Proof. We set $K' = K^{2}\big ( M(M+7)(K+1) + 1\big )$ .

Let $\eta : [0, L] \rightarrow X$ be a geodesic connecting $x$ to $y$ . Note that for

\begin{align*} \tau := \inf \Big \{ 0 \leqslant t\leqslant L : \operatorname {diam}\big (\pi _{A}(x) \cup \pi _{A}(\eta (t)) \big ) \gt K'+ K\Big \} \end{align*}

The $K$ -coarse Lipschitzness of $\pi _{A}$ and Inequality 2 imply

\begin{align*} \begin{aligned} \lim _{\epsilon \rightarrow 0+} \operatorname {diam}\big ( \pi _{A}\big (\eta (\tau )\big ) \cup \pi _{A}\big (\eta (\tau +\epsilon )\big ) \Big ) &\leqslant K, \\ \operatorname {diam}\big (\pi _{A}(x) \cup \pi _{A}(\eta (\tau )) \big ) \geqslant (K'+ K) - K &= K'. \end{aligned} \end{align*}

Hence, by replacing $y$ with $\eta (\tau )$ , we may assume $\operatorname {diam}\big (\pi _{A}(x) \cup \pi _{A}(\eta (t))\big ) \leqslant K' + K$ for $t \in [0, L]$ . If $d(\eta (t), A) \lt \frac {1}{M} d(x, A)$ for some $t \in [0, L]$ , then we are done; suppose not. We inductively take

\begin{align*} t_{0} := 0, \quad t_{i} := \min \left \{ t_{i-1} + \frac {1}{MK} d(x, A), \,\, L \right \} \quad ( i\gt 0). \end{align*}

The process halts at step $N$ when $t_{N}$ reaches $L$ . We then have

\begin{align*} d\big (\eta (t_{i-1}), \eta (t_{i})\big ) = t_{i} - t_{i-1} \leqslant \frac {1}{MK} d(x, A) \leqslant \frac {1}{K} d\big (\eta (t_{i-1}), A\big ) \end{align*}

for each $i$ . Using Inequality 2, we deduce

\begin{align*} \operatorname {diam} \big ( \pi _{A}(x) \cup \pi _{A}(y)\big ) \leqslant \sum _{i=1}^{N} \operatorname {diam} \big (\pi _{A}(\eta (t_{i-1})) \cup \pi _{A}(\eta (t_{i}))\big ) \leqslant NK. \end{align*}

Since the LHS is at least $K'$ , we have $N \geqslant K' /K \geqslant 2KM(M+7)(K+1) + 1$ .

Meanwhile, $t_{i} - t_{i-1} = d(x, A)/MK$ holds for $i \leqslant N-1$ . This implies

\begin{align*} d(x, y) \geqslant t_{N-1} - t_{0} \geqslant (N-1) \frac {1}{MK} d(x, A), \end{align*}

and considering the assumption $d(x, A) \geqslant K'\geqslant K$ we deduce

\begin{align*} \begin{aligned} d(x, y) - \operatorname {diam} \big (\pi _{A}(x) \cup \pi _{A}(y)\big ) &\geqslant (N-1) \frac {1}{MK} d(x, A) - (K'+ K)\\ &\geqslant MK(M+7)(K+1) \cdot \frac {1}{MK} d(x, A) - 2K'\\ & = (M+7) (K+1) d(x, A) -2K' \\ &\geqslant (M+1)(K+1) d(x, A)+ 4K. \end{aligned} \end{align*}

Now using Inequality 1 twice, we get

\begin{align*} \begin{aligned} d(y, A) &\geqslant \frac {1}{K+1} [d(y, \pi _{A}(y)) - 2K] \\ &\geqslant \frac {1}{K+1}[d(y, x) - d(x, \pi _{A}(x)) - \operatorname {diam} (\pi _{A}(x) \cup \pi _{A}(y)) - 2K] \\ &\geqslant \frac {1}{K+1} [d(x, y) - \operatorname {diam} (\pi _{A}(x) \cup \pi _{A}(y)) - (K+1) d(x, A) - 4K]\\ &\geqslant M d(x, A). \end{aligned} \end{align*}

Proof. Let $M:= 2K+4$ , let $K_{1} := K'(K, M)$ be as in Lemma 2.9, and let $K':= 9 MK_{1}$ .

Suppose to the contrary that Inequality 3 does not hold. Our goal is to find a triple $x', y', z'$ on $[x, y]$ , in order from left to right, such that

\begin{align*} \begin{aligned} d(y', A) &\gt \max \big ( Md(x', A), Md(z', A), K'\big ),\\ 4K_{1} & \geqslant \operatorname {diam}(\pi _{A}\{x', y', z'\}). \end{aligned} \end{align*}

If we find such triple, then we have

\begin{align*} \begin{aligned} d(y', A) &\gt \frac {K+1}{2K+4} \cdot M\cdot d(x', A) + \frac {K+1}{2K+4} \cdot M \cdot d(z', A) + \frac {2}{2K+4} \cdot K' \\ &\geqslant (K+1) d(x', A) + (K+1) d(z', A) + 18K_{1} \\ &\gt (K+1) d(x', A) + (K+1) d(z', A) + (2K_{1} + 4K). \end{aligned} \end{align*}

This will then lead to the contradiction

\begin{align*} \begin{aligned} d(x', z') &\leqslant d\big (x', \pi _{A}(x')\big ) + \operatorname {diam} \big (\pi _{A}(x')\cup \pi _{A}(z')\big ) + d\big (\pi _{A}(z'), z'\big ) \\ &\leqslant \big ( (K+1) d(x', A) + 2K \big ) + 4K_{1} + \big ( (K+1) d(z', A) + 2K \big ) \\ &\lt 2 d(y', A) - (K+1) d(x', A) - (K+1) d(z', A) - 4K \\ &\leqslant \left [ d(y', A) - d\big (x', \pi _{A}(x')\big ) \right ] + \left [ d(y', A) - d\big (z', \pi _{A}(z')\big ) \right ] \\ &\leqslant d(x', y') + d(y', z'). \end{aligned} \end{align*}

Let $\eta : [0, L] \rightarrow X$ be the geodesic connecting $p$ to $x$ and let $t_{0} = 0$ . Given $t_{i-1} \in [0, L)$ , we pick $t_{i} \in [t_{i-1}, L]$ such that

(4) \begin{equation} \operatorname {diam} \big (\pi _{A}(\eta (t_{i-1}))\cup \pi _{A}(\eta (t_{i}))\big ) \leqslant 2K_{1}, \quad d(\eta (t_{i}), A) \geqslant M d(\eta (t_{i-1}), A). \end{equation}

If such $t_{N}$ does not exist at step $N$ , we let $t_{N} = L$ and stop.

Recall that we are assuming

\begin{align*} d(\eta (t_{0}), A) \geqslant K'e^{-D_{1}/K'} d(x, A) + K'e^{-D_{2}/K'} d(y, A) + K' \geqslant K'. \end{align*}

Hence, $d(\eta (t_{i}), A) \geqslant M^{i} K' \geqslant K'$ for $i = 0, \ldots , N-1$ . ( $\ast$ ) Since $\eta$ is bounded, the process must stop at some $N$ . We always have $t_{N} = L$ and $\eta (t_{N}) = x$ . We discuss possible scenarios:

1. (i) $d\big (\pi _{A}(\eta (t_{N-1}))\cup \pi _{A}(\eta (t_{N}))\big ) \gt 2K_{1}$ . Recall Lemma 2.9: there exists $\tau \in [t_{N-1}, t_{N}]$ such that $\operatorname {diam} \big (\pi _{A}(\eta (t_{N-1})\cup \eta (\tau )\big ) \leqslant 2K_{1}$ and either $d(\eta (\tau ), A) \geqslant Md(\eta (t_{N-1}), A)$ or $d(\eta (\tau ), A) \leqslant \frac {1}{M} d(\eta (t_{N-1}), A)$ . Since the first possibility is excluded, we conclude that $d(\eta (\tau ), A) \leqslant{} \frac {1}{M}d(\eta (t_{N-1}), A)$ . There are two subcases.

   1. (a) $N \geqslant 2$ : in this case, $d(\eta (t_{N-1}), A) \geqslant M d(\eta (t_{N-2}), A)$ holds by our choice in Display 4. By ( $\ast$ ), we also know that $d(\eta (t_{N-1}), A) \geqslant K'$ . Lastly, $\pi _{A}\{\eta (t_{N-2}), \eta (t_{N-1}), \eta (\tau )\}$ has diameter at most $4K_{1}$ . Hence, we can take $x' = \eta (\tau )$ , $y' = \eta (t_{N-1})$ and $z' = \eta (t_{N-2})$ .

   2. (b) $N = 1$ : in this case, we have $d(\eta (\tau ), A) \leqslant \frac {1}{M} d(p, A)$ . We first pick $x' = \eta (\tau )$ and will pick $y'$ and $z'$ later.

2. (ii) $\operatorname {diam} \big (\pi _{A}(\eta (t_{N-1})), \pi _{A}(\eta (t_{N})\big ) \leqslant 2K_{1}$ . Then we have

   \begin{align*} D_{1} = \operatorname {diam} \big (\pi _{A}(\eta (0)), \pi _{A}(\eta (t_{N})\big ) \leqslant \sum _{i=1}^{N} \operatorname {diam} \big (\pi _{A}(\eta (t_{i-1})), \pi _{A}(\eta (t_{i}))\big ) \leqslant 2K_{1} N. \end{align*}
   Since $K'\geqslant 4K_{1}$ , $K' \geqslant e^{2}$ and $e\lt 4\lt 2K+4=M$ , we deduce
   \begin{align*} d(x, A) \leqslant \frac {1}{K'} e^{D_{1} / K'} d(p, A) \leqslant \frac {1}{K'} e^{N} d(p, A) \leqslant (2K+4)^{N-2} d(\eta (t_{0}) , A) \leqslant \frac {1}{M} d(\eta (t_{N-1}), A). \end{align*}
   Given this, when $N\geqslant 2$ , we can pick $x' = x=\eta (t_{N})$ , $y'= \eta (t_{N-1})$ and $z' = \eta (t_{N-2})$ and deduce a similar contradiction. When $N = 1$ , we set $x' = x$ .

So far, we have obtained either the desired triple $(x', y', z')$ , or a point $x' \in [x, p]$ such that

\begin{align*} \operatorname {diam}\big (\pi _{A}(x')\cup \pi _{A}(p)\big ) \leqslant 2K', \quad d(x', A) \leqslant \frac {1}{M} d(p, A). \end{align*}

A similar discussion on $[p, y]$ also gives either the desired triple, or a point $z' \in [p, y]$ such that $d(\pi _{A}(z'), \pi _{A}(p)) \leqslant 2K'$ and $d(z', A) \leqslant \frac {1}{M}d(p, A)$ . If we fall into the latter cases in both discussions, we let $y' = p$ and deduce the contradiction.

Proof. Without loss of generality, let $J = [a, b]$ , $J' = [c, d]$ and $\eta (a) = x$ , $\eta (b) = y$ . For each $s \in J'$ , we can pick $t_{s} \in J$ such that $d(\gamma (s), \eta (t_{s})) \lt K$ as $\gamma$ is coarsely contained in $A$ . Note that

\begin{align*} |t_{s_{1}}- t_{s_{2}}| \leqslant Kd \big ( \eta (t_{s_{1}}), \eta (t_{s_{2}}) \big ) + K^{2} \leqslant K d\big (\gamma (s_{1}), \gamma (s_{2}) \big ) + 3K^{2} = K|s_{1} - s_{2}| + 3K^{2}. \end{align*}

Similarly, $|t_{s_{1}} - t_{s_{2}}| \geqslant \frac {1}{K} |s_{1} - s_{2}| - 1 - 2K$ holds. Hence, $s \mapsto t_{s}$ is a $3K^{2}$ -quasi-isometric embedding.

It remains to show that $\{t_{s} : s \in J'\}$ is coarsely equivalent to $J$ . Note that $A$ is $3K$ -coarsely connected, as it is $K$ -coarsely contained in an $1$ -connected set $\gamma$ . It follows that $\eta ^{-1}(A)$ is $4K^{2}$ -coarsely connected subset of $[a, b]$ . Moreover, since $x$ and $y$ are $K$ -close to $A$ , we have $d(a, \eta ^{-1}(A)), d(b, \eta ^{-1}(A)) \lt 2K^{2}$ . Combined together, $[a, b]$ is $4K^{2}$ -coarsely contained in $\eta ^{-1}(A)$ .

Next, for each $p \in A$ there exists $s \in J'$ such that $d(\gamma (s), p) \lt K$ . This implies $d(\eta (t_{s}), p) \lt 2K$ and $\operatorname {diam}(t_{s}, \eta ^{-1}(p)) \lt 3K^{2}$ . Hence, $\eta ^{-1}(A)$ is $3K^{2}$ -coarsely contained in $\{t_{s} : s \in J'\}$ .

Proof. For simplicity, we focus on the case where $\kappa$ , $\eta$ have endpoints.

Let $\gamma : J' \rightarrow X$ be a geodesic that connects the endpoints of $\eta$ . Then $\gamma$ and $\eta$ are coarsely equivalent by Corollary 3.4. Lemma 2.2 tells us that $\pi _{\kappa }(\gamma )$ is coarsely equivalent to $\pi _{\kappa }(\eta )$ and hence large. By Lemma 3.1, there exist $u \lt u'$ in $J'$ such that $\pi _{\kappa }(\gamma )$ and $\gamma |_{[u, u']}$ are coarsely equivalent and such that $\gamma |_{(-\infty , u]}$ and $\gamma |_{[u', +\infty )}$ project onto $\kappa$ near $\gamma (u)$ and $\gamma (u')$ , respectively.

Note again that $\eta$ and $\gamma$ are fellow traveling by Corollary 3.4 and $\pi _{\kappa }$ is coarsely Lipschitz. This enables us to replace $\gamma$ with $\eta$ : there exist $s \lt s'$ in $J$ such that $\pi _{\kappa }(\eta )$ and $\eta |_{[s, s']}$ are coarsely equivalent and such that $\eta |_{(-\infty , s]}$ and $\eta |_{[s', +\infty )}$ project onto $\kappa$ near $\eta (s)$ and $\eta (s')$ , respectively.

Since $\pi _{\kappa }(\eta ) \subseteq \kappa$ and $\eta |_{[s, s']}$ are nearby, each point $\eta (t)$ in $\eta |_{[s, s']}$ is near a point $\kappa (s_{t})$ of $\kappa$ . This $\kappa (s_{t})$ projects onto $\eta$ near $\eta (t)$ . It follows that $\pi _{\eta }(\kappa )$ coarsely contains $\eta |_{[s, s']}$ and hence $\pi _{\kappa }(\eta )$ .

This implies that $\pi _{\eta }(\kappa )$ is also large, and we have another round: there exist $t \lt t'$ in $I$ such that $\pi _{\eta }(\kappa )$ and $\kappa |_{[t, t']}$ are coarsely equivalent. Moreover, $\pi _{\kappa }(\eta )$ coarsely contains $\pi _{\eta }(\kappa )$ . Hence, the two projections are coarsely equivalent, and

\begin{align*} \pi _{\eta }(\kappa ),\,\, \kappa |_{[t, t']}, \,\,\pi _{\kappa }(\eta ),\,\,\eta |_{[s, s']} \end{align*}

are all coarsely equivalent.

Proof of Lemma 2.7. Let $\eta$ and $\kappa$ denote the axes of $g$ and $h$ , i.e., $\eta : i \mapsto g^{i} o$ and $\kappa : j \mapsto h^{j} o$ . Let $\eta$ and $\kappa$ be $K$ -contracting axes for some $K\gt 0$ .

Suppose that $\pi _{\kappa }(\eta )$ has finite diameter, i.e., there exists $M$ such that

\begin{align*} \pi _{\kappa }(\eta ) \subseteq \big \{\kappa (-M), \kappa (-M+1), \ldots , \kappa (M-1), \kappa (M)\big \}. \end{align*}

Then for each $i \in \mathbb{Z}$ and $|j|\gt M+2K^{2}$ , the diameter of $\pi _{\kappa }(\eta (i) ) \cup \kappa (j)$ is greater than $K$ and $[\eta (i), \kappa (j)]$ is $2K$ -close to $\pi _{\kappa }(\eta (i))$ . This forces that

\begin{align*} d(\eta (i), \kappa (j)) \geqslant \inf _{|t| \leqslant M} d(\kappa (j), \kappa (t)) - 2K \geqslant \frac {1}{K} |j| - \frac {M}{K} - 3K. \end{align*}

Similarly, if $\pi _{\eta }(\kappa )$ has finite diameter, then there exists $M'$ such that $d(\eta (i), \kappa (j)) \geqslant \frac {1}{K} |i| - M'$ holds for all $j$ and $|i| \gt M'$ . Hence $d(g^{i} o, h^{j} o)$ is a proper function, and $g$ and $h$ are independent.

Now suppose that $\pi _{\kappa }(\eta )$ has infinite diameter. By Corollary 3.5, $\eta$ and $\kappa$ have subpaths $\eta '$ and $\kappa '$ , respectively, that are coarsely equivalent to $\pi _{\kappa }(\eta )$ , of infinite diameter. This means that $\eta$ and $\kappa$ are not independent.

Proof. For simplicity, let us assume that the domains of $\kappa$ and $\eta$ are closed intervals, say, $I = [t_{0}, t_{1}]$ and $J = [s_{0}, s_{1}]$ , respectively.

It suffices to show that $\pi _{\kappa }(\eta )$ and $\pi _{\eta }(\kappa )$ are both small. Suppose not. Then Corollary 3.5 provides $t \lt t'$ in $I$ and $s \lt s'$ in $J$ such that

\begin{align*} \pi _{\kappa }(\eta ),\,\, \pi _{\eta }(\kappa ), \,\, \kappa |_{[t, t']},\,\, \eta |_{[s. s']} \end{align*}

are all coarsely equivalent and large. Moreover, $\pi _{\eta }(\kappa (t_{0}))$ is near $\kappa (t)$ and $\pi _{\eta }(\kappa (t_{1}))$ is near $\kappa (t')$ . Similarly, $\pi _{\kappa }(\eta (s_{0}))$ is near $\eta (s)$ and $\pi _{\kappa }(\eta (s_{1}))$ is near $\eta (s')$ .

Since $\pi _{\kappa }(\eta )$ and $\pi _{\eta }(\kappa )$ are large, both $t' - t$ and $s' - s$ are large. Since $[\kappa (t), \kappa (t')]$ and $[\eta (s), \eta (s')]$ are coarsely equivalent, one of the following is true:

• – $\kappa (t)$ is near $\eta (s)$ and $\kappa (t')$ is near $\eta (s')$ ; or,

• – $\kappa (t)$ is near $\eta (s')$ and $\kappa (t')$ is near $\eta (s)$ .

This leads to the following contradictions:

• – If $\kappa (t)$ is near $\eta (s)$ , then $\eta (s_{0})$ projects onto $\kappa$ near $\kappa (t)$ . Since $t_{1} - t \geqslant t' - t$ is large, this projection cannot be near $\kappa (t_{1}) = y$ .

• – If $\kappa (t)$ is near $\eta (s')$ , then $\kappa (t_{0})$ projects onto $\eta$ near $\eta (s')$ . Since $s' - s_{0} \geqslant s' - s$ is large, this projection cannot be near $\eta (s_{0}) = x'$ .

Hence, $\pi _{\kappa }(\eta )$ and $\pi _{\eta }(\kappa )$ cannot be large and the conclusion follows.

Proof. Let $E = E(K, D)$ be as in Lemma 3.9 and let $L = 3KE + K^{2}$ . Our claim is that $(x, \kappa _{i})$ and $(\kappa _{i}, y)$ are $E$ -aligned for each $i$ . By symmetry, it suffices to prove the alignment of $(x, \kappa _{i})$ .

Let $\kappa$ be a $K$ -contracting axis whose domain is longer than $L$ . Then the endpoints of $\kappa$ are at least $3E$ -apart. Consequently, no point $p$ in $X$ satisfy the following at the same time:

\begin{align*} (p, \kappa ) \textrm {is}\, E-\textrm {aligned}, \quad (\kappa , p)\, \textrm {is}\, E-\textrm {aligned}. \end{align*}

From this observation, we inductively deduce

\begin{align*} (x, \kappa _{i}) \textrm {is} E-\textrm {aligned} \Rightarrow (\kappa _{i}, x) \textrm {is not} E-\textrm {aligned} \Rightarrow (x, \kappa _{i+1}) \textrm {is} E-\textrm {aligned}. \end{align*}

for $i=1, \ldots , n$ , where the latter implication follows from Lemma 3.9.

Proof. Let $E_{1} = E(K, D)$ and $L_{1} = L(K, D)$ be as in Proposition 3.10. Let $K_{1} = E_{1} + 8K$ and let $E = 10 K'(K, K_{1})$ , where $K'(K, K_{1})$ is as in Corollary 3.7. Let also $L = L_{1} + 101K(K+E) + 2K$ .

We will inductively prove a variant of the given statement, namely:

First, we know that $(\kappa _{1}, y)$ is $E_{1}$ -aligned by Proposition 3.10. Since $(x, \kappa _{1}, y)$ is $K_{1}$ -aligned and $\kappa _{1}$ is long enough, Corollary 3.7 provides a subsegment $\eta _{1} = [x_{1}', y_{1}']$ of $[x, y]$ that is $4K$ -coarsely contained in $\kappa _{1}$ and is $0.1E$ -fellow traveling with $\kappa _{1}$ . We then have

\begin{align*} d(x_{1}', y_{1}') \geqslant \operatorname {diam}(\kappa _{1}) - 0.2E \geqslant \frac {L}{K} - K - 0.2E \geqslant 100E. \end{align*}

If $n=1$ , this finishes the proof. If not, note that $y_{1}'$ is $4K$ -close to $\kappa _{1}$ . Lemma 2.2 implies that $(y_{1}', \kappa _{2})$ is $(D + 8K)$ -aligned, and hence $K_{1}$ -aligned. Now the induction hypothesis implies that $[y_{1}', y]$ has subsegments $\eta _{2}, \ldots , \eta _{n}$ , in order from left to right, that are longer than $100E$ and such that $\eta _{i}$ and $\kappa _{i}$ $0.1E$ -fellow travel for $i\geqslant 2$ . Then $\eta _{1}, \ldots , \eta _{n}$ become the desired subsegments.

Proof. (1) By Lemma 3.8. (2) Proposition 3.12 explains the first two claims in Item (ii). More explicitly, $[x, y]$ contains subsegments $[x_{1}', y_{1}']$ , $\ldots$ , $[x_{N}', y_{N}']$ , in order from left to right, such that:

1. (a) $[x_{i}', y_{i}']$ and $\kappa _{i}$ are $0.1E_{0}$ -coarsely equivalent;

2. (b) $d(x_{i}', x_{i}) \lt 0.1E_{0}$ and $d(y_{i}', y_{i}) \lt 0.1E_{0}$ ;

3. (c) $d(x_{i}', y_{i}') \gt 100E_{0}$

for each $i$ . This implies that

\begin{align*} \begin{aligned} d(x, y) &= d(x, x_{1}') + \sum _{i=1}^{N} d(x_{i}', y_{i}') + \sum _{i=1}^{N-1} d(y_{i}', x_{i+1}') + d(y_{N}', y) \\ &\geqslant d(x, x_{1}) + \sum _{i=1}^{N} d(x_{i}, y_{i}) + \sum _{i=1}^{N-1} d(y_{i}, x_{i+1}) + d(y_{N}, y) - 2 \sum _{i=1}^{N} \big (d(x_{i}, x_{i}') + d(y_{i}, y_{i}') \big )\\ &\geqslant d(x, x_{1}) + \sum _{i=1}^{N} d(x_{i}, y_{i}) + \sum _{i=1}^{N-1} d(y_{i}, x_{i+1}) + d(y_{N}, y) - E_{0} N \\ &\geqslant d(x, x_{1}) + \sum _{i=1}^{N-1} d(y_{i}, x_{i+1}) + d(y_{N}, y) + 50 E_{0} N. \end{aligned} \end{align*}

Proof. Since $\mu$ is non-elementary, the semigroup generated by $\textrm {supp} \mu$ contains independent strongly contracting isometries $a$ and $b$ . By taking suitable powers, we may assume that $a = \Pi (\alpha )$ and $b = \Pi (\beta )$ for some $\alpha , \beta \in (\textrm {supp} \mu )^{L_{0}}$ for some $L_{0}\gt 0$ . There exists $K_{0} \gt 0$ such that:

1. (i) $\Gamma ^{+}(\alpha )$ , $\Gamma ^{-}(\beta )$ are $K_{0}$ -contracting axes, and

2. (ii) $\operatorname {diam}(o \cup \pi _{\gamma }(\eta )) \lt K_{0}$ for distinct axes $\gamma$ , $\eta$ among

   \begin{align*} \Gamma ^{+}(\alpha ),\,\,\Gamma ^{-}(\alpha ),\,\,\Gamma ^{+}(\beta ), \,\,\Gamma ^{-}(\beta ). \end{align*}

The above statements still hold with the same $K_{0}$ when $\alpha$ and $\beta$ are replaced with their self-concatenations, thanks to Lemma 2.3 and Lemma 2.7. Let:

• – $K_{1} = E(K_{0}, K_{0}) \gt K_{0}$ be as in Lemma 3.9;

• – $K_{2} = E(K_{0}, K_{1}) \gt K_{1}$ , $L_{2} = L(K_{0}, K_{1})$ be as in Proposition 3.12;

• – $K_{3} = E(K_{0}, K_{2}) \gt K_{2}$ , $L_{3} = L(K_{0}, K_{2})$ be as in Proposition 3.12;

• – $L_{3}' = 3K_{0}K_{3}$ ;

• – $K_{4} = E(K_{0}, K_{0}, 0)$ , $L_{4} = L(K_{0}, K_{0})$ be as in Lemma 3.13.

By self-concatenating $\alpha$ and $\beta$ if necessary, we may assume that

\begin{align*} L_{0}\gt L_{2}+L_{3} + L_{3}' + L_{4}. \end{align*}

Since $\Gamma ^{+}(\alpha )$ is a $K_{0}$ -quasigeodesic whose domain $L_{0}$ -long, the endpoints of $\Gamma ^{+}(\alpha )$ are at least $(L_{0}/K_{1} - K_{1})$ -apart. Since $L_{0}$ is greater than $3K_{0} K_{3} \geqslant 2K_{0} K_{1} + K_{0}^{2}$ , the endpoints of $\Gamma ^{+}(\alpha )$ are $2K_{1}$ -far. In particular, no set $A \subseteq X$ can be simultaneously contained in the $K_{1}$ -neighborhoods of the two endpoints of $\Gamma ^{+}(\alpha )$ . Hence, the statements

\begin{align*} \big (x, \Gamma ^{\pm }(\alpha )\big )\,\, \textrm {is} K_{1}-\textrm {aligned}, \quad \big (\Gamma ^{\pm }(\alpha ), x\big )\,\, \textrm {is} K_{1}-\textrm {aligned} \end{align*}

are mutually exclusive for any $x \in X$ . Similarly, the statements

\begin{align*} \big (x, \Gamma ^{\pm }(\beta )\big )\,\, \textrm {is} K_{1}-\textrm {aligned}, \quad \big (\Gamma ^{\pm }(\beta ), x\big )\,\, \textrm {is} K_{1}-\textrm {aligned} \end{align*}

are mutually exclusive.

Let $S_{0}$ be the set of sequences of $NL_{0}$ isometries that are concatenations of $\alpha$ ’s and $\beta$ ’s, i.e.,

\begin{align*} S_{0}:= \left \{ (\phi _{1}, \ldots , \phi _{NL_{0}}) \in G^{NL_{0}} : (\phi _{L_{0}(i-1)+1}, \ldots , \phi _{L_{0}i}) \in \{\alpha , \beta \}\,\,\textrm {for} \,\, i =1, \ldots , N\right \}. \end{align*}

Note that $\#S_{0} = 2^{N}$ is greater than $N$ . We claim that for each $m\gt 0$ , the set

\begin{align*} S_{0}^{(m)} := \Big \{ m-\textrm {self-concatenations of} s \in S_{0}\Big \} = \Big \{ \big (\underbrace {s, \ldots , s}_{m \textrm {times}}\big ) : s \in S_{0}\Big \} \end{align*}

is a $((K_{0} L_{0} + K_{0})N + K_{2} + K_{4})$ -Schottky set.

Step 1: Investigating $\Gamma ^{m}(s)$ .

Pick $s =(\phi _{1}, \ldots , \phi _{NL_{0}})$ and $s' = (\phi _{1}', \ldots , \phi _{NL_{0}}')$ in $S_{0}$ . Recall the notation

\begin{align*} x_{nNL_{0} + i}(s) := (\phi _{1} \cdots \phi _{NL_{0}})^{n} \phi _{1} \cdots \phi _{i} o \end{align*}

for $n \in \mathbb{Z}$ and $i = 0, \ldots , NL_{0}-1$ . We now define “sub-axes”

\begin{align*} \begin{aligned} \Gamma _{i}(s) := \left (x_{L_{0}(i-1)}(s), \ldots , x_{L_{0}i}(s)\right ),\\ \Gamma _{-i}(s) := \left (x_{-L_{0}(i-1)}(s), \ldots , x_{-L_{0}i}(s)\right ) \end{aligned} \end{align*}

for each $i \gt 0$ . These are translates of $\Gamma ^{\pm }(\alpha )$ and $\Gamma ^{\pm }(\beta )$ . Our initial choices of $K_{0}$ and $L_{0}$ guarantee that:

• – $\Gamma _{i}(s)$ is a $K_{0}$ -contracting axis whose domain is longer than $L_{2}, L_{3}, L_{3}'$ and $L_{4}$ for each $i \in \mathbb{Z}$ ;

• – $(\Gamma _{i}(s), \Gamma _{i+1}(s))$ and $(\Gamma _{-i}(s'), \Gamma _{-(i+1)}(s'))$ are $K_{0}$ -aligned for each $i \gt 0$ . Moreover, $(\bar {\Gamma }_{-1}(s'), \Gamma _{1}(s))$ is $K_{0}$ -aligned.

Lemma 3.13 tells us that $\cup _{i \gt 0} \Gamma _{i}(s)$ is a $K_{4}$ -contracting axis. In particular, $\Gamma ^{m}(s)$ is a $K_{4}$ -contracting axis for each $m\gt 0$ . Similarly, $\Gamma ^{-m}(s')$ is a $K_{4}$ -contracting axis for each $m\gt 0$ .

Now note that the following sequence of sub-axes is $K_{0}$ -aligned:

\begin{align*} (\ldots , \bar {\Gamma }_{-2}(s'), \bar {\Gamma }_{-1}(s'), \Gamma _{1}(s), \Gamma _{2}(s), \ldots ). \end{align*}

Let $i \gt 0$ and let $p \in \Gamma _{-i} (s')$ . Then Proposition 3.12 tells us that $d\big (p, \Gamma _{1}(s)\big ) \lt d\big (p, \Gamma _{j}(s)\big )$ for each $j\gt 1$ and that $\big (p, \Gamma _{1}(s)\big )$ is $K_{2}$ -aligned. It follows that $\big (p, \cup _{i \gt 0} \Gamma _{i}(s)\big )$ is $K_{2}$ -aligned. For this reason (and its symmetric counterpart), $(\bar {\Gamma }^{-m}(s'), \Gamma ^{m}(s))$ is $K_{2}$ -aligned for each $m\gt 0$ .

Next, fix $x \in X$ and consider the condition

(5) \begin{equation} \big (x, \Gamma _{N}(s)\big )\,\,\textrm {is} K_{2}-\textrm {aligned}. \end{equation}

If Condition 5 holds, then for each $i\gt N$

\begin{align*} \big (x, \Gamma _{N}(s), \Gamma _{N+1}(s) \ldots , \Gamma _{i} (s)\big ) \end{align*}

is $K_{2}$ -aligned and $d(x, \Gamma _{N}(s)) \lt d(x, \Gamma _{i}(s))$ holds by Proposition 3.12. Hence, $\pi _{\cup _{i \gt 0} \Gamma _{i}(s)}(x)$ is contained in $\Gamma _{1}(s) \cup \cdots \cup \Gamma _{N}(s)$ . Meanwhile, recall that for each $i$ , $\Gamma _{i}(s)$ is a $K_{0}$ -quasigeodesic whose domain is $L_{0}$ -long. Hence, we have

\begin{align*} \operatorname {diam}(\Gamma _{i}(s)) \leqslant K_{0} \cdot (\textrm {length of the domain of} s) + K_{0} = K_{0} L_{0} + K_{0}. \end{align*}

Combining these ingredients, we observe that

\begin{align*} \operatorname {diam}\left (\pi _{\Gamma ^{m}(s)} (x) \cup o\right ) \leqslant \operatorname {diam}(\Gamma _{1}(s)) + \ldots + \operatorname {diam}(\Gamma _{N}(s)) \leqslant (K_{0}L_{0} + K_{0})N \end{align*}

holds for every $m\gt 0$ . For a similar reason, the condition

(6) \begin{equation} \big (x, \Gamma _{-N}(s)\big ) \,\,\textrm {is} K_{2}-\textrm {aligned} \end{equation}

implies $\operatorname {diam}\left ( \pi _{\Gamma ^{m}(s)}(x) \cup o \right ) \leqslant (K_{0} L_{0} + K_{0})N$ for all $m \lt 0$ . In summary,

Step 2. Comparing two distinct axes.

We now pick $m\gt 0$ and consider an element of $S_{0}^{(m)}$ which violates these conditions.

To show this, let $k \in \{1, \ldots , N\}$ be the first index such that $(\phi _{L_{0}(k-1)+1}, \ldots , \phi _{L_{0}k})$ and $(\phi _{L_{0}(k-1) + 1}', \ldots , \phi _{L_{0}k}')$ differ. Let us denote $x_{i}(s)$ by $x_{i}$ and $x_{i}(s')$ by $x_{i}'$ . Note that the path

\begin{align*} \left (x_{NL_{0}}, \,\, x_{NL_{0}-1}, \,\, \ldots , \,\,x_{(k-1)L_{0}} = x'_{(k-1)L_{0}},\,\, x'_{(k-1)L_{0}+1}, \,\, \ldots , \,\,x'_{NL_{0}}\right ) \end{align*}

is the concatenation of $K_{0}$ -aligned $K_{0}$ -contracting axes

\begin{align*} (\eta _{i})_{i=1}^{2(N-k+1)}:= \left (\bar {\Gamma }_{N}(s), \bar {\Gamma }_{N-1}(s), \ldots , \bar {\Gamma }_{k}(s), \Gamma _{k}(s'), \ldots , \Gamma _{N}(s')\right ). \end{align*}

Recall that $s$ violates Condition 5: $(\bar {\Gamma }_{N}(s), x) = (\eta _{1}, x)$ is not $K_{2}$ -aligned. Since $(\eta _{1}, \eta _{2})$ is $K_{0}$ -aligned, Lemma 3.9 tells us that $(x, \eta _{2})$ is $K_{1}$ -aligned. Then $(x, \eta _{2}, \ldots , \eta _{2(N-k+1)})$ is $K_{1}$ -aligned and Proposition 3.12 tells us that $(x, \eta _{2(N-k+1)}) = (x, \Gamma _{N}(s'))$ is $K_{2}$ -aligned. Hence, $s'$ satisfies Condition 5.

Similarly, by considering the $K_{0}$ -aligned sequence

\begin{align*} \big (\bar {\Gamma }_{N}(s), \,\,\bar {\Gamma }_{N-1}(s),\,\, \ldots ,\,\, \bar {\Gamma }_{1}(s), \,\,\Gamma _{-1}(s'), \,\,\Gamma _{-2}(s'),\,\, \ldots , \,\,\Gamma _{-N}(s') \big ), \end{align*}

we can deduce that $(x, \Gamma _{-N}(s'))$ is $K_{2}$ -aligned as desired.

A similar argument leads to the following.

Step 3: Summary.

We claim that $S_{0}^{(m)}$ is $((K_{0} L_{0} + K_{0})N + K_{2}+K_{4})$ -Schottky. The first and the third requirements for Schottky sets were already observed before, so it remains to discuss the second requirement. Considering Observation 3.20, it suffices to show that Condition 5 and Condition 6 are satisfied by all but at most 1 element of $S_{0}^{(m)}$ . Observation 3.21 and 3.22 imply that this is the case.

Given these observations, we can finish the proof by taking $K = (K_{0} L + K_{0} )N + K_{2} + K_{4}$ , $m =L$ and by taking any subset $S \subseteq S_{0}^{(m)}$ such that $\#S = N$ .