Proof It is routine to prove clause (1) by induction on k based on Proposition 2.1. Clause (2) is an easy corollary of (1). And clause (3) is trivial.

Proof If S contains an infinite strictly increasing sequence $(s_n)$ , then $(\phi (s_n))$ is an infinite strictly increasing sequence in T, contradicting that $(T,<)$ is well-founded.

Now we prove that $\rho _S(s)\le \rho _T(\phi (s))$ for $s\in S$ by induction on $\rho _S(s)$ . For the basis of induction, note that $\rho _S(s)=0\leq \rho _T(\phi (s))$ clearly holds; and for the inductive step, we have the following inequality:

$$ \begin{align*}\begin{array}{ll}\rho_S(s) &=\sup\{\rho_S(u)+1:s<u\wedge u\in S\}\cr &\le\sup\{\rho_T(\phi(u))+1:s<u\wedge u\in S\}\cr &\le\sup\{\rho_T(t)+1:\phi(s)<t\wedge t\in T\}=\rho_T(\phi(s)).\end{array}\end{align*} $$

This implies $\rho (S)\le \rho (T)$ .

Proof Assume for contradiction that $T_{\mathcal {G}}^X$ is ill-founded, then there exists an infinite sequence $(n,C_n),\,n<\omega $ in $T_{\mathcal {G}}^X$ with $C_n\supseteq C_{n+1}$ for each $n<\omega $ .

Let d be a compatible complete left-invariant metric on G.

Fix an $x_0\in C_0$ . Then we have $G_0\cdot x_0=C_0\supseteq C_1$ , so we can find a $g_0\in G_0$ such that $g_0\cdot x_0\in C_1$ . Inductively, we can find a $g_n\in G_n$ for each $n<\omega $ such that $g_ng_{n-1}\dots g_0\cdot x_0\in C_{n+1}$ . Put $h_n=g_0^{-1}\dots g_n^{-1}$ for each $n<\omega $ . Then, for any $n,p<\omega $ , we have $d(h_{n+p},h_n)=d(h_n^{-1}h_{n+p},1_G)=d(g_{n+1}^{-1}\dots g_{n+p}^{-1},1_G)\le \mathrm {diam}(G_{n+1})\to 0$ as $n\to \infty $ . It follows that $(h_n)$ is a d-Cauchy sequence in G, so it converges to some $h\in G$ .

Let $x_\infty =h^{-1}\cdot x_0$ . Since $h_n\to h$ , we have $h_n^{-1}\to h^{-1}$ , and hence $h_n^{-1}\cdot x_0\to h^{-1}\cdot x_0=x_\infty $ . Note that X is discrete, so there exists an integer N such that $h_n^{-1}\cdot x_0=x_\infty $ for any $n>N$ , thus $x_\infty =g_ng_{n-1}\dots g_0\cdot x_0\in C_{n+1}$ . This implies that $x_\infty \in \bigcap _nC_n$ and $G_n\cdot x_\infty =C_n$ for each $n<\omega $ .

Finally, define $G_{x_\infty }=\{g\in G:g\cdot x_\infty =x_\infty \}$ and put $f:G\to X$ as $f(g)=g\cdot x_\infty $ . Since f is continuous and $\{x_\infty \}$ is clopen in X, it follows that $G_{x_\infty }=f^{-1}(x_\infty )$ is a clopen subgroup of G. So there exists an $m<\omega $ such that $G_m\subseteq G_{x_\infty }$ . We now have $C_m=G_m\cdot x_\infty =\{x_\infty \}$ , which is a singleton $G_m$ -orbit, contradicting that $(m,C_m)\in T_{\mathcal {G}}^X$ .

Proof Note that $\theta $ is injective. Let us prove that $\phi (n,C)\in T_{\mathcal {F}}^Y$ holds for an arbitrary $(n,C)\in T_{\mathcal {E}}^X$ . Indeed, it suffices to note that since C is not a singleton, neither is $[\phi (C)]_{F_n}$ . This proves $\phi (n,C)\in T_{\mathcal {F}}^Y$ . The rest of the proof is trivial.

Proof $\rho (T|(n_i))\le \rho (T)$ follows from Proposition 2.3. We prove $\omega (\rho (T))\le \rho (T|(n_i))$ by induction on $\rho (T)$ .

First, if $\rho (T)<\omega $ , then $\rho (T)=\min \{n:L_n(T)=\emptyset \}$ , and hence $\rho (T|(n_i))=\min \{i:L_{n_i}(T)=\emptyset \}$ . So we have $\omega (\rho (T))=0\le \rho (T|(n_i))$ .

For $t\in T|(n_i)$ , note that $(T|(n_i))_t=\{u\in T|(n_i):t=u\vee t<u\}$ is a level-subtree of $T_t$ as well.

Case 1: If $\rho (T)$ is a limit ordinal, then $\omega (\rho (T))=\rho (T)$ . Proposition 2.1 implies that $\rho (T)=\sup \{\rho (T_t):t\in T\}$ . Since $\rho (T)$ is a limit ordinal and $\rho (T_t)$ is a successor ordinal, we have $\rho (T_t)<\rho (T)$ for $t\in T$ .

Subcase 1.1: If there is no maximum in $\{\omega (\rho (T_t)):t\in T\}$ , we have

$$ \begin{align*}\rho(T)=\sup\{\rho(T_t):t\in T\}=\sup\{\omega(\rho(T_t)):t\in T\}.\end{align*} $$

By the inductive hypothesis, we have $\omega (\rho (T_t))\le \rho ((T|(n_i))_t)$ . Proposition 2.3 gives $\rho ((T|(n_i))_t)\le \rho (T|(n_i))$ for each $t\in T$ , so we have $\rho (T|(n_i))=\rho (T)$ .

Subcase 1.2: Otherwise, let $\alpha =\max \{\omega (\rho (T_t)):t\in T\}$ . Since $\rho (T_t)<\rho (T)$ for $t\in T$ , we have $\rho (T)=\alpha +\omega $ . We can find a sequence $t_m,\,m<\omega $ in $L_0$ such that $\rho (T_{t_m})=\alpha +k_m$ with $\sup \{k_m:m<\omega \}=\omega $ . By Proposition 2.1, for each $m<\omega $ and $n<k_m$ we can find $t_m^n\in L_n(T)$ such that $t_m=t_m^0<t_m^1<\dots <t_m^{k_m-1}$ and $\rho (T_{t_m^n})=\alpha +(k_m-n)$ . For $k_m>n_0$ , let $i_m$ be the largest i such that $n_i<k_m$ , then $t_m^{n_j}\in L_j(T|(n_i))=L_{n_j}(T)$ for $j\le i_m$ . By the inductive hypothesis, $\alpha =\omega (\rho (T_{t_m^{n_j}}))\le \rho ((T|(n_i))_{t_m^{n_j}})$ . Since $\rho ((T|(n_i))_{t_m^{n_j}})\ge \rho ((T|(n_i))_{t_m^{n_{j+1}}})+1$ for each $j<i_m$ , we have $\rho ((T|(n_i))_{t_m^{n_0}})\ge \alpha +i_m$ . By the definition of $i_m$ , we have $\sup \{i_m:m<\omega \}=\omega $ . This gives $\rho (T|(n_i))=\alpha +\omega =\rho (T)$ .

Case 2: If $\rho (T)=\omega (\rho (T))+n$ with $1\le n<\omega $ , then there exists some $t_0\in L_0(T)$ such that $\rho (T)=\rho _T(t_0)+1$ . Since $\{u\in T|(n_i):t_0<u\}$ is a level-subtree of $\{u\in T:t_0<u\}$ and $\rho (\{u\in T:t_0<u\})=\rho _T(t_0)<\rho (T)$ , by the inductive hypothesis and Proposition 2.3, we have

$$ \begin{align*}\omega(\rho_T(t_0))=\omega(\rho(\{u\in T:t_0<u\}))\le\rho(\{u\in T|(n_i):t_0<u\})\le\rho(T|(n_i)).\end{align*} $$

It follows that $\omega (\rho (T))=\omega (\rho _T(t_0))\le \rho (T|(n_i))$ .

Proof (1) First, we consider the case where $(G_n')$ is a subsequence of $(G_n)$ , i.e., there is a strictly increasing sequence $(n_i)$ of natural numbers such that $G_i'=G_{n_i}$ for each $i<\omega $ .

We define $\psi :T_{\mathcal G'}^X\to T_{\mathcal {G}}^X$ as $\psi (i,C)=(n_i,C)$ . It is clear that $\psi $ is an order-preserving isomorphism from $T_{\mathcal G'}^X$ onto $T_{\mathcal {G}}^X|(n_i)$ . It follows that

$$ \begin{align*}\omega(\rho(T_{\mathcal{G}}^X))\le\rho(T_{\mathcal G'}^X)=\rho(T_{\mathcal{G}}^X|(n_i))\le\rho(T_{\mathcal{G}}^X).\end{align*} $$

So we have $\omega (\rho (T_{\mathcal {G}}^X))=\omega (\rho (T_{\mathcal G'}^X))$ .

(2) Since $(G_n),(G_n')\in \mathrm {dgnb}(G)$ , we can find two strictly increasing natural numbers $(n_i)$ and $(m_j)$ such that $n_0=0,m_0=0$ , and

$$ \begin{align*}G_0\supseteq G_{m_0}'\supseteq G_{n_1}\supseteq G_{m_1}'\supseteq G_{n_2}\supseteq\dots.\end{align*} $$

Define $H_{2i}=G_{n_i}$ and $H_{2i+1}=G_{m_i}'$ for each $i<\omega $ . Then $(H_k)\in \mathrm {dgnb}(G)$ . Put $\mathcal H=(H_k),\mathcal K=(G_{n_i})$ , and $\mathcal K'=(G_{m_i}')$ .

Note that $(G_{n_i})$ is a subsequence of $(G_n)$ and also a subsequence of $(H_k)$ . From (1), we obtain

$$ \begin{align*}\omega(\rho(T_{\mathcal{G}}^X))=\omega(\rho(T_{\mathcal{K}}^X))=\omega(\rho(T_{\mathcal{H}}^X)).\end{align*} $$

Similarly, we obtain

$$ \begin{align*}\omega(\rho(T_{\mathcal G'}^X))=\omega(\rho(T_{\mathcal K'}^X))=\omega(\rho(T_{\mathcal{H}}^X)).\end{align*} $$

So we have $\omega (\rho (T_{\mathcal {G}}^X))=\omega (\rho (T_{\mathcal G'}^X))$ .

Proof For any $(n,C)\in T_{\mathcal {F}}^Y$ , we construct $\psi (n,C)$ by induction on n such that $\psi (n,C)=(n,[x]_{E_n})$ for some $x\in X$ with $[\theta (x)]_{F_n}=C$ .

For $n=0$ , since $\theta $ is a surjection, we can find an $x\in X$ with $\theta (x)\in C$ . Then we let $\psi (0,C)=(0,[x]_{E_0})$ .

For $n>0$ , since C is an $F_n$ -equivalence class, there exists a unique $F_{n-1}$ -equivalence class $D\supseteq C$ . By the inductive hypothesis, we can find an $x'\in X$ such that $\psi (n-1,D)=(n-1,[x']_{E_{n-1}})$ with $[\theta (x')]_{F_{n-1}}=D$ . Since $\theta ([x']_{E_{n-1}})=[\theta (x')]_{F_{n-1}}=D\supseteq C$ , we can find an $x\in [x']_{E_{n-1}}$ such that $\theta (x)\in C$ . Then we put $\psi (n,C)=(n,[x]_{E_n})$ .

Since C is not a singleton, the equality $\theta ([x]_{E_n})=[\theta (x)]_{F_n}=C$ implies that $[x]_{E_n}$ is not a singleton. So $\psi (n,C)\in T_{\mathcal {E}}^X$ . From the construction, it is routine to check that $\psi :T_{\mathcal {F}}^Y\to T_{\mathcal {E}}^X$ is a Lipschitz embedding.

Finally, if $T_{\mathcal {E}}^X$ is well-founded, then by Proposition 2.3, $T_{\mathcal {F}}^Y$ is well-founded as well, and $\rho (T_{\mathcal {F}}^Y)\le \rho (T_{\mathcal {E}}^X)$ .

Proof (1) Note that $L_0(T^{X(\mathcal G)}_{\mathcal {G}})=\{(0,G/G_k):G\ne G_k\}$ . We point out that $(T^{X(\mathcal G)}_{\mathcal {G}})_{(0,G/G_k)}\cong T^{G/G_k}_{\mathcal {G}}$ for $G\ne G_k$ , and $(T^{X(\mathcal G)}_{\mathcal {G}})_{(0,G/G_k)}=T^{G/G_k}_{\mathcal {G}}=\emptyset $ for $G=G_k$ . So

$$ \begin{align*}\begin{array}{ll}\rho(\mathcal G)&=\rho(T^{X(\mathcal G)}_{\mathcal{G}})=\sup\{\rho((T^{X(\mathcal G)}_{\mathcal{G}})_{(0,G/G_k)}):k<\omega\}\cr &=\sup\{\rho(T^{G/G_k}_{\mathcal{G}}):k<\omega\}=\sup\{\rho^k(\mathcal G):k<\omega\}.\end{array}\end{align*} $$

(2) Given $k<\omega $ , we define $\theta :G/G_{k+1}\to G/G_k$ as $\theta (gG_{k+1})=gG_k$ for $g\in G$ . It is clear that $\theta $ is well-defined and is a surjection. Furthermore, for $n<\omega $ and $g\in G$ , we have

$$ \begin{align*}\begin{array}{ll}\theta(G_n\cdot gG_{k+1})&=\{\theta(hgG_{k+1}):h\in G_n\}\cr &=\{hgG_k:h\in G_n\}=G_n\cdot gG_k=G_n\cdot\theta(gG_{k+1}).\end{array}\end{align*} $$

From Lemma 3.8, we have

$$ \begin{align*}\rho(T_{\mathcal{G}}^{G/G_k})\le\rho(T_{\mathcal{G}}^{G/G_{k+1}}),\end{align*} $$

i.e., $(\rho ^k(\mathcal G))$ is increasing.

For each $k<\omega $ , since $T_{\mathcal {G}}^{G/G_k}$ is countable, $\rho ^k(\mathcal G)$ is countable as well. If $G=G_k$ , then $T^{G/G_k}_{\mathcal {G}}=\emptyset $ ; otherwise if $G\ne G_k$ , then $L_0(T^{G/G_k}_{\mathcal {G}})=\{(0,G/G_k)\}$ is a singleton. So $\rho ^k(\mathcal G)=\rho (T^{G/G_k}_{\mathcal {G}})$ is either $0$ or a successor ordinal.

Proof Fix an $x\in X$ . Since $\{x\}$ is clopen in X, by the continuity of the group action of G on X, we know $G_x$ is a clopen subgroup of G. So there is some $k<\omega $ such that $G_k\subseteq G_x$ . Then we can define $\theta :G/G_k\to X$ as $\theta (gG_k)=g\cdot x$ for $g\in G$ . Since X is a transitive G-space, $\theta $ is surjective and $\theta (G_n\cdot gG_k)=G_n\cdot \theta (gG_k)$ for each $n<\omega $ and $g\in G$ . By Lemma 3.8, we have $\rho (T_{\mathcal {G}}^X)\le \rho ^k(\mathcal G)$ .

Proof Note that $L_0(T_{\mathcal {G}}^X)=\{(0,G\cdot x):x\in X\wedge G\cdot x\ne \{x\}\}$ and $T_{\mathcal {G}}^{G\cdot x}\cong (T_{\mathcal {G}}^X)_{(0,G\cdot x)}$ for $G\cdot x\ne \{x\}$ , so we have

$$ \begin{align*}\rho(T_{\mathcal{G}}^X)=\sup\{\rho((T_{\mathcal{G}}^X)_{(0,G\cdot x)}):x\in X\}=\sup\{\rho(T_{\mathcal{G}}^{G\cdot x}):x\in X\}.\end{align*} $$

By Lemma 3.11, we have

$$ \begin{align*}\rho(T_{\mathcal{G}}^X)\le\sup\{\rho^k(\mathcal G):k<\omega\}=\rho(\mathcal G).\\[-35pt]\end{align*} $$

Proof The $(\Rightarrow )$ part follows from Lemma 3.3.

$(\Leftarrow )$ . Given a countable Polish G-space X. Following the arguments in the proof of Lemma 3.11, we can see that, for any $x\in X$ , there is a $k<\omega $ and a Lipschitz embedding from $T^{G\cdot x}_{\mathcal {G}}$ to $T_{\mathcal {G}}^{G/G_k}$ . Since $T_{\mathcal {G}}^{G/G_k}$ is well-founded, so is $T^{G\cdot x}_{\mathcal {G}}$ . By the arbitrariness of $x\in X$ , we conclude that $T^X_{\mathcal {G}}$ is also well-founded. Combining this with [Reference Malicki5, Theorem 6], it follows that G is CLI.

Proof By Corollary 3.12, we have $\rho (T^{X(\mathcal G)}_{\mathcal G'})\le \rho (\mathcal G')$ . Thus Lemma 3.7 implies

$$ \begin{align*}\omega(\rho(\mathcal G))=\omega(\rho(T^{X(\mathcal G)}_{\mathcal{G}}))=\omega(\rho(T^{X(\mathcal G)}_{\mathcal G'}))\le\omega(\rho(\mathcal G')),\end{align*} $$

and vice versa.

Proof (1) If $\mathrm {rank}(G)>0$ , then $\omega \cdot \mathrm {rank}(G)$ is a limit ordinal. By Lemma 3.10, $\rho ^k(\mathcal G)$ is either $0$ or a successor ordinal, so $\rho ^k(\mathcal G)<\omega \cdot \mathrm {rank}(G)$ for any $k<\omega $ .

(2) Clearly, $\rho (\mathcal G)=\omega \cdot \mathrm {rank}(G)+m$ for some $0<m<\omega $ . Again by Lemma 3.10, $(\rho ^k(\mathcal G))$ is increasing, so $\rho ^k(\mathcal G)=\omega \cdot \mathrm {rank}(G)+m$ for large enough $k<\omega $ .

Proof If $\mathrm {rank}(G)=0$ , then $\rho (\mathcal G)=\omega \cdot \mathrm {rank}(G)$ implies that $\rho ^k(\mathcal G)=0$ for any $k<\omega $ . So $T_{\mathcal {G}}^{G/G_k}=\emptyset $ , and hence $G_0\cdot G_k\notin T_{\mathcal {G}}^{G/G_k}$ . It follows that $G_0\cdot G_k=\{G_k\}$ , i.e., $G=G_0=G_k$ for any $k<\omega $ . This implies that $G=\{1_G\}$ . Then we can easily see that $T_{\mathcal G'}^{G/G_k'}=\emptyset $ for $k<\omega $ . So $\rho (\mathcal G')=\omega \cdot \mathrm { rank}(G)$ holds. And vice versa.

If $\mathrm {rank}(G)>0$ , assume for contradiction that $\rho (\mathcal G)=\omega \cdot \mathrm {rank}(G)$ , but $\rho (\mathcal G')>\omega \cdot \mathrm {rank}(G)$ . From Lemma 3.16, we have $\rho ^k(\mathcal G)<\omega \cdot \mathrm { rank}(G)$ for any $k<\omega $ , but $\rho ^l(\mathcal G')=\omega \cdot \mathrm {rank}(G)+m$ for some $0<m<\omega $ and large enough $l<\omega $ . From Lemmas 3.7 and 3.11, we can conclude that for any $l<\omega $ , there is some $k<\omega $ such that

$$ \begin{align*}\omega(\rho^l(\mathcal G'))=\omega(\rho(T_{\mathcal G'}^{G/G_l'}))=\omega(\rho(T_{\mathcal{G}}^{G/G_l'}))\le\rho(T_{\mathcal{G}}^{G/G_l'})\le\rho^k(\mathcal G).\end{align*} $$

A contradiction!

Proof Fix a sequence $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ .

(1) It follows from the first paragraph of the proof of Theorem 3.17.

(2) If G is L- $0$ -CLI, then $\mathrm {rank}(G)=0$ . So there is an $m<\omega $ such that $\rho (T_{\mathcal {G}}^{G/G_k})=\rho ^k(\mathcal G)=m$ for large enough $k<\omega $ . Then we have $L_m(T_{\mathcal {G}}^{G/G_k})=\emptyset $ . This implies that $G_m\cdot G_k=\{G_k\}$ . So $G_m\subseteq G_k$ for large enough $k<\omega $ , and thus $G_m=\{1_G\}$ . It follows that G is discrete.

On the other hand, if G is discrete, then there is an $m<\omega $ such that $G_m=\{1_G\}$ . Therefore, for any $k<\omega $ , we have $L_m(T_{\mathcal {G}}^{G/G_k})=\emptyset $ , and hence $\rho ^k(\mathcal G)\le m$ . This implies that $\mathrm {rank}(G)=0$ , i.e., G is L- $0$ -CLI.

(3) If G is $1$ -CLI, then Lemma 3.16 implies that $\rho ^k(\mathcal G)<\omega $ for any $k<\omega $ . So there is an $m_k<\omega $ such that $L_{m_k}(T_{\mathcal {G}}^{G/G_k})=\emptyset $ . It follows that, for $g\in G$ , $G_{m_k}\cdot gG_k=\{gG_k\}$ , so $g^{-1}G_{m_k}g\subseteq G_k$ . Put $U_k=\bigcup \{g^{-1}G_{m_k}g:g\in G\}\subseteq G_k$ . We can see that $(U_k)$ is a neighborhood basis of $1_G$ with $g^{-1}U_kg=U_k$ for all $g\in G$ . By Klee’s theorem (cf. [Reference Klee4] or [Reference Gao2, Exercise 2.1.4]), G is TSI.

On the other hand, if G is TSI, again by Klee’s theorem, we can find a neighborhood basis $(U_m)$ of $1_G$ with $g^{-1}U_mg=U_m$ for all $g\in G$ . For any $n<\omega $ , there is an $m_n<\omega $ such that $U_{m_n}\subseteq G_n$ . Let $V_n=U_{m_n}\cap U_{m_n}^{-1}$ and $G_n'=\bigcup _iV_n^i$ . Then $G_n'$ is an open normal subgroup of G with $G_n'\subseteq G_n$ . So $(G_n')\in \mathrm {dgnb}(G)$ . Let $\mathcal G'=(G_n')$ . Then $G_k'\cdot gG_k'=\{gG_k'\}$ for all $g\in G$ and $k<\omega $ , thus $L_k(T_{\mathcal G'}^{G/G_k'})=\emptyset $ . So $\rho ^k(\mathcal G')\le k<\omega $ , and hence G is $1$ -CLI.

Proof $(\Rightarrow )$ . Suppose G is L- $\alpha $ -CLI. Without loss of generality, we may assume that G is not $\alpha $ -CLI. Fix a sequence $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ . There exists an $m\ge 1$ such that $\rho (\mathcal G)=\omega \cdot \alpha +m$ , and thus we can pick a $k_0>m$ such that $\rho ^k(\mathcal G)=\omega \cdot \alpha +m$ for any $k\ge k_0$ . We will show that $G_{k_0}$ is $\alpha $ -CLI.

Let $H=G_{k_0}$ and $H_n=G_{n+k_0}$ for $n<\omega $ . Then $(H_n)\in \mathrm {dgnb}(H)$ . Put $\mathcal H=(H_n)$ . Given $k<\omega $ , define $\phi :T^{H/H_k}_{\mathcal {H}}\to (T^{G/G_{k+k_0}}_{\mathcal {G}})_{(k_0,G_{k_0}/G_{k+k_0})}$ as $\phi (n,C)=(n+k_0,C)$ . It is trivial to see that $\phi $ is an order-preserving isomorphism. From Proposition 2.2, since $k_0>m$ , we have

$$ \begin{align*}\rho^k(\mathcal H)=\rho(T^{H/H_k}_{\mathcal{H}})=\rho((T^{G/G_{k+k_0}}_{\mathcal{G}})_{(k_0,G_{k_0}/G_{k+k_0})})\le\omega\cdot\alpha.\end{align*} $$

So $\rho (\mathcal H)\le \omega \cdot \alpha $ , and hence $H=G_{k_0}$ is $\alpha $ -CLI.

$(\Leftarrow )$ . Suppose G is locally $\alpha $ -CLI. Let H be an open subgroup of G which is $\alpha $ -CLI, and let $\mathcal H=(H_n)\in \mathrm {dgnb}(H)$ . Then $\rho ^k(\mathcal H)\le \omega \cdot \alpha $ for $k<\omega $ .

Put $G_0=G$ and $G_n=H_{n-1}$ for $n\ge 1$ . Then $(G_n)\in \mathrm {dgnb}(G)$ . Put $\mathcal G=(G_n)$ . Given $g\in G$ and $k<\omega $ , by similar arguments in the $(\Rightarrow )$ part, we have $T^{H\cdot gH_k}_{\mathcal {H}}\cong (T^{G/G_{k+1}}_{\mathcal {G}})_{(1,G_1\cdot gG_{k+1})}$ . By Lemma 3.11, there exists an $l<\omega $ such that $\rho (T^{H\cdot gH_k}_{\mathcal {H}})\le \rho ^l(\mathcal H)$ . Therefore, by Proposition 2.1,

$$ \begin{align*}\begin{array}{ll}\rho^{k+1}(\mathcal G)&=\rho(T^{G/G_{k+1}}_{\mathcal{G}})\\[1pt]&\le\sup\{\rho((T^{G/G_{k+1}}_{\mathcal{G}})_{(1,G_1\cdot gG_{k+1})}):g\in G\}+1\\[3pt]&=\sup\{\rho(T^{H\cdot gH_k}_{\mathcal{H}}):g\in G\}+1\\[2pt]&\le\sup\{\rho^l(\mathcal H):l<\omega\}+1\\[2pt]&\le\omega\cdot\alpha+1.\end{array}\end{align*} $$

So $\rho (\mathcal G)\le \omega \cdot \alpha +1$ , and hence G is L- $\alpha $ -CLI.

Proof Let $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ , and put $H_n=H\cap G_n$ for $n<\omega $ . It is clear that $(H_n)\in \mathrm {dgnb}(H)$ . Put $\mathcal H=(H_n)$ . We only need to show that $\rho (\mathcal H)\le \rho (\mathcal G)$ .

Given $k<\omega $ , define $\theta :H/H_k\to G/G_k$ as $\theta (hH_k)=hG_k$ for $h\in H$ . By Proposition 3.4, there is a Lipschitz order-preserving map from $T^{H/H_k}_{\mathcal {H}}$ to $T^{G/G_k}_{\mathcal {G}}$ . So Proposition 2.3 implies

$$ \begin{align*}\rho^k(\mathcal H)=\rho(T^{H/H_k}_{\mathcal{H}})\le\rho(T^{G/G_k}_{\mathcal{G}})=\rho^k(\mathcal G).\end{align*} $$

Then we have $\rho (\mathcal H)\le \rho (\mathcal G)$ as desired.

Proof Let $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ , and put $H_n=G_n\cdot N=\{\hat gN:\hat g\in G_n\}$ for $n<\omega $ . It is clear that $H_0=G/N$ and $(H_n)\in \mathrm {dgnb}(G/N)$ . Put $\mathcal H=(H_n)$ . We only need to show that $\rho (\mathcal H)\le \rho (\mathcal G)$ .

Given $k<\omega $ , define $\theta :G/G_k\to (G/N)/H_k$ as $\theta (gG_k)=(gN)H_k$ for $g\in G$ . Note that for $n<\omega $ ,

$$ \begin{align*}\begin{array}{ll}\theta(G_n\cdot gG_k)&=\theta(\{\hat ggG_k:\hat g\in G_n\})\cr &=\{(\hat ggN)H_k:\hat g\in G_n\}\cr &=\{(\hat gN)(gN)H_k:\hat g\in G_n\}\cr &=\{(\hat gN)\theta(gG_k):\hat g\in G_n\}\cr &=\{\hat gN:\hat g\in G_n\}\theta(gG_k)=H_n\cdot\theta(gG_k).\end{array}\end{align*} $$

By Lemma 3.8, we have

$$ \begin{align*}\rho^k(\mathcal H)=\rho(T^{(G/N)/H_k}_{\mathcal{H}})\le\rho(T^{G/G_k}_{\mathcal{G}})=\rho^k(\mathcal G).\end{align*} $$

So $\rho (\mathcal H)\le \rho (\mathcal G)$ holds as desired.

Proof First, note that $[(x,y)]_{E_n\times F_n}=[x]_{E_n}\times [y]_{F_n}$ for all $(x,y)\in X\times Y$ and $n<\omega $ .

(1) For any sequences $(x_n),(y_n)$ in $X,Y$ , respectively, $((n,[(x_n,y_n)]_{E_n\times F_n}))$ is an infinite branch of $T_{\mathcal {E\times F}}^{X\times Y}$ iff either $((n,[x_n]_{E_n}))$ or $((n,[y_n]_{F_n}))$ is an infinite branch of $T_{\mathcal {E}}^X$ or $T_{\mathcal {F}}^Y$ , respectively.

(2) If $T_{\mathcal {E\times F}}^{X\times Y}$ is well-founded, by (1), $T_{\mathcal {E}}^X$ and $T_{\mathcal {F}}^Y$ are also well-founded. For all $(x,y)\in X\times Y$ and $n<\omega $ , note that

$$ \begin{align*}\begin{array}{ll} & (n,[(x,y)]_{E_n\times F_n})\in T^{X\times Y}_{\mathcal{E\times F}}\cr \iff & [(x,y)]_{E_n\times F_n}\ne\{(x,y)\}\cr \iff & [x]_{E_n}\ne\{x\}\vee[y]_{F_n}\ne\{y\}\cr \iff & (n,[x]_{E_n})\in T_{\mathcal{E}}^X\vee(n,[y]_{F_n})\in T_{\mathcal{F}}^Y.\end{array}\end{align*} $$

By Proposition 2.1, it is routine to prove

$$ \begin{align*}\rho((T_{\mathcal{E\times F}}^{X\times Y})_{(n,[(x,y)]_{E_n\times F_n})})=\max\{\rho((T_{\mathcal{E}}^X)_{(n,[x]_{E_n})}),\rho((T_{\mathcal{F}}^Y)_{(n,[y]_{F_n})})\}\end{align*} $$

by induction on $\rho ((T_{\mathcal {E\times F}}^{X\times Y})_{(n,[(x,y)]_{E_n\times F_n})})$ . Taking supremum on both sides of the above formula, we get

$$ \begin{align*}\rho(T_{\mathcal{E\times F}}^{X\times Y})=\max\{\rho(T_{\mathcal{E}}^X),\rho(T_{\mathcal{F}}^Y)\}.\\[-36pt]\end{align*} $$

Proof For any $k<\omega $ , put $X=G/G_k,Y=H/H_k$ , and define $E_n,F_n$ on X and Y for each $n<\omega $ , respectively, by

$$ \begin{align*}(\hat gG_k,\tilde gG_k)\in E_n\iff\exists g\in G_n\,(g\hat gG_k=\tilde gG_k)\quad(\forall\hat g,\tilde g\in G),\end{align*} $$

$$ \begin{align*}(\hat hH_k,\tilde hH_k)\in F_n\iff\exists h\in H_n\,(h\hat hH_k=\tilde hH_k)\quad(\forall\hat h,\tilde h\in H).\end{align*} $$

Then Lemma 4.3 gives $\rho ^k(\mathcal {G\times H})=\max \{\rho ^k(\mathcal G),\rho ^k(\mathcal H)\}$ . The rest follows trivially.

Proof Given $k<\omega $ , we put $Y=G^0/G^0_k\times \dots \times G^{k-1}/G^{k-1}_k$ and define $\theta :G/G_k\to Y$ as

$$ \begin{align*}\theta((g_i)G_k)=(g_0G^0_k,\dots,g_{k-1}G^{k-1}_k)\end{align*} $$

for $(g_i)\in G=\prod _iG^i$ . By the definition of $G_k$ , it is easy to see that $\theta $ is a bijection. Moreover, we put

$$ \begin{align*}H_n=\left\{\begin{array}{ll}\prod_{i<n}G^i_n\times\prod_{n\le i<k}G^i, & n<k,\cr \prod_{i<k}G^i_n, & n\ge k,\end{array} \right.\end{align*} $$

then $\theta $ is a reduction of $E^{G/G_k}_{G_n}$ to $E^Y_{H_n}$ for each $n<\omega $ . Put $\mathcal H=(H_n)$ . By Proposition 3.4, $(n,G_n\cdot (g_i)G_k)\mapsto (n,H_n\cdot \theta ((g_i)G_k))$ is an order-preserving isomorphism from $T^{G/G_k}_{\mathcal {G}}$ to $T^Y_{\mathcal {H}}$ . So $\rho ^k(\mathcal G)=\rho (T^{G/G_k}_{\mathcal {G}})=\rho (T^Y_{\mathcal {H}})$ .

For $n\ge k$ and $(g_i)\in G$ , we have

$$ \begin{align*}H_n\cdot\theta((g_i)G_k)=(G^0_n\cdot g_0G^0_k)\times\dots\times(G^{k-1}_n\cdot g_{k-1}G^{k-1}_k).\end{align*} $$

Lemma 4.3 gives

$$ \begin{align*}\rho((T^Y_{\mathcal{H}})_{(k,H_k\cdot\theta((g_i)G_k))})=\max\{\rho((T^{G^i/G^i_k}_{\mathcal G^i})_{(k,G^i_k\cdot g_iG^i_k)}):i<k\}.\end{align*} $$

Hence, by Proposition 2.2(1),

$$ \begin{align*}\begin{array}{ll}\rho^k(\mathcal G)&=\rho(T^{G/G_k}_{\mathcal{G}})=\rho(T^Y_{\mathcal{H}})\\[2pt]&\le\sup\{\rho((T^Y_{\mathcal{H}})_{(k,H_k\cdot\theta((g_i)G_k))}):(g_i)\in G\}+k\\[2pt]&=\sup\{\max\{\rho((T^{G^i/G^i_k}_{\mathcal G^i})_{(k,G^i_k\cdot g_iG^i_k)}):i<k\}:(g_i)\in G\}+k\\[2pt]&=\max\{\sup\{\rho((T^{G^i/G^i_k}_{\mathcal G^i})_{(k,G^i_k\cdot gG^i_k)}):g\in G^i\}:i<k\}+k\\[2pt]&\le\max\{\rho(T^{G^i/G^i_k}_{\mathcal G^i}):i<k\}+k\\[2pt]&=\max\{\rho^k(\mathcal G^i):i<k\}+k.\end{array}\\[-37pt]\end{align*} $$

Proof Fix a $\mathcal G^i=(G^i_n)\in \mathrm {dgnb}(G^i)$ for each $i<\omega $ . Put

$$ \begin{align*}G_n=\prod_{i<n}G^i_n\times\prod_{i\ge n}G^i\quad(\forall n<\omega).\end{align*} $$

(1) Assume G is $\alpha $ -CLI. Since each $G^i$ is topologically isomorphic to a closed subgroup of G, by Theorem 4.1, $G^i$ is also $\alpha $ -CLI.

On the other hand, assume all $G^i$ are $\alpha $ -CLI. Now by Lemma 3.16(1), $\rho ^k(\mathcal G^i)<\omega \cdot \alpha $ holds for all $i,k<\omega $ . Then Lemma 4.5 implies

$$ \begin{align*}\rho^k(\mathcal G)\le\max\{\rho^k(\mathcal G^i):i<k\}+k<\omega\cdot\alpha\end{align*} $$

holds for each $k<\omega $ . Consequently, G is $\alpha $ -CLI.

(2) Assume G is L- $\alpha $ -CLI. Since each $G^i$ is topologically isomorphic to a closed subgroup of G, we can see $G^i$ is L- $\alpha $ -CLI too. Moreover, by Theorem 3.21, there is an open subgroup H of G which is $\alpha $ -CLI. Hence, $G_n$ is a clopen subgroup of H for some $n<\omega $ , from which we can see this $G_n$ is also $\alpha $ -CLI. Now by (1), we conclude that for all $i\ge n$ , $G^i$ is $\alpha $ -CLI.

On the other hand, assume all $G^i$ are L- $\alpha $ -CLI, and there is an $m<\omega $ such that $G^i$ is $\alpha $ -CLI for each $i\ge m$ . Then (1) implies that $\prod _{i\ge m}G^i$ is $\alpha $ -CLI. Note that $G=G^0\times \dots \times G^{m-1}\times \prod _{i\ge m}G^i$ . By Corollary 4.4, we have

$$ \begin{align*}\mathrm{rank}(G)=\max\{\mathrm{rank}(G^0),\dots,\mathrm{rank}(G^{m-1}),\mathrm{rank}(\prod_{i\ge m}G^i)\}\le\alpha,\end{align*} $$

i.e., G is L- $\alpha $ -CLI.

Proof Note that all $G^i$ are $(\alpha +1)$ -CLI but not $\alpha $ -CLI.

Question 4.8. Let $0<\alpha <\omega _1$ , and let G be an $(\alpha +1)$ -CLI group. Can we find a sequence of L- $\alpha $ -CLI groups $G^i$ such that G is isomorphic to a closed subgroup of $\prod _iG^i$ ?

Proof For brevity, we denote $\Lambda \wr G$ by H. Let $\lambda _i,\,i<\omega $ be an enumeration of $\Lambda $ without repetition. Note that the underlying space of $\Lambda \wr G$ is $\Lambda \times G^\Lambda $ . For $(\lambda ,\chi )\in \Lambda \wr G$ , put $\pi _\Lambda (\lambda ,\chi )=\lambda $ and $\pi _G^i(\lambda ,\chi )=\chi (\lambda _i)$ .

Let $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ . Since G is not $\alpha $ -CLI, $G\ne \{1_G\}$ . Without loss of generality, we can assume that $G\ne G_1$ . Put $H_0=H$ and

$$ \begin{align*}H_{n+1}=\{(1_\Lambda,\chi):\chi\in G^\Lambda\wedge\forall i<n\,(\chi(\lambda_i)\in G_n)\}\end{align*} $$

for $n<\omega $ . Then $\mathcal H=(H_n)\in \mathrm {dgnb}(H)$ . Note that the open subgroup $H_1=\{1_\Lambda \}\times G^\Lambda $ is topologically isomorphic to $G^\omega $ . By Theorem 4.6, $H_1$ is $(\alpha +1)$ -CLI. So we can see that H is L- $(\alpha +1)$ -CLI from Theorem 3.21.

Since G is $(\alpha +1)$ -CLI but not $\alpha $ -CLI, $\omega \cdot \alpha <\rho (\mathcal G)\le \omega \cdot (\alpha +1)$ . By Lemma 3.16, there exist $1\le m,k<\omega $ such that $\rho ^k(\mathcal G)=\omega \cdot \alpha +m$ . To see that H is not $(\alpha +1)$ -CLI, we will show that $\rho ^{k+1}(\mathcal H)>\omega \cdot (\alpha +1)$ as follows.

For any $(\lambda _l,\hat \chi )\in H$ and $(1_\Lambda ,\tilde \chi )\in H_{k+1}$ , note that $(\lambda _l,\hat \chi )(1_\Lambda ,\tilde \chi )=(\lambda _l,\hat \chi \tilde \chi _l)$ , where $\tilde \chi _l(\lambda )=\tilde \chi (\lambda _l^{-1}\lambda )$ for $\lambda \in \Lambda $ . It follows that

$$ \begin{align*}(\lambda_l,\hat\chi)H_{k+1}=\{(\lambda_l,\chi):\chi\in G^\Lambda\wedge\forall i<k\,(\chi(\lambda_l\lambda_i)\in\hat\chi(\lambda_l\lambda_i)G_k)\}.\end{align*} $$

There is a unique $l_i<\omega $ such that $\lambda _{l_i}=\lambda _l\lambda _i$ for $i<k$ . It is clear that $\pi _\Lambda ((\lambda _l,\hat \chi )H_{k+1})=\{\lambda _l\}$ and for $j<\omega $ ,

$$ \begin{align*}\pi_G^j((\lambda_l,\hat\chi)H_{k+1})=\left\{\begin{array}{ll}\hat\chi(\lambda_{l_i})G_k, & j=l_i,i<k,\cr G, & \mbox{otherwise.}\end{array} \right.\end{align*} $$

Let $m_l=\max \{l_i:i<k\}$ . Then it is clear that $m_l\ge k-1$ for $l<\omega $ and $\sup \{m_l:l<\omega \}=\omega $ . Note that $\pi _G^{m_l}(H_{m_l+1}\cdot (\lambda _l,\hat \chi )H_{k+1})=G/G_k$ is not a singleton, and so $(m_l+1,H_{m_l+1}\cdot (\lambda _l,\hat \chi )H_{k+1})\in T^{H/H_{k+1}}_{\mathcal {H}}$ . Also note that

$$ \begin{align*}(T^{H/H_{k+1}}_{\mathcal{H}})_{(m_l+2,H_{m_l+2}\cdot(\lambda_l,\hat\chi)H_{k+1})}\cong T^{H_{m_l+2}\cdot(\lambda_l,\hat\chi)H_{k+1}}_{\mathcal H'}, \textrm{ and}\end{align*} $$

$$ \begin{align*}(T^{G/G_k}_{\mathcal{G}})_{(m_l+1,G_{m_l+1}\cdot gG_k)}\cong T^{G_{m_l+1}\cdot gG_k}_{\mathcal G'},\end{align*} $$

where $\mathcal H'=(H_{n+m_l+2})$ and $\mathcal G'=(G_{n+m_l+1})$ for $n<\omega $ . Define $\theta :H/H_{k+1}\to G/G_k$ as $\theta (C)=\pi _G^{m_l}(C)$ . Now we restrict $\theta $ as a function that maps $H_{m_l+2}\cdot (\lambda _l,\hat \chi )H_{k+1}$ onto $G_{m_l+1}\cdot \hat \chi (\lambda _{m_l})G_k$ . Then apply Lemma 3.8 and Proposition 2.2 to obtain the following:

$$ \begin{align*}\begin{array}{ll}&\rho((T^{H/H_{k+1}}_{\mathcal{H}})_{(m_l+1,H_{m_l+1}\cdot(\lambda_l,\bar\chi)H_{k+1})})\cr \ge &\sup\{\rho((T^{H/H_{k+1}}_{\mathcal{H}})_{(m_l+2,H_{m_l+2}\cdot(\lambda_l,\hat\chi)H_{k+1})}):\forall i<m_l\,(\bar\chi(\lambda_i)=\hat\chi(\lambda_i))\}+1\\ \ge &\sup\{\rho((T^{G/G_k}_{\mathcal{G}})_{(m_l+1,G_{m_l+1}\cdot gG_k)}):g\in G\}+1\\ \ge &\omega(\rho(T^{G/G_k}_{\mathcal{G}}))+1\\ = &\omega\cdot\alpha+1.\end{array}\end{align*} $$

This implies $\rho (T^{H/H_{k+1}}_{\mathcal {H}})\ge \omega \cdot \alpha +m_l+2$ for all $l<\omega $ , and hence

$$ \begin{align*}\rho^{k+1}(\mathcal H)=\rho(T^{H/H_{k+1}}_{\mathcal{H}})\ge\omega\cdot\alpha+\omega=\omega\cdot(\alpha+1).\end{align*} $$

Since $\rho ^{k+1}(\mathcal H)$ is a successor ordinal, we have $\rho ^{k+1}(\mathcal H)>\omega \cdot (\alpha +1)$ .

Proof For each $(g_i)\in G$ , the subspace $(g_i)H$ of $(G,\tau )$ is homeomorphic to H, so it is Polish. Let $D^i$ be a subset of $G^i$ which meets every coset of $H^i$ at exactly one point. Note that $D^i$ is countable for all $i<\omega $ . We define

$$ \begin{align*}D=\{(g_i)\in\prod_iD^i:\forall^\infty i\,(g_i=1_{G^i})\}.\end{align*} $$

It is clear that $G/H=\{(g_i)H:(g_i)\in D\}$ is countable. So $(G,\tau )$ is a sum of countably many Polish spaces, thus is a Polish space.

For $(g_i),(h_i)\in G$ and U open in H with $1_H\in H$ , there exists an $m<\omega $ such that $g_i,h_i\in H^i$ for $i>m$ and $U^0\times \dots \times U^m\times \prod _{i>m}H^i\subseteq U$ , where $U^i$ is an open subset of $H^i$ with $1_{H^i}\in U^i$ for each $i\le m$ . We can find open neighborhoods $V^i$ and $W^i$ of $1_{H^i}$ with $(g_iV^i)(h_iW^i)^{-1}\subseteq g_ih_i^{-1}U^i$ for $i\le m$ . Now let $V=V^0\times \dots \times V^m\times \prod _{i\ge m}H^i$ and $W=W^0\times \dots \times W^m\times \prod _{i\ge m}H^i$ , then V and W are open neighborhoods of $1_H$ , and $((g_i)V)((h_i)W)^{-1}\subseteq (g_ih_i^{-1})U$ . So $(G,\tau )$ is a Polish group.

Finally, suppose $\tau '$ is another group topology on G such that H is an open subgroup of G. Then for each $(g_i)\in G$ , the subspace $(g_i)H$ of $(G,\tau ')$ is homeomorphic to H, so the restrictions of $\tau $ and $\tau '$ on $(g_i)H$ are the same. Hence, $\tau =\tau '$ .

Proof Put $G_0=G$ and for $n\ge 1$ , let

$$ \begin{align*}G_n=\prod_{i<n-1}G^i_n\times\prod_{i\ge n-1}G^i_1.\end{align*} $$

It is clear that $G_1=\prod _iG^i_1$ and $\mathcal G=(G_n)\in \mathrm {dgnb}(G)$ .

Given $j<\omega $ , define $\theta :G/G_1\to G^j/G^j_1$ as $\theta ((g_i)G_1)=g_jG^j_1$ for $(g_i)\in G$ . Applying Lemma 3.8 to the restriction of $\theta $ as in the proof of Theorem 4.9, we have

$$ \begin{align*}\begin{array}{ll}\rho(T^{G/G_1}_{\mathcal{G}}) &=\sup\{\rho((T^{G/G_1}_{\mathcal{G}})_{(1,G_1\cdot(g_i)G_1)}):(g_i)\in G\}+1\\ &\ge\sup\{\rho((T^{G^j/G^j_1}_{\mathcal G^j})_{(1,G^j_1\cdot gG^j_1)}):g\in G^j\}+1\\ &=\rho(T^{G^j/G^j_1}_{\mathcal G^j})=\rho^1(\mathcal G^j).\end{array}\end{align*} $$

Therefore,

$$ \begin{align*}\rho^1(\mathcal G)=\rho(T^{G/G_1}_{\mathcal{G}})\ge\sup\{\rho^1(\mathcal G^j):j<\omega\}=\omega\cdot\alpha.\end{align*} $$

Since $\rho ^1(\mathcal G)$ is a non-limit ordinal and $\alpha>0$ , we have $\rho ^1(\mathcal G)>\omega \cdot \alpha $ . So G is not $\alpha $ -CLI.

Proof We construct $G_\alpha $ and $H_\alpha $ by induction on $\alpha $ . From Corollary 4.7 and Theorem 4.9, we only need to consider the case where $\alpha $ is a limit ordinal.

Let $(\alpha _i)$ be a sequence of ordinals less than $\alpha $ with $\sup \{\alpha _i:i<\omega \}=\alpha $ . By the inductive hypothesis, we can find a non-archimedean CLI Polish group $G^i$ for each $i<\omega $ such that $G^i$ is L- $\alpha _i$ -CLI but not $\alpha _i$ -CLI. It is clear that $\mathrm {rank}(G^i)=\alpha _i<\alpha $ .

Put $H_\alpha =\prod _iG^i$ . Theorem 4.6(1) implies that $H_\alpha $ is $\alpha $ -CLI. By Theorem 4.1, $\mathrm {rank}(H_\alpha )\ge \mathrm {rank}(G^i)$ for each $i<\omega $ . So $\mathrm {rank}(H_\alpha )=\alpha $ .

For $i<\omega $ , let $\mathcal G^i=(G^i_n)\in \mathrm {dgnb}(G^i)$ . By Lemma 3.16, there exist $0<k_i<\omega $ such that $\omega (\rho ^{k_i}(\mathcal G^i))=\omega \cdot \alpha _i$ . Put $H^i_0=G^i$ and $H^i_{n+1}=G^i_{n+k_i}$ for $n<\omega $ . Then $\mathcal H^i=(H^i_n)\in \mathrm {dgnb}(G^i)$ and $H^i_1=G^i_{k_i}$ . By Lemma 3.7, we have

$$ \begin{align*}\omega(\rho^1(\mathcal H^i))=\omega(\rho(T^{G^i/H^i_1}_{\mathcal H^i}))=\omega(\rho(T^{G^i/G^i_{k_i}}_{\mathcal G^i}))=\omega(\rho^{k_i}(\mathcal G^i))=\omega\cdot\alpha_i.\end{align*} $$

So $\sup \{\rho ^1(\mathcal H^i):i<\omega \}=\omega \cdot \alpha $ . Now we let

$$ \begin{align*}G_\alpha=\{(g_i)\in\prod_iG^i:\forall^\infty i\,(g_i\in G^i_{k_i})\}\end{align*} $$

be equipped with the unique group topology making $\prod _iG^i_{k_i}$ an open subgroup of $G_\alpha $ . By Lemma 4.11, $G_\alpha $ is not $\alpha $ -CLI. It is clear that the open subgroup $\prod _iG^i_{k_i}$ is $\alpha $ -CLI, so $G_\alpha $ is L- $\alpha $ -CLI, and hence $\mathrm {rank}(G_\alpha )=\alpha $ .