Proof. Let $p=p_1qp_2$ be decomposition of p into subpaths $p_1$ , q and $p_2$ . For any $H_\lambda $ -coset $xH_\lambda \in S(q_-,q_+;D)$ , there exists a geodesic path $\alpha $ in $\Gamma (G,X\sqcup \mathcal {H})$ from $q_-$ to $q_+$ that essentially penetrates $xH_\lambda $ . Since $p_1\alpha p_2$ is a geodesic path in $\Gamma (G,X\sqcup \mathcal {H})$ from $p_-$ to $p_+$ that essentially penetrates $xH_\lambda $ , we have $xH_\lambda \in S(p_-,p_+;D)$ .

Proof. Let $x=p_{in}(B)$ and $y=q_{in}(B)$ for brevity, and let e be the edge in $\Gamma (G,X\sqcup \mathcal {H})$ from x to y whose label is in $H_\lambda $ . Since $p,q$ are geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ and penetrates B, e is an isolated component in the geodesic triangle $p_{[o,x]}e(q_{[o,y]})^{-1}$ by Remark 2.32. This implies $\widehat {d}_\lambda (x,y)\le 3C$ by Proposition 2.30.

Proof. Let $p=p_1ap_2bp_3$ be decomposition of p into subpaths such that $a,b$ are components of p corresponding to $C_0,C_1$ , respectively. By $a_+\in C_0$ and $b_-\in C_1$ , we have $d_{X\cup \mathcal {H}}(C_0,C_1) \le d_{X\cup \mathcal {H}}(a_+,b_-)$ . Suppose for contradiction that there exist $x\in C_0, y\in C_1$ such that $d_{X\cup \mathcal {H}}(x,y) < d_{X\cup \mathcal {H}}(a_+,b_-)$ . By $x,a_-\in C_0$ and $y,b_+\in C_1$ , we have $d_{X\cup \mathcal {H}}(a_-,x)\le 1$ and $d_{X\cup \mathcal {H}}(y,b_+) \le 1$ . This implies

$$ \begin{align*} d_{X\cup\mathcal{H}}(a_-,b_+) &\le d_{X\cup\mathcal{H}}(a_-,x)+d_{X\cup\mathcal{H}}(x,y)+d_{X\cup\mathcal{H}}(y,b_+)\\ &< 1+d_{X\cup\mathcal{H}}(a_+,b_-)+1 =d_{X\cup\mathcal{H}}(a_-,b_+), \end{align*} $$

which is a contradiction. Hence, we also have $d_{X\cup \mathcal {H}}(a_+,b_-) \le d_{X\cup \mathcal {H}}(C_0,C_1)$ .

Proof. Let $x_0=p_{out}(C_0), x_1=p_{in}(C_1), y_0=q_{out}(C_0), y_1=q_{in}(C_1)$ for brevity, and let $e_0,e_1$ be edges in $\Gamma (G,X\sqcup \mathcal {H})$ such that $e_0$ is from $x_0$ to $y_0$ with its label in $H_{\lambda _0}$ and $e_1$ is from $x_1$ to $y_1$ with its label in $H_{\lambda _1}$ . Since the subpaths $p_{[x_0,x_1]},q_{[y_0,y_1]}$ don’t penetrate $C_0$ nor $C_1$ by Remark 2.32 and we have $C_0\neq C_1$ , $e_0$ and $e_1$ are isolated components of the geodesic quadrilateral $e_0 q_{[y_0,y_1]} e_1^{-1} (p_{[x_0,x_1]})^{-1}$ . This implies $\widehat {d}_{\lambda _0}(x_0,y_0)\le 4C$ and $\widehat {d}_{\lambda _1}(x_1,y_1)\le 4C$ by Proposition 2.30.

Proof. Let q penetrate $C_i$ and $j\in \mathbb {N}$ satisfy $j<i$ . By $C_j\in S(o,x;D)$ , their exists a geodesic path p in $\Gamma (G,X\sqcup \mathcal {H})$ from o to x that essentially penetrates $C_j$ . Note that p penetrates $C_i$ by Lemma 2.38. Let $p=p_1 a p_2$ , $q=q_1 b q_2$ be decomposition of $p,q$ into subpaths such that $a,b$ are $H_\lambda $ -component of $p,q$ corresponding to $C_i$ , respectively. Let e be the edge from $a_-$ to $b_+$ in $\Gamma (G,X\sqcup \mathcal {H})$ whose label is in $H_\lambda $ . Since we have $d_{X\cup \mathcal {H}}(o,a_-)=d_{X\cup \mathcal {H}}(o,b_-)=d_{X\cup \mathcal {H}}(o,C_i)$ by Lemma 2.36, the path $p_1eq_2$ from o to y is geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ and essentially penetrates $C_j$ . This implies $C_j \in S(o,y;D)$ .

Proof. If $x\in X$ and $x\neq 1$ , then the edge in $\Gamma (G,X\sqcup \mathcal {H})$ from $1$ to x with the label $x\in X$ is geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ . Since this path consisting of one edge has no $H_\lambda $ -component for any $\lambda \in \Lambda $ , we have $S(1,x;D)=\emptyset $ by Lemma 2.38. This implies $x\in Y$ , hence $X\subset Y$ . We will check the two conditions in Proposition 3.6 considering $\Gamma =\Gamma (G,Y\sqcup \mathcal {H})$ . Note that $\Gamma (G,X\sqcup \mathcal {H})$ is a subgraph of $\Gamma (G,Y\sqcup \mathcal {H})$ by $X\subset Y$ . For each pair $(x,y)$ of elements of G, fix a geodesic path $\gamma _{x,y}$ in $\Gamma (G,X\sqcup \mathcal {H})$ from x to y and define $\mathcal {L}(x,y)=\gamma _{x,y}$ . Since $\Gamma (G,X\sqcup \mathcal {H})$ is $\delta _X$ -hyperbolic with $\delta _X\in \mathbb {N}$ (see Definition 2.27 (1)), Proposition 3.6 (1) is satisfied with $h=\delta _X$ . Next, let $x,y\in G$ satisfy $d_{Y\cup \mathcal {H}}(x,y)\le 1$ , then $S(x,y;D)=\emptyset $ . Hence, for any vertex $z\in \gamma _{x,y}$ , we have $S(x,z;D)=\emptyset $ by Lemma 3.1. This implies $d_{Y\cup \mathcal {H}}(x,z)\le 1$ . Hence, the diameter of $p_{x,y}$ in $\Gamma (G,Y\sqcup \mathcal {H})$ is at most 2, which verifies Proposition 3.6 (2). Since both conditions in Proposition 3.6 are satisfied with $h=\max \{\delta _X,2\}$ , the graph $\Gamma (G,Y\sqcup \mathcal {H})$ is $k(h)$ -hyperbolic. Also, let $x,y \in G$ and let $\alpha $ be a geodesic path in $\Gamma (G,X\sqcup \mathcal {H})$ from x to y, and $\beta $ be a geodesic path in $\Gamma (G, Y\sqcup \mathcal {H})$ from x to y. By Proposition 3.6, the Hausdorff distance between $\beta $ and $\gamma _{x,y}$ in $\Gamma (G,Y\sqcup \mathcal {H})$ is at most $k(h)$ and by $\delta _X$ -hyperbolicity of $\Gamma (G,X\sqcup \mathcal {H})$ , the Hausdorff distance between $\alpha $ and $\gamma _{x,y}$ in $\Gamma (G,Y\sqcup \mathcal {H})$ is at most $\delta _X$ . Thus, the statement (b) holds with $M_X=k(h)+\delta _X$ .

Proof. For each $i \in \mathbb {N}$ , let $[o,g_i]$ be a geodesic path in $\Gamma (G,Y\sqcup \mathcal {H})$ from o to $g_i$ and $w_i \in [o,g_i]$ be a vertex satisfying $d_{Y\cup \mathcal {H}} (o,w_i)=R$ . By $(g_i,g_j)_o^{Y\cup \mathcal {H}}\ge R$ , we have $d_{Y\cup \mathcal {H}}(w_i,w_j) \le \delta _Y$ for any $i,j \in \mathbb {N}$ . By Lemma 3.7 (b), for each $i\in \mathbb {N}$ , there exists a vertex $v_i \in p_i$ such that $d_{Y\cup \mathcal {H}}(v_i,w_i) \le M_X$ . We have for any $i,j\in \mathbb {N}$ ,

$$ \begin{align*} &d_{Y\cup\mathcal{H}}(v_i,v_j) \le d_{Y\cup\mathcal{H}}(v_i,w_i) + d_{Y\cup\mathcal{H}}(w_i,w_j) + d_{Y\cup\mathcal{H}}(w_j,v_j) \le \delta_Y + 2M_X \\ \mathrm{and}~~~~~ &d_{Y\cup\mathcal{H}}(o, v_i) \ge d_{Y\cup\mathcal{H}}(o, w_i) - d_{Y\cup\mathcal{H}}(w_i,v_i) \ge R-M_X.\\[-34pt] \end{align*} $$

Proof. Take geodesic paths $[o,x],[o,y]$ in $\Gamma (G,Y\sqcup \mathcal {H})$ . By Lemma 3.7 (b), there exist vertices $a\in [o,x], b \in [o,y]$ such that $d_{Y\cup \mathcal {H}}(a,v) \le M_X$ and $d_{Y\cup \mathcal {H}}(b,w) \le M_X$ . We have

$$ \begin{align*} d_{Y\cup\mathcal{H}}(a,b) &\le d_{Y\cup\mathcal{H}}(a,v) + d_{Y\cup\mathcal{H}}(v,w) + d_{Y\cup\mathcal{H}}(w,b) \le K + 2M_X, \\ d_{Y\cup\mathcal{H}}(o,a) &\ge d_{Y\cup\mathcal{H}}(o,v)- d_{Y\cup\mathcal{H}}(a,v) \ge R-M_X, \\ \mathrm{and}~~~~~ d_{Y\cup\mathcal{H}}(o,b) &\ge d_{Y\cup\mathcal{H}}(o,v)- (d_{Y\cup\mathcal{H}}(v,w) + d_{Y\cup\mathcal{H}}(w,b)) \ge R-(K+M_X). \end{align*} $$

This implies $(a,b)_o^{Y\cup \mathcal {H}} \ge d_{Y\cup \mathcal {H}}(o,a)-d_{Y\cup \mathcal {H}}(a,b) \ge R-(K+3M_X)$ . Note $(x,a)_o^{Y\cup \mathcal {H}}=d_{Y\cup \mathcal {H}}(o,a)$ and $(y,b)_o^{Y\cup \mathcal {H}}=d_{Y\cup \mathcal {H}}(o,b)$ since $[o,a], [o,b]$ are geodesic in $\Gamma (Y\cup \mathcal {H})$ . Hence,

$$ \begin{align*} (x,y)_o^{Y\cup\mathcal{H}} &\ge \min\{(x,a)_o^{Y\cup\mathcal{H}}, (a,b)_o^{Y\cup\mathcal{H}}, (b,y)_o^{Y\cup\mathcal{H}} \}-2\delta_Y \\ &\ge R-(K+3M_X) - 2\delta_Y.\\[-34pt] \end{align*} $$

Proof. Let $i \in \{1,\cdots ,n\}$ satisfy $3d_{Y\cup \mathcal {H}}(x,y)+1 \le n-i$ , and take a geodesic path r in $\Gamma (G,X\sqcup \mathcal {H})$ from y to x. Since the path $qr$ is from o to x and composed of two geodesic segments, one of q or r penetrates $C_i$ by Lemma 2.35. Suppose for contradiction that r penetrates $C_i$ . This implies $C_j\in S(x,y;D)$ for any $j\in \{i+1,\cdots ,n\}$ by applying Lemma 3.5 to $r^{-1}$ and $S(x,o;D)=\{C_n \preceq \cdots \preceq C_1\}$ . This and equation (5) imply

$$\begin{align*}3d_{Y\cup\mathcal{H}}(x,y)+1 \le n-i \le |S(x,y;D)| \le 3d_{Y\cup\mathcal{H}}(x,y). \end{align*}$$

This is a contradiction. Hence, q penetrates $C_i$ .

Proof. Let $\gamma =(x_0,x_1,\cdots )$ and $B\in S(\gamma ;D)$ . There exists $n\in \mathbb {N}$ such that $B\in S(x_0,x_n;D)$ . By Lemma 2.38, $\gamma _{[x_0,x_n]}$ penetrates B, hence so does $\gamma $ . If $\gamma $ penetrates an $H_\lambda $ -coset more than once, then $\gamma $ can be shortened by an edge whose label is in $H_\lambda $ , which contradicts that $\gamma $ is geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ .

Proof. Let $\gamma =(x_0,x_1,\cdots )$ . Take $n\in \mathbb {N}$ satisfying $d_{X\cup \mathcal {H}}(\gamma _-,\gamma _{out}(B))<d_{X\cup \mathcal {H}}(\gamma _-,x_n)$ , then we have $d_{X\cup \mathcal {H}}(\gamma _-,B)=d_{X\cup \mathcal {H}}(\gamma _-,(\gamma _{[x_0,x_n]})_{in}(B))=d_{X\cup \mathcal {H}}(\gamma _-,\gamma _{in}(B))$ by Lemma 2.36.

Proof. By $d_{X\cup \mathcal {H}}(x_0,\gamma _{out}(C_{N+1})) \le d_{X\cup \mathcal {H}}(x_0,x_\ell )$ , the subpath $\gamma _{[x_0,x_\ell ]}$ penetrates $C_{N+1}$ . Take $k\in \mathbb {N}$ such that $\{C_1,\cdots ,C_{N+1}\}\subset S(x_0,x_k;D)$ . By applying Lemma 3.5 to $x_0,x_k,\gamma _{[x_0,x_\ell ]}$ , we have $\{C_1,\cdots ,C_N\}\subset S(x_0,x_\ell ;D)$ . Let $S(x_0,x_\ell ;D)=\{B_1\preceq \cdots \preceq B_N\preceq \cdots \}$ . We have $B_1\preceq C_1$ by $C_1\in S(x_0,x_\ell ;D)$ and by minimality of $B_1$ in $(S(x_0,x_\ell ;D),\preceq )$ . On the other hand, by minimality of $C_1$ in $(S(\gamma ;D),\preceq )$ , we also have $C_1\preceq B_1$ , hence $B_1=C_1$ . Repeating the same argument inductively, we can see $B_n=C_n$ for any $n\in \{1,\cdots ,N\}$ .

Proof. (i) $\Rightarrow $ (ii) and (v) $\Rightarrow $ (vi) are trivial. (iii) $\Rightarrow $ (iv) follows from Remark 3.13. (i) $\Rightarrow $ (v) and (ii) $\Rightarrow $ (vi) follow from the definition of the Gromov product (see equation (1)). (iii) $\Leftrightarrow $ (v) and (iv) $\Leftrightarrow $ (vi) follow from equation (5). We are left to show (iv) $\Rightarrow $ (iii) and (v) $\Rightarrow $ (i).

$\underline {\mathrm {(iv)}\Rightarrow \mathrm {(iii)}}$ By Remark 3.13, we have

$$\begin{align*}\lim_{k\to\infty} |S(x_0,x_{n_k};D)|=\sup_{k\in\mathbb{N}} |S(x_0,x_{n_k};D)| =\sup_{n\in\mathbb{N}} |S(x_0,x_n;D)|=\lim_{n\to\infty} |S(x_0,x_n;D)|. \end{align*}$$

Thus, $\lim _{k\to \infty } |S(x_0,x_{n_k};D)| = \infty $ implies $|S(\gamma ;D)|=\lim _{n\to \infty } |S(x_0,x_n;D)| = \infty $ .

$\underline {\mathrm {(v)}\Rightarrow \mathrm {(i)}}$ For any $R\in \mathbb {N}$ , define $R_1=R+3M_X+2\delta _Y$ . By (v), there exists $N\in \mathbb {N}$ such that $d_{Y\cup \mathcal {H}}(x_0,x_N) \ge R_1$ . Hence, for any $n,m\ge N$ , we have $(x_n,x_m)_{x_0}^{Y\cup \mathcal {H}}\ge R_1-3M_X-2\delta _Y=R$ by Lemma 3.10 applied to $o=x_0, x=x_n, y=x_m, v=x_N, w=x_N$ . This implies $\liminf _{n,m\to \infty }(x_n,x_m)_{x_0}^{Y\cup \mathcal {H}} \ge \inf _{n,m\ge N} (x_n,x_m)_{x_0}^{Y\cup \mathcal {H}} \ge R$ for any $R\in \mathbb {N}$ . Thus, $\lim _{n,m\to \infty }(x_n,x_m)_{x_0}^{Y\cup \mathcal {H}}=\infty $ .

Proof. Let $(g_n)_{n=1}^\infty $ be a sequence of elements of G that converges to $\xi $ . For each $n\in \mathbb {N}$ , take a geodesic path $\gamma _n$ in $\Gamma (G,X\sqcup \mathcal {H})$ from o to $g_n$ . Since $\lim _{i,j\to \infty } (g_i,g_j)_o^{Y\cup \mathcal {H}} = \infty $ , there exists a subsequence $(g_{0k})_{k=1}^\infty $ of $(g_n)_{n=1}^\infty $ such that

(6) $$ \begin{align} \inf_{k,\ell \in\mathbb{N}}(g_{0k},g_{0\ell})_o^{Y\cup\mathcal{H}} \ge 2(3(\delta_Y + 2M_X)+4) + 1 + M_X. \end{align} $$

By Lemma 3.9 and equation (6), there exist vertices $v_{0k} \in \gamma _{0k}$ for each $k \in \mathbb {N}$ satisfying $d_{Y\cup \mathcal {H}}(v_{0k},v_{0\ell }) \le \delta _Y + 2M_X$ and $d_{Y\cup \mathcal {H}}(o, v_{0\ell }) \ge 2(3(\delta _Y + 2M_X)+4) + 1$ for any $k,\ell \in \mathbb {N}$ . This and equation (5) imply for any $k,\ell \in \mathbb {N}$ ,

(7) $$ \begin{align} 3d_{Y\cup\mathcal{H}}(v_{0k},v_{0\ell}) + 4 \le 3(\delta_Y + 2M_X) + 4 \le |S(o,v_{0\ell};D)|. \end{align} $$

Let $m=|S(o,v_{01};D)|$ and $S(o,v_{01};D)=\{C_1\preceq C_2\preceq \cdots \preceq C_m\}$ . Since equation (7) implies $3d_{Y\cup \mathcal {H}}(v_{0k},v_{01})+1 \le m-3$ , the path $\gamma _{0k}$ penetrates $C_1$ , $C_2$ , and $C_3$ for any $k\in \mathbb {N}$ by Lemma 3.11. Let $C_3$ be an $H_\lambda $ -coset. For each $k \in \mathbb {N}$ , let $a_{0k-}$ and $a_{0k+}$ be the entrance and exit points of $\gamma _{0k}$ in $C_3$ . Since $\gamma _{0k}$ is geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ for each $k\in \mathbb {N}$ , we have

$$\begin{align*}\widehat{d}_{\lambda}(a_{01-},a_{0k-}) \le 3C \end{align*}$$

by Lemma 3.2. Since the relative metric $\widehat {d}_{\lambda }$ is locally finite (see Definition 2.27), the set $\{a_{0k-} \mid k\in \mathbb {N}\}$ is finite. Hence, there exist $a_1 \in \{a_{0k-} \mid k\in \mathbb {N}\}$ and a subsequence $(g_{1'k})_{k=1}^\infty $ of $(g_{0k})_{k=1}^\infty $ such that $a_1 \in \gamma _{1'k}$ for any $k\in \mathbb {N}$ . Note $S(o,a_1;D) \ge 1$ since we have $C_1 \in S(o,a_1;D)$ by Lemma 3.5. By $\lim _{k,\ell \to \infty } (g_{1'k},g_{1'\ell })_o^{Y\cup \mathcal {H}} = \infty $ , there exists a subsequence $(g_{1k})_{k=1}^\infty $ of $(g_{1'k})_{k=1}^\infty $ such that

(8) $$ \begin{align} \inf_{k,\ell \in\mathbb{N}}(g_{1k},g_{1\ell})_o^{Y\cup\mathcal{H}} \ge 2(3(\delta_Y + 2M_X) + |S(o,a_1;D)|+4) + 1 + M_X. \end{align} $$

Note $a_1 \in \gamma _{1k}$ for any $k\in \mathbb {N}$ . By the same argument as $(g_{0k})_{k=1}^\infty $ , we can see that there exist $a_2 \in G$ and a subsequence $(g_{2'k})_{k=1}^\infty $ of $(g_{1k})_{k=1}^\infty $ such that $a_2 \in \gamma _{2'k}$ for any $k\in \mathbb {N}$ and $|S(o,a_2;D)| \ge |S(o,a_1;D)| +1$ . The latter inequality comes from the term $|S(o,a_1;D)|+4$ in equation (8), which corresponds to $4$ in equation (6). By repeating this argument, we can see that there exist $a_1,a_2,\cdots \in G$ and a sequence of subsequences $(g_{1k})_{k=1}^\infty \supset (g_{2k})_{k=1}^\infty \supset \cdots $ satisfying (i) and (ii) below for any $n\in \mathbb {N}$ .

1. (i) $\{a_1,\cdots ,a_n\}\subset \gamma _{nk}$ for any $k\in \mathbb {N}$ .

2. (ii) $|S(o,a_{n+1};D)| \ge |S(o,a_n;D)|+1$ .

Take the diagonal sequence $(g_{kk})_{k=1}^\infty $ , then for any $n\in \mathbb {N}$ , $(g_{kk})_{k=n}^\infty $ is a subsequence of $(g_{nk})_{k=1}^\infty $ . Hence, $\{a_1,\cdots ,a_n\}\subset \gamma _{nn}$ for any $n\in \mathbb {N}$ . Note that (ii) and Lemma 3.1 imply $d_{X\cup \mathcal {H}}(o,a_n) < d_{X\cup \mathcal {H}}(o,a_{n+1})$ for any $n\in \mathbb {N}$ . Define the path $\gamma \colon [0,\infty ) \to \Gamma (G,X\sqcup \mathcal {H})$ by

$$\begin{align*}\gamma=\bigcup_{n=1}^\infty \gamma_{nn[a_{n-1},a_n]}, \end{align*}$$

where we define $a_0$ by $a_0=o$ for convenience. the path $\gamma $ is a geodesic ray since for any $n \in \mathbb {N}$ , we have

$$\begin{align*}\sum_{i=1}^n |\gamma_{ii[a_{i-1},a_i]}| = \sum_{i=1}^n d_{X\cup\mathcal{H}}(a_{i-1},a_i) = \sum_{i=1}^n |\gamma_{nn[a_{i-1},a_i]}| = |\gamma_{nn[o,a_n]}| =d_{X\cup\mathcal{H}}(o,a_n). \end{align*}$$

Since we have $\lim _{n\to \infty } |S(o,a_n;D)| = \infty $ by (ii) and $(a_n)_{n=1}^\infty $ is a subsequence of $\gamma $ , the path $\gamma $ converges to infinity in $\Gamma (G,Y\sqcup \mathcal {H})$ by Lemma 3.20 and we have $Y\text{-}{\lim}\ \gamma = Y\text{-}{\lim} _{n\to \infty } a_n$ .

Finally, we will show $Y\text{-}{\lim}\ \gamma = \xi \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ . Since we have $\xi = Y\text{-}{\lim} _{n\to \infty } g_n =Y\text{-}{\lim} _{n\to \infty } g_{nn}$ , it’s enough to show $Y\text{-}{\lim} _{n\to \infty } a_n=Y\text{-}{\lim} _{n\to \infty } g_{nn} \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ . For any $R\in \mathbb {N}$ , there exists $N\in \mathbb {N}$ such that $d_{Y\cup \mathcal {H}}(o,a_N)\ge R+3M_X+2\delta _Y$ since we have $\lim _{n\to \infty } d_{Y\cup \mathcal {H}}(o,a_n)=\infty $ by Lemma 3.20. Hence, for any $m,n\ge N$ , we have $(a_m,g_{nn})_o^{Y\cup \mathcal {H}} \ge R$ by Lemma 3.10 applied to $x=a_m, y=g_{nn}, v=a_N, w=a_N$ . This implies $\lim _{m,n\to \infty } (a_m,g_{nn})_o^{Y\cup \mathcal {H}} = \infty $ , hence $Y\text{-}{\lim} _{n\to \infty } a_n=Y\text{-}{\lim} _{n\to \infty } g_{nn}$ in $\partial \Gamma (G,Y\sqcup \mathcal {H})$ .

Proof. Given $N\in \mathbb {N}$ , define $R\in \mathbb {N}$ by

(9) $$ \begin{align} R=2(3(\delta_Y+2M_X)+N+2)+1+M_X+2\delta_Y. \end{align} $$

Let $\eta \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ satisfy $(\xi ,\eta )_o^{Y\cup \mathcal {H}}> R$ (see equation (2)) and $\beta $ be a geodesic ray in $\Gamma (G,X\sqcup \mathcal {H})$ from o such that $Y\text{-}{\lim}\ \beta =\eta $ . Let $\alpha =(x_0,x_1,\cdots )$ and $\beta =(y_0,y_1,\cdots )$ . It’s not difficult to see from equation (2) that $(\xi ,\eta )_o^{Y\cup \mathcal {H}}> R$ implies

$$\begin{align*}\liminf_{i,j\to\infty} (x_i,y_j)_o^{Y\cup\mathcal{H}}> R-2\delta_Y. \end{align*}$$

Hence, there exists $k\in \mathbb {N}$ such that $(x_k,y_k)_o^{Y\cup \mathcal {H}}> R-2\delta _Y$ . By Lemma 3.9 applied to $o,x_k,y_k,\alpha _{[o,x_k]},\beta _{[o,y_k]}$ , there exist $\ell ,m$ with $\ell ,m \le k$ such that $d_{Y\cup \mathcal {H}} (x_\ell ,y_m) \le \delta _Y+2M_X$ and $d_{Y\cup \mathcal {H}} (o,x_\ell ) \ge R-2\delta _Y-M_X$ . This implies, together with equation (5) in Theorem 2.41 and equation (9),

(10) $$ \begin{align} 3d_{Y\cup\mathcal{H}} (x_\ell,y_m)+N+2 \le 3(\delta_Y+2M_X)+N+2 \le |S(o,x_\ell;D)|. \end{align} $$

Let $L=|S(o,x_\ell ;D)|$ and $S(o,x_\ell ;D)=\{B_1\preceq \cdots \preceq B_L\}$ . By Lemma 3.11, $\beta _{[y_0,y_m]}$ penetrates $B_n$ for any $n\in \{1,\cdots ,N\}$ since equation (10) implies $3d_{Y\cup \mathcal {H}} (x_\ell ,y_m)+1\le L-N$ . Meanwhile, suppose $d_{X\cup \mathcal {H}}(o,x_\ell ) \le d_{X\cup \mathcal {H}}(o,\alpha _{out}(C_{N+1}))$ for contradiction, then we have $\alpha _{[o,x_\ell ]} \subset \alpha _{[o,\alpha _{out}(C_{N+1})]}$ . This implies $|S(o,x_\ell ;D)| \le |S(o,\alpha _{out}(C_{N+1});D)| \le N+1$ by Lemma 3.1. This contradicts that we get $N+2\le |S(o,x_\ell ;D)|$ by equation (10). Thus, we have $d_{X\cup \mathcal {H}}(o,\alpha _{out}(C_{N+1})) < d_{X\cup \mathcal {H}}(o,x_\ell )$ . Hence, we have $B_n=C_n$ for any $n\in \{1,\cdots ,N\}$ by Lemma 3.18. Thus, $\beta $ penetrates $C_n$ for any $n\in \{1,\cdots ,N\}$ . Finally, since the set $V=\{\eta \in \partial \Gamma (G,Y\sqcup \mathcal {H}) \mid (\xi ,\eta )_o^{Y\cup \mathcal {H}}> R \}$ is a neighborhood of $\xi $ , there exists an open set U of $\partial \Gamma (G,Y\sqcup \mathcal {H})$ such that $\xi \in U$ and $U\subset V$ . This U satisfies the statement for N.

Proof. Let $\alpha =(x_0,x_1,\cdots )$ , $\beta =(y_0,y_1,\cdots )$ , and $S(\alpha ;D)=\{C_1 \preceq C_2\preceq \cdots \}$ . By Proposition 3.23, $\beta $ penetrates $C_n$ for any $n\in \mathbb {N}$ since $Y\text{-}{\lim}\ \beta $ is obviously contained in any open neighborhood of $Y\text{-}{\lim}\ \alpha $ . For any $n\in \mathbb {N}$ , there exists $k\in \mathbb {N}$ such that $\{C_1, \cdots ,C_{n+1}\} \subset S(o,x_k;D)$ . Let $\ell \in \mathbb {N}$ satisfy $d_{X\cup \mathcal {H}}(o,\beta _{out}(C_{n+1}))\le d_{X\cup \mathcal {H}}(o,y_\ell )$ . By Lemma 3.5, we have $\{C_1, \cdots ,C_n\} \subset S(o,y_\ell ;D)\subset S(\beta ;D)$ . This implies $S(\alpha ;D) \subset S(\beta ;D)$ . Similarly, we can also see $S(\beta ;D) \subset S(\alpha ;D)$ .

Proof. Define the map $f \colon \mathbb {N}\to \mathbb {Z}$ by $f(n)=d_\Gamma (y_n,y_0)-d_\Gamma (y_n,x)$ . Since $\beta $ is geodesic, we have $f(n)=n-d_\Gamma (y_n,x), \forall n\in \mathbb {N}$ . This and $|d_\Gamma (y_{n+1},x)-d_\Gamma (y_n,x)|\le 1, \forall n\in \mathbb {N}$ imply that f is non-decreasing. The map f is also bounded above by $d_\Gamma (y_0,x)$ . Hence, there exists $k\in \mathbb {N}$ such that $f(n)=f(k)$ for any n with $n\ge k$ . Since this implies $d_\Gamma (y_n,x)=d_\Gamma (y_n,y_k)+d_\Gamma (y_k,x)$ for any n with $n\ge k$ , this k satisfies the statement.

Proof. Let $\beta =(y_0,y_1,\cdots )$ . By applying Lemma 3.27 to $\Gamma (G,X\sqcup \mathcal {H})$ , there exist $k\in \mathbb {N}$ and a geodesic path p in $\Gamma (G,X\sqcup \mathcal {H})$ from $\alpha _-$ to $y_k$ such that the path $\gamma $ defined by $\gamma =p\beta _{[y_k,\infty )}$ is a geodesic ray in $\Gamma (G,X\sqcup \mathcal {H})$ . By $\gamma _{[y_k,\infty )}=\beta _{[y_k,\infty )}$ , we have $Y\text{-}{\lim}\ \gamma =Y\text{-}{\lim}\ \beta =Y\text{-}{\lim}\ \alpha $ . This implies $S(\gamma ;D)=S(\alpha ;D)$ by Corollary 3.24. Hence, $\gamma $ penetrates $C_n$ for any $n\in \mathbb {N}$ by Lemma 3.14. Note that $|S(\gamma ;D)|=\infty $ implies $\lim _{n\to \infty } d_{X\cup \mathcal {H}}(\gamma _-, \gamma _{in}(C_n))=\infty $ . Hence, there exists $N\ge 2$ such that $d_{X\cup \mathcal {H}}(\gamma _-, \gamma _{in}(C_{N-1}))>d_{X\cup \mathcal {H}}(\gamma _-, y_k)$ . By $\gamma _{[y_k,\infty )}=\beta _{[y_k,\infty )}$ , the path $\beta $ penetrates $C_n$ for any $n\ge N-1$ . Let $n\ge N$ and $C_n$ be an $H_\lambda $ -coset. By $C_{n-1}\preceq C_n$ and $\gamma _-=\alpha _-$ , we have $d_{X\cup \mathcal {H}}(\gamma _-, C_{n-1})<d_{X\cup \mathcal {H}}(\gamma _-, C_n)$ . This implies

$$ \begin{align*} d_{X\cup\mathcal{H}}(\beta_-, C_n)-d_{X\cup\mathcal{H}}(\beta_-, C_{n-1}) &=d_{X\cup\mathcal{H}}(y_k, \beta_{in}(C_n))-d_{X\cup\mathcal{H}}(y_k, \beta_{in}(C_{n-1})) \\ &=d_{X\cup\mathcal{H}}(y_k, \gamma_{in}(C_n))-d_{X\cup\mathcal{H}}(y_k, \gamma_{in}(C_{n-1})) \\ &=d_{X\cup\mathcal{H}}(\gamma_-, C_n)-d_{X\cup\mathcal{H}}(\gamma_-, C_{n-1})>0. \end{align*} $$

Hence, we have $\widehat {d}_{\lambda }(\alpha _{in}(C_n), \beta _{in}(C_n)) \le 4C$ by applying Lemma 3.4 to $C_{n-1},C_n,\alpha ,\beta $ . Similarly, we have $\widehat {d}_{\lambda }(\alpha _{out}(C_n), \beta _{out}(C_n)) \le 4C$ by applying Lemma 3.4 to $C_n,C_{n+1},\alpha ,\beta $ .

Proof. Let $\alpha =(x_0,x_1,\cdots )$ . For any open neighborhood U of $Y\text{-}{\lim}\ \alpha $ in $\partial \Gamma (G,Y\sqcup \mathcal {H})$ , there exists $R\in \mathbb {N}$ such that

(12) $$ \begin{align} \{ \eta \in \partial\Gamma(G,Y\sqcup\mathcal{H}) \mid (Y\text{-}{\lim}\ \alpha, \eta)_o^{Y\cup\mathcal{H}}\ge R \}\subset U \end{align} $$

by Proposition 2.19. We define $R_1=R+\delta _X+3M_X+2\delta _Y$ . Since we have $\lim _{n\to \infty }d_{Y\cup \mathcal {H}}(o,x_n)=\infty $ by Lemma 3.20, there exists $N\in \mathbb {N}$ such that

$$\begin{align*}d_{Y\cup\mathcal{H}}(o,x_{N}) \ge R_1. \end{align*}$$

By Proposition 2.19, there exists an open neighborhood V of $X\text{-}{\lim}\ \alpha $ in $\partial \Gamma (G,X\sqcup \mathcal {H})$ such that

(13) $$ \begin{align} V \subset \{ \eta' \in \partial\Gamma(G,X\sqcup\mathcal{H}) \mid (X\text{-}{\lim}\ \alpha, \eta')_o^{X\cup\mathcal{H}}>N+2\delta_X \}. \end{align} $$

We show that V satisfies the statement. Let $\beta =(y_0,y_1,\cdots )$ be a geodesic ray in $\Gamma (G,X\sqcup \mathcal {H})$ from o converging to infinity in $\Gamma (G,Y\sqcup \mathcal {H})$ such that $X\text{-}{\lim}\ \beta \in V$ . By equation (13), it’s not difficult to see $\liminf _{i,j\to \infty }(x_i,y_j)_o^{X\cup \mathcal {H}}>N$ . Since $\alpha $ and $\beta $ are geodesic rays in $\Gamma (G,X\sqcup \mathcal {H})$ from o, this implies

$$\begin{align*}d_{Y\cup\mathcal{H}}(x_N,y_N) \le d_{X\cup\mathcal{H}}(x_N,y_N) \le \delta_X. \end{align*}$$

Hence, for any $n,m\ge N$ , we have $(x_{n},y_{m})_o^{Y\cup \mathcal {H}} \ge R_1-(\delta _X+3M_X)-2\delta _Y=R$ by applying Lemma 3.10 to $\alpha _{[o,x_n]}, \beta _{[o,y_m]},x=x_n,y=y_m,v=x_N,w=y_N$ . This implies

$$\begin{align*}(Y\text{-}{\lim}\ \alpha,Y\text{-}{\lim}\ \beta)_o^{Y\cup\mathcal{H}} \ge \liminf_{i,j\to\infty} (x_i,y_j)_o^{Y\cup\mathcal{H}} \ge R. \end{align*}$$

Hence, we have $Y\text{-}{\lim}\ \beta \in U$ by equation (12).

Proof. Fix a vertex $o\in \Gamma $ , and take the map $D \colon (\Gamma \cup \partial \Gamma )^2 \to [0,\infty )$ and constants $\varepsilon ,\varepsilon '>0$ as in Proposition 2.21. We define the map $\widetilde {D} \colon (V(\Gamma )\cup \partial \Gamma )^2 \to [0,\infty )$ by $\widetilde {D}(x,y)=D(x,y)$ if $x\neq y$ and $\widetilde {D}(x,y)=0$ if $x=y$ . By Proposition 2.21, it’s not difficult to see that $\widetilde {D}$ is a metric and the metric topology of $\widetilde {D}$ coincides with the relative topology of $\mathcal {O}_\Gamma $ on $V(\Gamma )\cup \partial \Gamma $ (see Proposition 2.19). Here, we used discreteness of $(V(\Gamma ), \mathcal {O}_\Gamma |_{V(\Gamma )})$ since the metric topology of $\widetilde {D}$ on $V(\Gamma )$ is discrete by Remark 2.22. We will show that $\widetilde {D}$ is complete. Let $(x_n)_{n=1}^\infty $ be a Cauchy sequence of $(V(\Gamma )\cup \partial \Gamma ,\widetilde {D})$ . Since $V(\Gamma )$ is dense in $V(\Gamma )\cup \partial \Gamma $ , we can take for each $n\in \mathbb {N}$ , a vertex $y_n\in V(\Gamma )$ such that $\widetilde {D}(x_n,y_n)\le \frac {1}{n}$ . The sequence $(y_n)_{n=1}^\infty $ in $V(\Gamma )$ is also a Cauchy sequence in the metric $\widetilde {D}$ and it’s enough to show that $(y_n)_{n=1}^\infty $ is convergent. If there exists a constant subsequence $(y_{n_k})_{k=1}^\infty $ (i.e., $y_{n_k}\equiv y$ for some $y\in V(\Gamma )$ ), then $(y_n)_{n=1}^\infty $ converges to y in $\widetilde {D}$ . If there is no constant subsequence of $(y_n)_{n=1}^\infty $ , then there exists a subsequence $(y_{n_k})_{k=1}^\infty $ whose elements are all distinct. This implies $\widetilde {D}(y_{n_k},y_{n_\ell })=D(y_{n_k},y_{n_\ell })$ for any $k\neq \ell $ . Since this implies $\lim _{k,\ell \to \infty } (y_{n_k},y_{n_\ell })_o^\Gamma =\infty $ , the sequence $(y_{n_k})_{k=1}^\infty $ converges to some $y \in \partial \Gamma $ in $\mathcal {O}_\Gamma $ . Hence, $(y_n)_{n=1}^\infty $ converges to y as well. Thus, $V(\Gamma )\cup \partial \Gamma $ is completely metrizable with $\widetilde {D}$ . If $V(\Gamma )$ is countable, $V(\Gamma )$ is a countable dense subset of $V(\Gamma )\cup \partial \Gamma $ . Hence, $V(\Gamma )\cup \partial \Gamma $ is Polish.

Proof. Note that $(X\sqcup \mathcal {H})^{\mathbb {N}}$ is metrizable. Suppose that a sequence $(w_i)_{i=1}^\infty $ in $\mathcal {G}(\xi )$ converges to $w\in (X\sqcup \mathcal {H})^{\mathbb {N}}$ . It’s straightforward to see that $\gamma ^w$ is geodesic in $\Gamma (G,X\sqcup \mathcal {H})$ . We will first show $|S(\gamma ^w;D)|=\infty $ . Let $S(1,\xi ;D)=\{C_1\preceq C_2 \preceq \cdots \}$ (see Definition 3.25). Since $w_i\in \mathcal {G}(\xi )$ implies $S(\gamma ^{w_i};D)=S(1,\xi ;D)$ for any $i\in \mathbb {N}$ , we have for any $i,n\in \mathbb {N}$ ,

(14) $$ \begin{align} d_{X\cup\mathcal{H}}(1,\gamma^{w_i}_{out}(C_n)) =d_{X\cup\mathcal{H}}(1,\gamma^{w_i}_{in}(C_n))+1 =d_{X\cup\mathcal{H}}(1,C_n)+1 \end{align} $$

by Lemma 3.15. For any $N\in \mathbb {N}$ , we define k by $k=d_{X\cup \mathcal {H}}(1,C_{N+1})+1$ . Since $(w_i)_{i=1}^\infty $ converges to w, there exists $I\in \mathbb {N}$ such that $x_{w_I,m}=x_{w,m}$ for any $m\in \{1,\cdots ,k\}$ (see Definition 4.2). This implies $ \{C_1,\cdots ,C_N\} \subset S(1,x_{w_I,k};D)=S(1,x_{w,k};D) \subset S(\gamma ^w;D) $ by equation (14) and Lemma 3.18. Hence, we have $|S(\gamma ^w;D)|=\infty $ since $N\in \mathbb {N}$ is arbitrary. Since $(w_i)_{i=1}\infty $ converges to w, $(X\text{-}{\lim}\ \gamma ^{w_i})_{i=1}\infty $ converges to $X\text{-}{\lim}\ \gamma ^w$ in $\partial \Gamma (G,X\sqcup \mathcal {H})$ . Hence, $(Y\text{-}{\lim}\ \gamma ^{w_i})_{i=1}\infty $ converges to $Y\text{-}{\lim}\ \gamma ^w$ in $\partial \Gamma (G,Y\sqcup \mathcal {H})$ by Proposition 3.29. This implies $Y\text{-}{\lim}\ \gamma ^w=\xi $ by $Y\text{-}{\lim}\ \gamma ^{w_i}=\xi , \forall i\in \mathbb {N}$ . Thus, we have $w\in \mathcal {G}(\xi )$ .

Proof. For each $n\in \mathbb {N}$ , we define the element $s_{\xi ,n}\in X\sqcup \mathcal {H}$ and the subset $\mathcal {G}(\xi )_n$ of $\mathcal {G}(\xi )$ inductively as follows:

$$ \begin{align*} s_{\xi,1}&=\min \{ s_1 \in (X\sqcup\mathcal{H},\le) \mid \exists w\in \mathcal{G}(\xi) ~\mathrm{s.t.}~ w=(s_1,s_2,\cdots) \}, \\ \mathcal{G}(\xi)_1&=\{ w\in \mathcal{G}(\xi) \mid w=(s_{\xi,1},s_2,\cdots) \}, \\ s_{\xi,n+1}&=\min \{ s_{n+1} \in (X\sqcup\mathcal{H},\le) \mid \exists w\in \mathcal{G}(\xi)_n ~\mathrm{s.t.}~ w=(s_{\xi,1},\cdots,s_{\xi,n},s_{n+1},\cdots) \}, \\ \mathcal{G}(\xi)_{n+1}&=\{ w\in \mathcal{G}(\xi)_n \mid w=(s_{\xi,1},\cdots,s_{\xi,n},s_{\xi,n+1},\cdots) \}. \end{align*} $$

Note that each $\mathcal {G}(\xi )_n$ is nonempty since $\mathcal {G}(\xi )$ is nonempty. We define the element $w_\xi \in (X\sqcup \mathcal {H})^{\mathbb {N}}$ by $w_\xi =(s_{\xi ,1},s_{\xi ,2},s_{\xi ,3},\cdots )$ and take an element $w_n \in \mathcal {G}(\xi )_n$ for each $n\in \mathbb {N}$ . Since $(w_n)_{n=1}\infty $ converges to $w_\xi $ in $(X\sqcup \mathcal {H})^{\mathbb {N}}$ , we have $w_\xi \in \mathcal {G}(\xi )$ by Lemma 4.5. Since $w_\xi \in \bigcap _{n=1}^\infty \mathcal {G}(\xi )_n$ , we have $w_\xi \le _{\mathrm {lex}} w$ for any $w\in \mathcal {G}(\xi )$ .

Proof. This follows since we have $\xi =Y\text{-}{\lim}\ \gamma ^{\Phi (\xi )}$ for any $\xi \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ .

Proof. It’s enough to show that for any $\xi \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ and any $k\in \mathbb {N}$ , there exists an open neighborhood U of $\xi $ in $\partial \Gamma (G,Y\sqcup \mathcal {H})$ such that $s_{\eta ,n}=s_{\xi ,n}$ for any $\eta \in U$ and any $n\in \{1,\cdots ,k\}$ . Indeed, given $\xi \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ , the open sets $\{V_k\}_{k=1}^\infty $ defined by $V_k=\{(s_n)_{n=1}^\infty \in (X\sqcup \mathcal {H})^{\mathbb {N}} \mid s_n=s_{\xi ,n}, \forall n\in \{1,\cdots ,k\} \}$ form a neighborhood basis of $\Phi (\xi )$ . Hence, for any open neighborhood V of $\Phi (\xi )$ , there exists $k\in \mathbb {N}$ such that $V_k \subset V$ . For this k, we will be able to take an open neighborhood U of $\xi $ such that $\Phi (U)\subset V_k$ . This will imply continuity of $\Phi $ at $\xi $ . Hence, we will get continuity of $\Phi $ since $\xi \in \partial \Gamma (G,Y\sqcup \mathcal {H})$ is arbitrary.

Let $S(1,\xi ;D)=\{C_1\preceq C_2 \preceq \cdots \}$ (see Definition 3.25). By $\lim _{i\to \infty }d_{X\cup \mathcal {H}}(1,C_i)=\infty $ , there exists $N\in \mathbb {N}$ such that $d_{X\cup \mathcal {H}}(1,C_N)> k$ . By Proposition 3.23, there exists an open neighborhood U of $\xi $ in $\partial \Gamma (G,Y\sqcup \mathcal {H})$ such that for any $\eta \in U$ and any $w \in \mathcal {G}(\eta )$ , the path $\gamma ^w$ penetrates $C_i$ for any $i\in \{1,\cdots ,N\}$ . Let $\eta \in U$ . We define m by $m=d_{X\cup \mathcal {H}}(1,C_N)$ for brevity and let $C_N$ be an $H_\lambda $ -coset. We claim