align*}z(n)=\left\{\begin{array}{ll}1_G, & n=2k,\\ h, & n=2k+1.\end{array}\right.\end{align*} $$

It is clear that $\gamma _mz\eta _m\rightrightarrows 1_{G^\omega }$ . From Lemma 5.2, for any $x,y\in G^\omega $ , if $xy^{-1}=z$ , we have $x\leftrightsquigarrow y$ . But it is trivial that $[x]\ne [y]$ .

Proof. From Theorem 5.16, we only need to prove that if G is l.m.-unbalanced then it is not distal.

Let $x,y\in G^\omega $ with $x\leftrightsquigarrow y$ but $[x]\ne [y]$ . By Lemma 5.2, there exist two sequences $(\gamma _m),(\eta _m)$ in $c(G)$ such that $\gamma _mxy^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ . Since $\gamma _m,\eta _m\in c(G)$ for each m, we can find $g_m,v_m\in G$ such that $g_m=\lim _n\gamma _m(n)$ and $v_m=\lim _n\eta _m(n)$ .

Let $B\subseteq G$ be an open neighborhood of $1_G$ with $\overline {B}$ compact. There is a large enough $k\in \omega $ such that $\gamma _k(n)x(n)y(n)^{-1}\eta _k(n)\in B\subseteq \overline {B}$ for all $n\in \omega $ . Thus there exist a $p\in \overline {B}$ and a strictly increasing natural numbers $n_0<n_1<\cdots $ such that $\lim _i\gamma _k(n_i)x(n_i)y(n_i)^{-1}\eta _k(n_i)=p$ . By $[x]\ne [y]$ , we have $xy^{-1}\notin c(G)$ , thus $\gamma _kxy^{-1}\eta _k\notin c(G)$ . So there exist a $q\in \overline {B}$ with $q\ne p$ and another strictly increasing natural numbers $l_0<l_1<\cdots $ such that $\lim _i\gamma _k(l_i)x(l_i)y(l_i)^{-1}\eta _k(l_i)=q$ . Therefore, we have

$$ \begin{align*}\lim_ix(n_i)y(n_i)^{-1}=g_k^{-1}pv_k^{-1},\quad\lim_ix(l_i)y(l_i)^{-1}=g_k^{-1}qv_k^{-1}.\end{align*} $$

By the property of uniformly convergence of $\gamma _mxy^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ , we have

$$ \begin{align*}\begin{array}{ll}\lim_mg_mg_k^{-1}pv_k^{-1}v_m &\!\!\! =\lim_m\lim_i\gamma_m(n_i)x(n_i)y(n_i)^{-1}\eta_m(n_i)\\ &\!\!\!=\lim_i\lim_m\gamma_m(n_i)x(n_i)y(n_i)^{-1}\eta_m(n_i)\\ &\!\!\!=\lim_i1_G=1_G.\end{array}\end{align*} $$

Similarly, we have $\lim _mg_mg_k^{-1}qv_k^{-1}v_m=1_G$ . In the end, we set $h=g_k^{-1}pq^{-1}g_k\ne 1_G$ . It follows that

$$ \begin{align*}\lim_mg_mhg_m^{-1}=\lim_m(g_mg_k^{-1}pv_k^{-1}v_m)(g_mg_k^{-1}qv_k^{-1}v_m)^{-1}=1_G.\end{align*} $$

So G is not distal.

Proof. It follows from Theorems 5.4, 5.5, 5.16, and 5.17.

Proof. Let d be a compatible metric on G and $d_u$ the supremum metric on $G^\omega $ . Since G is locally compact, we can find an $r>0$ with $\overline {B(1_G,r)}$ is compact. Let $V=\{\gamma \in c(G):d_u(1_{c(G)},\gamma )<r\}$ .

Assume for contradiction that there exist $x,y\in G^\omega $ with $x\leftrightsquigarrow _{c(G)}^2 y$ but $[x]\ne [y]$ , then there exist $g^x,g^y\in c(G)$ such that $g^xx\leftrightsquigarrow _V^1g^yy$ . Following the arguments of $\Rightarrow $ part in the proof of Lemma 5.2, we can find two sequences $(\gamma _m)$ and $(\eta _m)$ in V such that $\gamma _mg^xxy^{-1}(g^y)^{-1}\eta _m\rightrightarrows 1_{G^\omega }$ . Let $g_m=\lim _n\gamma _m(n)$ . Following the arguments in the proof of Theorem 5.17, there exists $h\ne 1_G$ such that $g_mhg_m^{-1}\to 1_G$ .

Since $\gamma _m\in V$ , we have $g_m\in \overline {B(1_G,r)}$ . There exists a subsequence of $(g_m)$ converging to some $g\in G$ , so $ghg^{-1}=1_G$ , i.e., $h=1_G$ . A contradiction!

Proof. Let $d\leq 1$ be a compatible metric on G and $d_u$ the supremum metric on $G^\omega $ . For $\varepsilon>0$ , we define $V_\varepsilon =\{\gamma \in c(G):d_u(1_{c(G)},\gamma )<\varepsilon \}$ .

Let $x,y,x',y'\in G^\omega $ , $m\in \omega $ , and $\alpha <\omega _1$ . If

1. (1) $x\leftrightarrow _{V_\varepsilon }^\alpha y$ ,

2. (2) $x(n)=x'(n),y(n)=y'(n)$ for $n\ge m$ , and

3. (3) $x'(y')^{-1},y'(x')^{-1}\in \overline {V_{\varepsilon /2}}$ ,

then we claim that $x'\leftrightsquigarrow _{V_\varepsilon }^\alpha y'$ .

To see this, we prove by induction on $\alpha $ . It is trivial that (3) implies $x'\leftrightsquigarrow _{V_\varepsilon }^0 y'$ . For any $\delta>0$ and any open $U\subseteq G^\omega $ with $x'\in U$ or $y'\in U$ , without loss of generality, we can assume that $x'\in U$ , and $U=U_0\times \cdots \times U_l\times G^\omega $ with $l\ge m$ and $U_0,\dots ,U_l$ open in G. By $x\leftrightarrow _{V_\varepsilon }^\alpha y$ , there exist $\gamma ^x,\gamma ^y\in V_{\varepsilon }$ such that $\gamma ^xx\leftrightarrow _{V_{\delta /2}}^\beta \gamma ^yy$ for $\beta <\alpha $ . We put

$$ \begin{align*}(\gamma^x)'(n)=\left\{\begin{array}{ll}1_G, & n\le l,\\ \gamma^x(n), & n>l,\end{array}\right. \quad(\gamma^y)'(n)=\left\{\begin{array}{ll}x'(n)y'(n)^{-1}, & n\le l,\\ \gamma^y(n), & n>l.\end{array}\right.\end{align*} $$

Then we have $(\gamma ^x)',(\gamma ^y)'\in V_\varepsilon $ and $(\gamma ^x)'x',(\gamma ^y)'y'\in U$ . Note that $\gamma ^xx\leftrightarrow _{V_{\delta /2}}^0\gamma ^yy$ gives $\gamma ^xx(\gamma ^yy)^{-1},\gamma ^yy(\gamma ^xx)^{-1}\in \overline {V_{\delta /2}}$ . Since

$$ \begin{align*}(\gamma^x)'(n)x'(n)((\gamma^y)'(n)y'(n))^{-1}=\left\{\begin{array}{ll}1_G , & n\le l,\\ \gamma^x(n)x(n)(\gamma^y(n)y(n))^{-1}, & n>l,\end{array}\right.\end{align*} $$

we have $(\gamma ^x)'x'((\gamma ^y)'y')^{-1}\in \overline {V_{\delta /2}}$ . Similarly, $(\gamma ^y)'y'((\gamma ^x)'x')^{-1}\in \overline {V_{\delta /2}}$ . Since $\gamma ^xx\leftrightarrow _{V_{\delta /2}}^\beta \gamma ^yy$ implies $\gamma ^xx\leftrightarrow _{V_\delta }^\beta \gamma ^yy$ for $\beta <\alpha $ . By the induction hypothesis, $(\gamma ^x)'x'\leftrightsquigarrow _{V_\delta }^\beta (\gamma ^y)'y'$ for $\beta <\alpha $ . Thus $(\gamma ^x)'$ and $(\gamma ^y)'$ witness that $x'\leftrightsquigarrow _{V_\varepsilon }^\alpha y'$ .

In the end, by setting $V_{\varepsilon /2}=c(G)$ , it follows that $x\leftrightarrow _{c(G)}^\alpha y$ implies $x\leftrightsquigarrow _{c(G)}^\alpha y$ .

Proof. Let d be a compatible metric on G, and let $d_u$ be the supremum metric on $G^\omega $ . By the continuity of $\phi $ , we can see that $\phi ^\omega :G^\omega \to H^\omega $ is continuous, and $\phi ^\omega \upharpoonright c(G)$ is also continuous from $c(G)$ to $c(H)$ .

We prove by induction on $\alpha $ . If $\alpha =0$ , note that $x\leftrightarrow _V^0 y$ iff $xy^{-1},yx^{-1}\in \overline V$ . It follows from the continuity of $\phi ^\omega $ that $\phi ^\omega (x)\phi ^\omega (y)^{-1}=\phi ^\omega (xy^{-1})\in \overline W$ and $\phi ^\omega (y)\phi ^\omega (x)^{-1}=\phi ^\omega (yx^{-1})\in \overline W$ , and hence $\phi ^\omega (x)\in \overline {W\phi ^\omega (y)}$ and $\phi ^\omega (y)\in \overline {W\phi ^\omega (x)}$ , i.e., $\phi ^\omega (x)\leftrightarrow _W^0\phi ^\omega (y)$ .

For $\alpha>0$ , we let $W'\subseteq c(H)$ be an open neighborhood of $1_{c(H)}$ . By the continuity of $\phi ^\omega \upharpoonright c(G)$ , we can find an open neighborhood $V'$ of $1_{c(G)}$ such that $\phi ^\omega (V')\subseteq W'$ . Note that $x\leftrightarrow _V^\alpha y$ , so there exist $\gamma ^x,\gamma ^y\in V$ such that $\gamma ^xx\leftrightarrow _{V'}^\beta \gamma ^yy$ for $\beta <\alpha $ . By induction hypothesis, we have $\phi ^\omega (\gamma ^x)\phi ^\omega (x)=\phi ^\omega (\gamma ^xx)\leftrightarrow _{W'}^\beta \phi ^\omega (\gamma ^yy)=\phi ^\omega (\gamma ^y)\phi ^\omega (y)$ . Therefore, $\phi ^\omega (\gamma ^x)$ and $\phi ^\omega (\gamma ^y)$ witness that $\phi ^\omega (x)\leftrightarrow _W^\alpha \phi ^\omega (y)$ .

Proof. Let $\rho $ be a compatible metric on $\Lambda \wr G$ , $\rho _u$ the supremum metric on $(\Lambda \wr G)^\omega $ . For any $\varepsilon>0$ , we define

$$ \begin{align*}V_\varepsilon=\{\gamma\in c(\Lambda\wr G):\rho_u(1_{c(\Lambda\wr G)},\gamma)<\varepsilon\}.\end{align*} $$

( $\Rightarrow $ ). Suppose G is $\alpha $ -l.m.-unbalanced. Then there exists an $a\in G^\omega \setminus c(G)$ such that $a\leftrightsquigarrow _{c(G)}^\alpha 1_{G^\omega }$ , and hence $a\leftrightarrow _{c(G)}^\alpha 1_{G^\omega }$ . For any $n\in \omega $ , let

$$ \begin{align*}\chi_n(\lambda)=\left\{\begin{array}{ll}a(n), & \lambda=1_\Lambda,\\ 1_G, & \lambda\ne 1_\Lambda.\end{array}\right.\end{align*} $$

We define $z\in (\Lambda \wr G)^\omega $ as $z(n)=(1_\Lambda ,\chi _n)$ for $n\in \omega $ . It is trivial that $z\notin c(\Lambda \wr G)$ . To see that $\Lambda \wr G$ is $(\alpha +1)$ -l.m.-unbalanced, by Lemma 5.23, we only need to show that $z\leftrightarrow _{c(\Lambda \wr G)}^{\alpha +1}1_{(\Lambda \wr G)^\omega }$ .

Note that the underlying topology of $\Lambda \wr G$ is the product topology on $\Lambda \times G^\Lambda $ . Thus for any given $\varepsilon>0$ , there exists a $\lambda _0\in \Lambda $ such that, for any $\chi \in G^\Lambda $ , if $\chi (\lambda )=1_G$ for all $\lambda \ne \lambda _0$ , then $\rho (1_{\Lambda \wr G},(1_\Lambda ,\chi ))<\varepsilon /2$ . Define

$$ \begin{align*}H=\{(1_\Lambda,\chi)\in\Lambda\wr G:\forall\lambda\in\Lambda\,(\lambda\ne\lambda_0\Rightarrow\chi(\lambda)=1_G)\}.\end{align*} $$

It is clear that H is a closed subgroup of $\lambda \wr G$ . Because $\rho _u(1_{(\Lambda \wr G)^\omega },x)\le \varepsilon /2<\varepsilon $ for all $x\in H^\omega $ , we have

$$ \begin{align*}c(H)=(c(\Lambda\wr G)\cap H^\omega)\subseteq V_\varepsilon.\end{align*} $$

We define $f:G\to H$ as: for $g\in G$ , $f(g)=(1_\Lambda ,\chi ^g)$ with

$$ \begin{align*}\chi^g(\lambda)=\left\{\begin{array}{ll}g, & \lambda=\lambda_0,\\ 1_G, & \lambda\ne\lambda_0.\end{array}\right.\end{align*} $$

It is easy to see that f is a topological isomorphism from G to H, and hence a continuous homomorphism from G to $\Lambda \wr G$ . Note that $f^\omega (c(G))\subseteq c(H)\subseteq V_\varepsilon $ . Then Lemma 5.24 gives $f^\omega (a)\leftrightarrow _{V_\varepsilon }^\alpha 1_{(\Lambda \wr G)^\omega }$ .

In $\Lambda \wr G$ , for $n\in \omega $ , we have

$$ \begin{align*}(\lambda_0,1_{G^\Lambda})z(n)(\lambda_0,1_{G^\Lambda})^{-1}=(1_\Lambda,\chi^{a(n)})=f(a(n)).\end{align*} $$

Define $\gamma _0\in c(\Lambda \wr G)$ as $\gamma _0(n)=(\lambda _0,1_{G^\Lambda })$ for $n\in \omega $ , then $\gamma _0z\gamma _0^{-1}=f^\omega (a)$ . It follows that $\gamma _0z=f^\omega (a)\gamma _0\leftrightarrow _{V_\varepsilon }^\alpha \gamma _0$ . From Definition 5.22, we can see that $\gamma _0z\leftrightarrow _{V_\varepsilon }^\beta \gamma _0$ for $0<\beta \le \alpha $ . To prove for $\beta =0$ , because $f^\omega (a),f^\omega (a)^{-1}\in H^\omega $ , we have $\rho _u(1_{(\Lambda \wr G)^\omega },f^\omega (a))<\varepsilon $ and $\rho _u(1_{(\Lambda \wr G)^\omega },f^\omega (a)^{-1})<\varepsilon $ . So $f^\omega (a),f^\omega (a)^{-1}\in \overline {V_\varepsilon }$ , then we have $f^\omega (a)\gamma _0\in \overline {V_\varepsilon \gamma _0}$ and $\gamma _0\in \overline {V_\varepsilon f^\omega (a)\gamma _0}$ , and hence $\gamma _0z=f^\omega (a)\gamma _0\leftrightarrow _{V_\varepsilon }^0\gamma _0$ . It follows that $z\leftrightarrow _{c(\Lambda \wr G)}^{\alpha +1}1_{(\Lambda \wr G)^\omega }$ .

( $\Leftarrow $ ). On the other hand, suppose $\Lambda \wr G$ is $(\alpha +1)$ -l.m.-unbalanced. Then there exists $z\in (\Lambda \wr G)^\omega \setminus c(\Lambda \wr G)$ such that $z\leftrightarrow _{c(\Lambda \wr G)}^{\alpha +1}1_{(\Lambda \wr G)^\omega }$ . Thus for any $\varepsilon>0$ , there exist $\gamma _\varepsilon ^z,\gamma _\varepsilon \in c(\Lambda \wr G)$ such that $\gamma _\varepsilon ^zz\leftrightarrow _{V_\varepsilon }^\alpha \gamma _\varepsilon $ and $\gamma _\varepsilon ^zz\leftrightarrow _{V_\varepsilon }^0\gamma _\varepsilon $ . Then we have $\gamma _\varepsilon ^zz\gamma _\varepsilon ^{-1}\leftrightarrow _{V_\varepsilon }^\alpha 1_{(\Lambda \wr G)^\omega }$ and $\gamma _\varepsilon ^zz\gamma _\varepsilon ^{-1}\in \overline {V_\varepsilon }$ .

Fix a small enough $\varepsilon _0>0$ such that, for any $\gamma \in \overline {V_{\varepsilon _0}}$ and any $n\in \omega $ , $\gamma (n)=(1_\Lambda ,\chi )$ for some $\chi \in G^\Lambda $ , i.e., $\overline {V_{\varepsilon _0}}$ is a subset of the subgroup $(\{1_\Lambda \}\times G^\Lambda )^\omega $ of $(\Lambda \wr G)^\omega $ . So we can write $\gamma _{\varepsilon _0}^z(n)z(n)\gamma _{\varepsilon _0}(n)^{-1}=(1_\Lambda ,\xi _n)$ with $\xi _n\in G^\Lambda $ for each $n\in \omega $ . Note that $\gamma _{\varepsilon _0}^zz\gamma _{\varepsilon _0}^{-1}\notin c(\Lambda \wr G)$ , we can find a $\lambda _0\in \Lambda $ such that $\lim _n\xi _n(\lambda _0)$ diverges.

Now we define $\pi :(\{1_\Lambda \}\times G^\Lambda )\to G$ as $\pi (1_\Lambda ,\chi )=\chi (\lambda _0)$ for $\chi \in G^\Lambda $ . It is clear that $\pi $ is a continuous homomorphism. Note that $\gamma _{\varepsilon _0}^zz\gamma _{\varepsilon _0}^{-1}\in \overline {V_{\varepsilon _0}}\subseteq (\{1_\Lambda \}\times G^\Lambda )^\omega $ . Then $\pi ((\gamma _{\varepsilon _0}^zz\gamma _{\varepsilon _0}^{-1})(n))=\xi _n(\lambda _0)$ for all $n\in \omega $ , so $\pi ^\omega (\gamma _{\varepsilon _0}^zz\gamma _{\varepsilon _0}^{-1})\notin c(G)$ . By the continuity of $\pi :(\{1_\Lambda \}\times G^\Lambda )\to G$ , we have $\pi ^\omega (c(\{1_\Lambda \}\times G^\Lambda ))\subseteq c(G)$ , and hence $\pi ^\omega (V_{\varepsilon _0})\subseteq c(G)$ . Then Lemma 5.24 gives $\pi ^\omega (\gamma _{\varepsilon _0}^zz\gamma _{\varepsilon _0}^{-1})\leftrightarrow _{c(G)}^\alpha 1_{G^\omega }$ . Therefore, G is $\alpha $ -l.m.-unbalanced.

Proof. It is trivial that G is topologically isomorphic to a closed subgroup of $\Lambda \wr G$ , so $E(G)\le _BE(\Lambda \wr G)$ . It follows from Theorems 5.15 and 5.25 that $E(\Lambda \wr G)\not \le _BE(G)$ .