Proof. The lemma follows from unraveling the definitions. For instance, in case (ii), we have $\sigma '(\psi _{a,b}(a)) = \sigma '(a) = \sigma (a)$ , $\sigma '(\psi _{a,b}(b)) = \sigma '(ab) = \sigma (a)t = \sigma (b)$ , and $\sigma '(\psi _{a,b}(c)) = \sigma '(c) = \sigma (c)$ for all $c \not = a,b$ , which shows that $\sigma '\psi _{a,b} = \sigma $ .

Proof of Lemma 3.2

By contradiction, we assume that u, v, and $\{\sigma _j\}_{j \in J}$ are counterexamples for the lemma. Moreover, we suppose that $\ell $ is as small as possible.

Let us write $u = au'$ and $v=bv'$ , where $a,b \in {\mathcal A}$ . Since $\sigma _j(u)$ is a prefix of $\sigma _j(v)$ , we have that for every $j \in J$ ,

(8) $$ \begin{align} \text{one of the words in}\ \{\sigma_j(a), \sigma_j(b)\}\ \text{is a prefix of the other.} \end{align} $$

We consider two cases. First, we suppose that $\ell _a = \ell _b$ . In this case, (8) implies that $\sigma _j(a) = \sigma _j(b)$ for every $j\in J$ . Hence, we can use (i) of Lemma 3.1 to decompose each $\sigma _j$ as $\sigma _j'\phi _{a,b}$ , where $\sigma ^{\prime }_j$ is the restriction of $\sigma _j$ to $({\mathcal A}\setminus \{b\})^+$ . Since $\phi _{a,b}$ is letter-onto and $\ell _c = |\sigma ^{\prime }_j(c)|$ for every $j \in J$ , $c\in {\mathcal A}\setminus \{b\}$ , the conclusion of the lemma holds, contrary to our assumptions.

This leaves to consider the case in which $\ell _a \not = \ell _b$ . We only do the case $\ell _a < \ell _b$ as the other is similar. Then, by (8), for every $j \in J$ , there exists a non-empty word $t_j \in {\mathcal B}_j^{\ell _b-\ell _a}$ of length $\ell _b-\ell _a$ such that $\sigma _j(b) = \sigma _j(a)t_j$ . Thus, we can use (ii) of Lemma 3.1 to write, for any $j \in J$ , $\sigma _j = \sigma ^{\prime }_j\psi _{a,b}$ , where $\sigma ^{\prime }_j$ is defined as in (5).

Let $\tilde {u} = \psi _{a,b}(u')$ and $\tilde {v} = b\psi _{a,b}(v')$ . We want now to prove that $\tilde {u}$ , $\tilde {v}$ , and $\{\sigma ^{\prime }_j : j \in J\}$ satisfy the hypothesis of the lemma. First, we observe that for every $j \in J$ ,

(9) $$ \begin{align} \text{if }c\not=b, \text{ then } |\sigma^{\prime}_j(c)| = \ell_c, \text{ and } |\sigma^{\prime}_j(b)| = |t_j| = \ell_b-\ell_a. \end{align} $$

Therefore, $\{\sigma ^{\prime }_j\}_{j\in J}$ satisfy (7). Also, since $\psi _{a,b}(c)$ never starts with b, we have that

(10) $$ \begin{align} \tilde{u}, \tilde{v}\ \text{start with different letters.} \end{align} $$

Furthermore, by using the symbol $\leq _p$ to denote the prefix relation, we can compute:

$$ \begin{align*} \sigma_j(a)\sigma^{\prime}_j(\tilde{u}) &= \sigma_j(a)\sigma_j(u') = \sigma_j(u) \leq_p \sigma_j(v) = \sigma^{\prime}_j(\psi_{a,b}(v)) = \sigma^{\prime}_j(a)\sigma^{\prime}_j(\tilde{v}). \end{align*} $$

This and the fact that $\sigma _j(a)$ is equal to $\sigma ^{\prime }_j(a)$ imply that

(11) $$ \begin{align} \sigma^{\prime}_j(\tilde{u}) \text{ is a prefix of } \sigma^{\prime}_j(\tilde{v}) \text{ for every } j \in J. \end{align} $$

Finally, we note

(12) $$ \begin{align} |\tilde{u}| \geq |u|-1 \geq \sum_{c\in{\mathcal A}}\ell_c - \ell_a =: \ell'. \end{align} $$

We conclude from (9), (10), (11), and (12) that $\tilde {u}$ , $\tilde {v}$ , and $\{\sigma ^{\prime }_j : j \in J\}$ satisfy the hypothesis of this lemma. Since $\ell ' < \ell $ , the minimality of $\ell $ implies that there exist a letter-onto morphism $q'\colon {\mathcal A}^+\to {\mathcal C}^+$ , with $\#{\mathcal C} < \#{\mathcal A}$ , and morphisms $\{p_j\colon {\mathcal C}^+\to {\mathcal B}_j^+\}_{j\in J}$ satisfying $\sigma ^{\prime }_j=p_jq'$ and a property analogous to (7). However, then $q := q'\psi _{a,b}$ is also letter-onto and the morphisms $\{p_j\}_{j\in J}$ satisfy $\sigma _j = p_jq$ and a property analogous to (7). Thus, the conclusion of the lemma holds for $\{\sigma _j\}_{j\in J}$ , contrary to our assumptions.

Proof of Lemma 3.4

Let us write $u = au'$ and $v = bv'$ . We first consider the case in which $s=1$ . In this situation, u and v start with different letters, so Lemma 3.2 can be applied (with the index set J chosen as a singleton) to obtain a decomposition ${\mathcal A}^+ \overset {q}{\to } {\mathcal C}^+ \overset {p}{\to } {\mathcal B}^+$ such that q is letter-onto, $\#{\mathcal C} < \#{\mathcal A}$ , and $\sigma = pq$ . Since ${\mathcal C}$ has strictly fewer elements than ${\mathcal A}$ , we have $|p|_1 < |\sigma |_1$ . Hence, the conclusion of the lemma holds in this case.

We now assume that $s\not =1$ . In this case, t and s are non-empty, so we can use (iii) of Lemma 3.1 to factorize $\sigma = \sigma '\theta _{a, \tilde {a}}$ , where $\tilde {a}$ is a letter not in ${\mathcal A}$ and $\sigma '$ is defined as in (6). We set $\tilde {u} = a\theta _{a,\tilde {a}}(u')$ and $\tilde {v} = \theta _{a,\tilde {a}}(v)$ . Our plan is to use Lemma 3.2 with $\tilde {u}$ , $\tilde {v}$ and $\sigma '$ .

Observe that $\theta _{a,\tilde {a}}(c)$ never starts with a, so

(13) $$ \begin{align} \tilde{u}, \tilde{v}\ \text{start with different letters.} \end{align} $$

Also, by using, as in the previous proof, the symbol $\leq _p$ to denote the prefix relation,

$$ \begin{align*} s\sigma'(\tilde{u}) = s\sigma'(a) \sigma'(\theta_{a,\tilde{a}}(u')) = st\sigma(u') = \sigma(u) \leq_p s\sigma(v) = s\sigma'(\theta_{a,\tilde{a}}(v)) = s\sigma'(\tilde{v}), \end{align*} $$

which implies that

(14) $$ \begin{align} \sigma'(\tilde{u})\ \text{is a prefix of}\ \sigma'(\tilde{v}). \end{align} $$

Finally, we use (6) to compute:

(15) $$ \begin{align} |\tilde{u}| \geq |u|-1 \geq |\sigma|_1 +|s|-1 \geq |\sigma|_1 = |\sigma'|_1. \end{align} $$

We conclude, by (13), (14), and (15) that Lemma 3.2 can be applied with $\tilde {u}$ , $\tilde {v}$ , and $\sigma '$ (and J as a singleton). Thus, there exist morphisms $q'\colon ({\mathcal A}\cup \{\tilde {a}\})^+\to {\mathcal C}^+$ and $p\colon {\mathcal C}^+\to {\mathcal B}^+$ such that $\#{\mathcal C} < \#({\mathcal A}\cup \{\tilde {a}\})$ , $q'$ is letter-onto, and $\sigma '=pq'$ . Then, $\#{\mathcal C} \leq \#{\mathcal A}$ , $q := q'\theta _{a, \tilde {a}}$ is letter-onto, and $\sigma = p q'\theta _{a, \tilde {a}} = pq$ . Moreover, since $\theta _{a, \tilde {a}}$ is not the identity function, we have $|p|_1 < |\sigma |_1$ .

Proof. By contradiction, we suppose that the lemma does not hold for $\phi $ and $\tau $ and, moreover, that $|\phi |_1$ as small as possible.

That $\phi ({\mathcal A})^+ \cap \tau ({\mathcal B}^{++})$ is non-empty means that there exist $u = u_1\ldots u_n \in {\mathcal A}^+$ and $w = w_1\ldots w_m \in {\mathcal B}^{++}$ with $\phi (u) = \tau (w)$ . If $m = 1$ , then, since $w \in {\mathcal B}^{++}$ , we have $\#{\mathcal B} = \{v_1\}$ and the conclusion of the lemma trivially holds for ${\mathcal D} = \{a\in {\mathcal C}:a\ \text {occurs in}\ \tau (w_1)\}$ , $q\colon {\mathcal B}^+\to {\mathcal D}^+$ , $w_1\mapsto \tau (w_1)$ , and $p\colon {\mathcal D}^+\to {\mathcal C}^+$ the inclusion map, contradicting our initial assumption. Therefore, $m \geq 2$ and $\{1,\ldots ,m-1\}$ is non-empty.

Let $k \in \{1,\ldots ,m-1\}$ . We define $i_k$ as the smallest number in $\{1,\ldots ,n\}$ for which $|\tau (w_1\ldots w_k)| < |\phi (u_1\ldots u_{i_k})|$ holds. Since $|\phi (u_1)| \leq |\phi |_1 \leq \ell \leq |\tau (w_1\ldots w_k)|$ , $i_k$ is at least $2$ and, thus, $|\phi (u_1\ldots u_{i_k-1})| \leq |\tau (w_1\ldots w_k)|$ by minimality of $i_k$ . Hence, there exists a decomposition $\phi (u_{i_k}) = s_kt_k$ such that $t_k$ is non-empty and

(16) $$ \begin{align} t_k\phi(u_{i_k+1}\ldots u_n) = \tau(w_{k+1}\ldots w_m). \end{align} $$

Our next objective is to use Lemma 3.4 to prove that $s_k$ and $u_k$ have a very particular form.

Proof. To prove this, we suppose that it is not true, that is, there exists $k\in \{1,\ldots ,m-1\}$ such that

(17) $$ \begin{align} s_k \not= 1 \quad \text{or}\quad u_1 \not= u_{i_k}. \end{align} $$

Let $\tilde {u} := u_{i_k}\ldots u_{i_k+|\phi |^2_1-1}$ and $\tilde {v} := u_1\ldots u_{|\phi |^3_1}$ . We are going to check the hypothesis of Lemma 3.4 for $\tilde {u}$ , $\tilde {v}$ and $\phi $ .

First, we observe that, since $\phi (u)=\tau (v)$ , we have that $\phi (\tilde {v})$ is a prefix of $\tau (v)$ . Moreover, given that $|\phi (\tilde {v})| \leq |\phi |_1^4 \leq \ell $ and that $\tau $ is $\ell $ -proper, $\phi (\tilde {v})$ is a prefix of $\tau (b)$ for every $b \in {\mathcal B}$ . In particular,

(18) $$ \begin{align} \phi(\tilde{v})\ \text{is a prefix of}\ \tau(w_k). \end{align} $$

Second, from (16) and the inequalities $|t_k\phi (u_{i_k+1}\ldots u_{i_k+|\phi |^2_1-1})| \leq |\phi |_1^3 \leq \ell \leq |\tau (w_k)|$ , we deduce that $t_k\phi (u_{i_k+1}\ldots u_{i_k+|\phi |^2_1-1})$ is a prefix of $\tau (w_k)$ . Therefore,

(19) $$ \begin{align} \phi(\tilde{u}) = s_kt_k\phi(u_{i_k+1}\ldots u_{i_k+|\phi|^2_1-1})\ \text{is a prefix of}\ s_k\tau(w_k). \end{align} $$

We conclude from (18), (19), and the inequality $|\phi (\tilde {u})| \leq |\phi |_1^3 = |\tilde {v}| \leq |s_k\phi (\tilde {v})|$ that

$$ \begin{align*} \phi(\tilde{u})\ \text{is a prefix of}\ s_k\phi(\tilde{v}). \end{align*} $$

This, the inequality $|\tilde {u}| \geq |\phi |_1+|s_k|$ and (17) allow us to use Lemma 3.4 with $\tilde {u}$ , $\tilde {v}$ , and $\phi $ and obtain morphisms ${\mathcal A}^+ \overset {\tilde {q}}{\longrightarrow } \tilde {{\mathcal A}}^+ \overset {\tilde {\phi }}{\longrightarrow } {\mathcal C}^+$ such that $\#\tilde {{\mathcal A}} \leq \#{\mathcal A}$ , $\phi = \tilde {\phi }\tilde {q}$ and $|\tilde {\phi }|_1 < |\phi |_1$ . Then, $\ell \geq |\phi |_1^4> |\tilde {\phi }|_1^4$ and $\tilde {\phi }(\tilde {{\mathcal A}}^+) \cap \tau ({\mathcal B}^{++})$ contains the element $\tilde {\phi }(\tilde {q}(u)) = \tau (w)$ , and so $\tau $ and $\tilde {\phi }$ satisfy the hypothesis of this lemma. Therefore, by the minimality of $|\phi |_1$ , there exists a decomposition ${\mathcal B}^+ \overset {q}{\to } {\mathcal D}^+ \overset {p}{\to } {\mathcal C}^+$ of $\tau $ satisfying conditions (i)–(iii) of this lemma, contrary to our assumptions.

An argument similar to the one used in the proof of the previous claim gives us that

(20) $$ \begin{align} u_n = u_{i_k-1}\quad \text{for every}\ k\in\{1,\ldots,m-1\}. \end{align} $$

We refer the reader to Remark 3.5 for further details.

Now we can finish the proof. First, from (16) and the first part of the claim, we get that $\tau (w_k) = \phi (u_{i_{k-1}}\ldots u_{i_k-1})$ for $k \in \{2,\ldots ,m-1\}$ , $\tau (w_1) = \phi (u_1\ldots u_{i_1-1})$ and $\tau (w_m) = \phi (u_{i_{m-1}}\ldots u_n)$ . Being $w \in {\mathcal B}^{++}$ , these equations imply that each $\tau (b)$ , $b \in {\mathcal B}$ , can be written as a concatenation $x_1\ldots x_N$ , with $x_j \in \phi ({\mathcal A})$ . Moreover, by the second part of the claim and (20), we can choose this decomposition so that $x_1 = u_1$ and $x_N = u_n$ . This defines (maybe non-unique) morphisms ${\mathcal B}^+ \overset {q}{\longrightarrow } {\mathcal D}_1^+ \overset {p_1}{\longrightarrow } {\mathcal C}^+$ such that $\tau = p_1q$ , $\#{\mathcal D}_1 \leq \#\{\phi (u_1),\ldots ,\phi (u_n)\} \leq \#{\mathcal A}$ , and q is proper. If we define ${\mathcal D}$ as the set of letters $d \in {\mathcal D}_1$ that occur in some $w \in q({\mathcal B})$ , and p as the restriction of $p_1$ to ${\mathcal D}$ , then we obtain a decomposition ${\mathcal B}^+ \overset {q}{\longrightarrow } {\mathcal D}^+ \overset {p}{\longrightarrow } {\mathcal C}^+$ that still satisfies the previous properties, but in which q is letter-onto. Hence, p and q met conditions (i), (ii), and (iii).

Proof. We start by proving that $X_{\boldsymbol {\tau }}^{(n)} \subseteq \bigcup _{k\in {\mathbb Z}}T^k\phi _n(X_{\boldsymbol {\sigma }}^{(n)})$ . Let $y \in X_{\boldsymbol {\tau }}^{(n)}$ and $\ell \in {\mathbb N}$ . There exist $N>n$ and $b \in {\mathcal B}_n$ such that $y_{[-\ell ,\ell ]}$ occurs in $\tau _{[n,N)}(b)$ . In addition, since $\phi _N$ is letter-onto, there exists $a \in {\mathcal A}_N$ for which b occurs in $\phi _N(a)$ . Then, $y_{[-\ell ,\ell ]}$ occurs in $\tau _{[n,N)}\phi _N(b)$ and, consequently, also in $\phi _n\sigma _{[n,N)}(b)$ as $\tau _{[n,N)}\phi _N = \phi _n\sigma _{[n,N)}$ . Hence, by taking the limit $\ell \to \infty $ , we can find $(k',x) \in {\mathbb Z}\times X_{\boldsymbol {\sigma }}^{(n)}$ such that $y = T^{k'}\phi _n(x)$ . Therefore, $y \in \bigcup _{k\in {\mathbb Z}}T^k\phi _n(X_{\boldsymbol {\sigma }}^{(n)})$ . To prove the other inclusion, we use Lemma 2.6 to compute:

$$ \begin{align*} \phi_n(X_{\boldsymbol{\sigma}}^{(n)}) &= \bigcap_{N>n}\bigcup_{k\in{\mathbb Z}}T^k\phi_n\sigma_{[n,N)}({\mathcal A}_N^{\mathbb Z}) = \bigcap_{N>n}\bigcup_{k\in{\mathbb Z}}T^k\tau_{[n,N)}\phi_N({\mathcal A}_N^{\mathbb Z}) \\&\subseteq \bigcap_{N>n}\bigcup_{k\in{\mathbb Z}}T^k\tau_{[n,N)}({\mathcal B}_N^{\mathbb Z}) = X_{\boldsymbol{\tau}}^{(n)}.\\[-46pt] \end{align*} $$

Proof. Let us write $\boldsymbol {\sigma } = ({\mathcal A}_{n+1}^+\to {\mathcal A}_n^+)_{n\in {\mathbb N}}$ and $\boldsymbol {\tau } = ({\mathcal B}_{n+1}^+\to {\mathcal B}_n^+)_{n\in {\mathbb N}}$ . Up to contractions, we can suppose that for every $n \geq 1$ , $\#{\mathcal A}_n = \mathrm {AR}(\boldsymbol {\sigma })$ and that $\tau _n$ is $|\phi _n|^4_1$ -proper (for the last property, we used that $\boldsymbol {\tau }$ is everywhere growing and proper).

Using that $\phi _{n+1}$ is letter-onto, we can compute:

$$ \begin{align*} \tau_n({\mathcal B}_{n+1}^{++}) \supseteq \tau_n(\phi_{n+1}({\mathcal A}_{n+1}^{++})) = \phi_n(\sigma_n({\mathcal A}_{n+1}^{++})) \subseteq \phi_n({\mathcal A}_n^+), \end{align*} $$

where in the middle step, we used the commutativity property of $\boldsymbol {\phi }$ . We deduce that

$$ \begin{align*} \tau_n({\mathcal B}_{n+1}^{++}) \cap \phi_n({\mathcal A}_n^+) \not= \emptyset\quad \text{for every}\ n \in {\mathbb N}. \end{align*} $$

This and the fact that $\tau _n$ is a $|\phi _n|^4_1$ -proper morphism allow us to use Lemma 3.6 to find morphisms ${\mathcal B}_{n+1}^+ \overset {q_{n+1}}{\longrightarrow } {\mathcal D}_{n+1}^+ \overset {p_n}{\longrightarrow } {\mathcal B}_n^+$ such that

$$ \begin{align*} \text{(i)}\ \#{\mathcal D}_{n+1} \leq \#{\mathcal A}_n,\quad \text{(ii)}\ \tau_n = p_nq_{n+1},\quad \text{(iii)}\ q_{n+1}\ \text{is letter-onto and proper.} \end{align*} $$

We define $\nu _0 := p_0$ , the morphisms $\nu _n := q_np_n \colon {\mathcal D}_{n+1}^+ \to {\mathcal D}_n^+$ and $\psi _n := q_n\phi _n\colon {\mathcal A}_n^+\to {\mathcal D}_n^+$ , $n \geq 1$ , and the sequences $\boldsymbol {\nu } = (\nu _n)_{n\in {\mathbb N}}$ and $\boldsymbol {\psi } = (\psi _n)_{n\in {\mathbb N}}$ , where $\psi _0 := \phi _0$ . We are going to show that these objects satisfy the conclusion of the proposition.

We start by observing that it follows from the definitions that the diagram below commutes for all $n\geq 1$ :

In particular, $\nu _n\nu _{n+1} = q_n\tau _np_{n+1}$ , so $\langle \nu _{[n,n+1]} \rangle \geq \langle \tau _n\rangle $ . Being $\boldsymbol {\tau }$ everywhere growing, this implies that $\boldsymbol {\nu }$ has the same property. We also observe that condition (iii) implies that $\nu _n = q_np_n$ is letter-onto and proper. Altogether, these arguments prove that, up to contracting the first levels, $\boldsymbol {\nu }$ is everywhere growing and proper.

Next, we note that $\boldsymbol {\nu }$ and $\boldsymbol {\tau }$ are equivalent as both are contractions of $(p_0, q_1, p_1, q_2, \ldots )$ . This implies, by Lemma 2.3, that $\boldsymbol {\nu }$ is recognizable if $\boldsymbol {\tau }$ is recognizable. Further, by condition (i), $\boldsymbol {\nu }$ has alphabet rank at most $\mathrm {AR}(\boldsymbol {\sigma })$ .

It is only left to prove that $\boldsymbol {\psi }$ is a letter-onto and proper factor. By unraveling the definitions we can compute:

$$ \begin{align*} \psi_0 = \phi_0 = \tau_0\phi_1 = p_0q_1\phi_1 = \nu_0\psi_1, \end{align*} $$

and from the diagram, we have $\sigma _n\psi _n = \psi _{n+1}\tau _n$ for all $n\geq 1$ . Therefore, $\boldsymbol {\psi }$ is a factor. Finally, since $q_n$ is letter-onto and proper by condition (iii) and $\boldsymbol {\phi }$ was assumed to be letter-onto, $\psi _n = q_n\phi _n$ is letter-onto and proper.

Proof. By Lemma 2.5, $\tilde {\tau }_n$ is letter-onto mapping $\tilde {{\mathcal A}}_{n+1}^+$ into $\tilde {{\mathcal A}}_n$ . Moreover, that lemma also gives that for every $x \in X_{\boldsymbol {\tau }}^{(n)}$ and $N>n$ , there exists a $\tau _{[n,N)}$ -factorization $(k',x')$ of x in $X_{\boldsymbol {\tau }}^{(N)}$ . This together with the inclusion $X_{\boldsymbol {\tau }}^{(N)} \subseteq \tilde {{\mathcal A}}_N^{\mathbb Z}$ imply that

$$ \begin{align*} Z := \bigcap_{N>n}\bigcup_{k\in{\mathbb Z}}T^k\tau_{[n,N)}(\tilde{{\mathcal A}}_N^{\mathbb Z}) \supseteq X_{\boldsymbol{\tau}}^{(n)}. \end{align*} $$

Now, $\boldsymbol {\tilde {\tau }}$ is everywhere growing and proper, so we can apply Lemma 2.6 to obtain that $X_{\boldsymbol {\tilde {\tau }}}^{(n)} = Z \supseteq X_{\boldsymbol {\tau }}^{(n)}$ . Since it is clear that $X_{\boldsymbol {\tilde {\tau }}}^{(n)}\subseteq X_{\boldsymbol {\tau }}^{(n)}$ as $\tilde {{\mathcal A}}_N \subseteq {\mathcal A}_N$ for every N, we conclude that $X_{\boldsymbol {\tilde {\tau }}}^{(n)} = X_{\boldsymbol {\tau }}^{(n)}$ .

If $\boldsymbol {\tau }$ is letter-onto, then ${\mathcal A}_n \subseteq {\mathcal L}(\bigcup _{k\in {\mathbb Z}}T^k\tau _{[n,N)}({\mathcal A}_N^{\mathbb Z}))$ for every $N>n$ , and hence, by the formula in Lemma 2.6, ${\mathcal A}_n \subseteq {\mathcal L}(X_{\boldsymbol {\tau }}^{(n)})$ .

Proof. Thanks to Lemma 4.6, we can assume without loss of generality that $\boldsymbol {\sigma }$ is letter-onto. Moreover, in this case,

(22) $$ \begin{align} {\mathcal A}_n \subseteq {\mathcal L}(X_{\boldsymbol{\sigma}}^{(n)})\quad \text{for every}\ n \in {\mathbb N}. \end{align} $$

Let us write $\boldsymbol {\sigma } = (\sigma _n\colon {\mathcal A}_{n+1}^+\to {\mathcal A}_n^+)_{n\in {\mathbb N}}$ . By contracting $\boldsymbol {\sigma }$ , we can further assume that $\sigma _0$ is r-proper and $\pi $ has radius r. Then, Lemma 2.4 gives us a proper morphism $\tau \colon {\mathcal A}_1^+\to {\mathcal B}^+$ , where ${\mathcal B}$ is the alphabet of Y, such that

(23) $$ \begin{align} \pi(\sigma_0(x)) = \tau(x)\ \text{for all}\ x \in X_{\boldsymbol{\sigma}}^{(1)}\quad \text{and}\quad |\sigma_0(a)| = |\tau(a)|\ \text{for every}\ a \in {\mathcal A}_1. \end{align} $$

In particular, $\pi (\sigma _{[0,n)}(x)) = \tau \sigma _{[1,n)}(x)$ and $|\sigma _{[0,n)}(a)| = |\tau \sigma _{[1,n)}(a)|$ for all $n \in {\mathbb N}$ , $x \in X_{\boldsymbol {\sigma }}^{(n)}$ , and $a \in {\mathcal A}_n$ , so (23) holds for any contraction of $\boldsymbol {\sigma }$ .

We define $\boldsymbol {\tilde {\sigma }} = (\tau , \sigma _1, \sigma _2, \ldots )$ and observe this is a letter-onto, everywhere growing and proper sequence generating Y. This and that Y is aperiodic allow us to use Proposition 4.4 and obtain, after a contraction, a letter-onto factor $\boldsymbol {\tilde {\phi }}\colon \boldsymbol {\tilde {\sigma }}\to \boldsymbol {\tilde {\tau }}$ , where $\tilde {\phi }_0 = \tilde {\sigma }_0 = \tau $ and $\boldsymbol {\tilde {\tau }}$ is a letter-onto, everywhere growing, proper, and recognizable directive sequence generating Y. The sequence $\boldsymbol {\tilde {\tau }}$ has all the properties required by the theorem but having alphabet rank bounded by K. To overcome this, we use Proposition 4.5 with $\boldsymbol {\tilde {\phi }}$ and do more contractions to obtain a letter-onto and proper factor $\boldsymbol {\phi }\colon \boldsymbol {\tilde {\sigma }} \to \boldsymbol {\tau }$ such that $\phi _0 = \tilde {\phi }_0 = \tau $ and $\boldsymbol {\tau }$ is a letter-onto, everywhere growing, proper, and recognizable directive sequence generating Y and satisfying $\mathrm {AR}(\boldsymbol {\tau }) \leq \mathrm {AR}(\tilde {\boldsymbol {\sigma }}) = \mathrm {AR}(\boldsymbol {\sigma })$ .

It is left to prove the last part of the theorem. Observe that since $\boldsymbol {\tilde {\sigma }}$ and $\boldsymbol {\sigma }$ differ only at their first coordinate, $\boldsymbol {\phi }$ is also a factor from $\boldsymbol {\sigma }$ to $\boldsymbol {\tau }$ . Further, by (23) and the fact that $\phi _0 = \tau $ , we have $\pi (\sigma _0(x)) = \tau (x) = \phi _0(x)$ and $|\sigma _0(a)| = |\phi _0(a)|$ for every $x \in X_{\boldsymbol {\sigma }}^{(1)}$ and $a \in {\mathcal A}_1$ .

Proof. We can use Theorem 1.3 to obtain an everywhere growing, proper, and recognizable directive sequence $\boldsymbol {\tau } = (\tau _n\colon {\mathcal B}_{n+1}^+\to {\mathcal B}_n^+)_{n\in {\mathbb N}}$ generating X and having alphabet rank at most K. Due to Lemma 4.6, we can assume that $\boldsymbol {\tau }$ is letter-onto. In particular, ${\mathcal B}_n \subseteq {\mathcal L}(X_{\boldsymbol {\tau }}^{(n)})$ for every $n \in {\mathbb N}$ .

We claim that $X_{\boldsymbol {\tau }}^{(n)}$ is minimal. Indeed, if $Y \subseteq X_{\boldsymbol {\tau }}^{(n)}$ is a subshift, then $\tau _{[0,n)}(Y)$ is closed (as $\tau _{[0,n)}\colon X_{\boldsymbol {\tau }}^{(n)}\to X_{\boldsymbol {\tau }}$ is continuous), so $\bigcup _{k\in {\mathbb Z}}T^k\tau _{[0,n)}(Y) = \bigcup _{|k|\leq |\tau _{[0,n)}|}T^k\tau _{[0,n)}(Y)$ is a subshift in $X_{\boldsymbol {\tau }}$ which, by minimality, is equal to it. Thus, any point $x \in X_{\boldsymbol {\tau }}^{(n)}$ has a $\tau _{[0,n)}$ -factorization $(k,z)$ with $z \in Y$ . The recognizability property of $(X_{\boldsymbol {\tau }}^{(n)},\tau _{[0,n)})$ then implies that $Y = X_{\boldsymbol {\tau }}^{(n)}$ .

Now, we prove that for any $n\in {\mathbb N}$ , there exists $N>n$ such that $\tau _{[n,N)}$ is positive. This would imply that the topological rank of X is at most K and hence would complete the proof. Let $n \in {\mathbb N}$ and R be a constant of recognizability for $(X_{\boldsymbol {\tau }}^{(n)},\tau _{[0,n)})$ . Since $X_{\boldsymbol {\tau }}^{(n)}$ is minimal, there exists a constant $L\geq 1$ such that two consecutive occurrences of a word $w \in {\mathcal L}(X_{\boldsymbol {\tau }}^{(n)})\cap {\mathcal B}_n^{2R+1}$ in a point $x \in X_{\boldsymbol {\tau }}^{(n)}$ are separated by at most L. Let $N> n$ be big enough so that $\langle \tau _{[0,N)}\rangle \geq L+2R$ . Then, for all $a \in {\mathcal B}_N \subseteq {\mathcal L}(X_{\boldsymbol {\tau }}^{(N)})$ and $w \in {\mathcal L}(X_{\boldsymbol {\tau }}^{(n)})\cap {\mathcal B}_n^{2R+1}$ , w occurs at a position $i \in \{R,R+1,\ldots ,|\tau _{[0,N)}(a)|-R\}$ of $\tau _{[0,N)}(a)$ . Since R is a recognizability constant for $(X_{\boldsymbol {\tau }}^{(n)},\tau _{[0,n)})$ , we deduce that for all $a \in {\mathcal B}_N$ and $b \in {\mathcal B}_n$ , b occurs in $\tau _{[n,N)}(a)$ . Thus, $\tau _{[n,N)}$ is positive.

Proof. We are going to prove that X is generated by an everywhere growing and proper directive sequence $\boldsymbol {\tau }$ of finite alphabet rank. This would imply, by Corollary 4.7, that the topological rank of X is finite. Let $\boldsymbol {\sigma } = (\sigma _n\colon {\mathcal A}_{n+1}^+\to {\mathcal A}_n^+)_{n\in {\mathbb N}}$ be an everywhere growing directive sequence of finite alphabet rank generating X. We contract $\boldsymbol {\tau }$ in a way such that $\#{\mathcal A}_n \leq K$ for every $n\geq 1$ .

We are going to inductively define subshifts $X_n$ , $n\in {\mathbb N}$ . We start with $X_0 := X$ . We now assume that $X_n$ is defined for some $n \in {\mathbb N}$ . Then the set $X^{\prime }_{n+1} = \{x \in X_{\boldsymbol {\sigma }}^{(n+1}: \sigma _n(x) \in X_n\}$ is a subshift. We define $X_{n+1}$ as any minimal subshift contained in $X^{\prime }_{n+1}$ . It follows from the definition of $X_{n+1}$ that $\bigcup _{k\in {\mathbb Z}}T^k\sigma _n(X_{n+1}) \subseteq X_n$ . Being $X_n$ minimal,

(24) $$ \begin{align} \bigcup_{k\in{\mathbb Z}}T^k\sigma_n(X_{n+1}) = X_n. \end{align} $$

Let $\tilde {{\mathcal A}}_n = {\mathcal A}_n \cap {\mathcal L}(X_n)$ . Equation (24) and the fact that $\boldsymbol {\sigma }$ is everywhere growing allow us to assume without loss of generality that, after a contraction of $\boldsymbol {\sigma }$ , the following holds for every $n\in {\mathbb N}$ :

(25) $$ \begin{align} \text{if}\ a\in\tilde{{\mathcal A}}_{n+1}\ \text{and}\ w \in {\mathcal L}(X_n)\ \text{has length}\ 3, \text{then}\ w\ \text{occurs twice in}\ \sigma_n(a). \end{align} $$

Let us fix a word $w_n = a_nb_nc_b \in {\mathcal L}(X_n)$ of length $3$ . Then, by (25), we can decompose $\sigma _n(a) = u_n(a)v_n(a)$ in a way such that

(26) $$ \begin{align} u_n(a)\ \text{ends with}\ a_n, v_n(a)\ \text{starts with}\ b_nc_n, \text{and}\ |v_n(a)| \geq 2. \end{align} $$

To define $\boldsymbol {\tau }$ , we need to introduce additional notation first. Let ${\mathcal B}_n$ be the alphabet consisting of tuples ${a\brack b}$ such that $ab \in {\mathcal L}(X_n)$ . Also, if $w = w_1\ldots w_{|w|} \in {\mathcal L}(X_n)$ has length $|w|\geq 2$ , then $\chi _n(w) := {w_1\brack w_2} {w_2\brack w_3}\ldots {w_{|w|-1}\brack w_{|w|}} \in {\mathcal B}_n^+$ , and if $w' = {w_1\brack w_2}\ldots {w_{|w|-1}\brack w_{|w|}} \in {\mathcal B}_0^+$ , then $\eta (w') := w_1\ldots w_{|w|-1} \in {\mathcal A}_0^+$ . Observe that $\eta \colon {\mathcal B}_0^+\to {\mathcal A}_0^+$ is a morphism.

We now define $\boldsymbol {\tau }$ . Let $\tau _n\colon {\mathcal B}_{n+1}^+\to {\mathcal B}_n^+$ be the unique morphism such that $\tau _n({a\brack b}) = \chi _n(v_n(a)u_n(a)b_n)$ for every ${a\brack b} \in {\mathcal B}_{n+1}$ . Observe that since $v_n(a)u_n(a)b_n \in {\mathcal L}(X_n)$ , it is indeed the case that $\tau _n({a\brack b})\in {\mathcal B}_n^+$ . We set $\boldsymbol {\tau } = (\eta \tau _0,\tau _1,\tau _2,\ldots )$ .

It follows from (26) that for every $n \in {\mathbb N}$ and ${a\brack b} \in {\mathcal B}_{n+1}$ , $\tau _n({a\brack b})$ starts with ${b_n\brack c_n}$ and ends with ${a_n \brack b_n}$ . Thus, $\boldsymbol {\tau }$ is proper. Moreover, since $|v_n(a)| \geq 2$ , we have $|v_n(a)u_n(a)b_n| \geq 3$ and thus $|\tau _n({a\brack b})| \geq 2$ . Therefore, $\langle \tau _n\rangle \geq 2$ and $\boldsymbol {\tau }$ is everywhere growing. Also, $\#{\mathcal B}_n \leq \#{\mathcal A}_n^2 \leq K^2$ for every $n \in {\mathbb N}$ , so the alphabet rank of $\boldsymbol {\tau }$ is finite.

It remains to prove that $X = X_{\boldsymbol {\tau }}$ . By minimality, it is enough to prove that $X \supseteq X_{\boldsymbol {\tau }}$ . Observe that since $\tau _n\chi _{n+1}(ab) = \chi _n(v_n(a)u_n(b)b_n)$ , the word $\tau _n\chi _{n+1}(ab)$ occurs in $\chi _n\sigma _n(ab)$ . Moreover, for every $w = w_1\ldots w_{|w|} \in {\mathcal L}(X_{\boldsymbol {\sigma }}^{(n)})$ , $\tau _n\chi _{n+1}(w)$ occurs in $\chi _n\sigma _n(w)$ . Then, by using the symbol $\sqsubseteq $ to denote the ‘subword’ relation, we can write for every $n \in {\mathbb N}$ and $ab \in {\mathcal L}(X_{\boldsymbol {\sigma }}^{(n)})$ :

$$ \begin{align*} \tau_{[0,n)}\chi_n(ab) &\sqsubseteq \tau_{[0,n-1)}\chi_{n-1}\sigma_{n-1}(ab) \\[3pt] &\sqsubseteq \tau_{[0,n-2)}\chi_{n-2}\sigma_{[n-2,n)}(ab) \sqsubseteq \cdots \sqsubseteq \chi_0\sigma_{[0,n)}(ab). \end{align*} $$

Hence, $\eta \tau _{[0,n)}({a\brack b}) \sqsubseteq \eta \chi _0\sigma _{[0,n)}(ab) \sqsubseteq \sigma _{[0,n)}(ab)$ . We conclude that $X_{\boldsymbol {\tau }} \subseteq X_{\boldsymbol {\sigma }} = X$ .

Proof. By Theorem 1.1, $(X,T)$ is generated by a proper and primitive directive sequence $\boldsymbol {\sigma }$ of alphabet rank equal to K. In particular, $\boldsymbol {\sigma }$ is everywhere growing and proper, so we can use Theorem 1.3 to obtain an everywhere growing, proper, and recognizable directive sequence $\boldsymbol {\tau } = (\tau _n\colon {\mathcal B}_{n+1}^+\to {\mathcal B}_n^+)_{n\geq 0}$ generating $(Y,T)$ and having alphabet rank at most K. Then, the hypothesis of Corollary 4.7 holds for $(Y,T)$ , and thus the topological rank of $(Y,T)$ is at most K.

Proof. We start by using Theorem 1.3 with the identity function $\mathrm {id}\colon (X,T)\to (X,T)$ to obtain a letter-onto, everywhere growing, proper, and recognizable directive sequence $\boldsymbol {\sigma _L}$ of alphabet rank at most K generating X. By doing a contraction, we can assume that $\boldsymbol {\sigma _L}$ has exact alphabet rank at most K.

By Theorem 1.3 applied to $\pi _{L-1}$ and $\boldsymbol {\sigma _L}$ , there exists, after a contraction of $\boldsymbol {\sigma _L}$ , a letter-onto factor $\boldsymbol {\phi _{L-1}}\colon \boldsymbol {\sigma _L}\to \boldsymbol {\sigma _{L-1}}$ , where $\boldsymbol {\sigma _{L-1}}$ is letter-onto, everywhere growing, proper, recognizable, has alphabet rank at most K, generates $X_{L-1}$ , and, if $\phi _{L-1,0}$ and $\sigma _{L,0}$ are the first coordinates of $\boldsymbol {\phi _{L-1}}$ and $\boldsymbol {\sigma _{L}}$ , respectively, then $\pi _{L-1}(\sigma _{L,0}(x)) = \phi _{L-1,0}(x)$ for every $x \in X_{\boldsymbol {\sigma _L}}^{(1)}$ and $|\sigma _{L,0}(a)| = |\phi _{L-1,0}(a)|$ for every letter a in the domain of $\sigma _{L,0}$ . By contracting these sequences, we can also suppose that $\boldsymbol {\sigma _{L-1}}$ has exact alphabet rank at most K. The same procedure applies to $\pi _{L-2}$ and $\boldsymbol {\sigma _{L-1}}$ . Thus, by continuing in this way, we obtain for every $j=0,\ldots ,L-1$ a letter-onto factor $\boldsymbol {\phi _j}\colon \boldsymbol {\sigma _{j+1}}\to \boldsymbol {\sigma _j}$ such that the following holds.

• • $\boldsymbol {\sigma _j}$ is letter-onto, everywhere growing, proper, recognizable, has exact alphabet rank at most K, generates $X_j$ , $\pi _{j}(\sigma _{j+1,0}(x)) = \phi _{j,0}(x)$ for every $x \in X_{\boldsymbol {\sigma _{j+1}}}^{(1)}$ , and $|\sigma _{j+1,0}(a)| = |\phi _{j,0}(a)|$ for every $a \in {\mathcal A}_{j+1,1}$ .

Here, we are using the notation $\boldsymbol {\sigma _j} = (\sigma _{j,n}\colon {\mathcal A}_{j,n+1}^+\to {\mathcal A}_{j,n}^+)_{n\in {\mathbb N}}$ , $\boldsymbol {\phi _j} = (\phi _{j,n}\colon {\mathcal A}_{j+1,n}^+\to {\mathcal A}_{j,n}^+)_{n\in {\mathbb N}}$ and $X_j^{(n)} = X_{\boldsymbol {\sigma _j}}^{(n)}$ . We note that:

1. (△₁ ) for every $x \in X_{j+1}^{(1)}$ , $\pi _j(\sigma _{j+1,0}(x)) = \phi _{j,0}(x) = \sigma _{j,0}\phi _{j,1}(x)$ since $\phi _{j,0} = \sigma _{j,0}\phi _{j,1}$ ;

2. (△₂ ) $X_j^{(1)} = \bigcup _{k\in {\mathbb Z}}T^k\phi _{j,1}(X_{j+1}^{(1)})$ by Lemma 4.3.

Hence, the following diagram commutes:

Proof. Let us assume that $(X_{j+1}^{(1)}, \phi _{j,1})$ is recognizable, and let, for $i=0,1$ , $x^i \in X_{j+1}^{(1)}$ such that $y = \pi _j(x^0) = \pi _j(x^1)$ . We have to show that $x^0 = x^1$ . First, we use Lemma 2.5 to find a centered $\sigma _{j+1,0}$ -factorization $(k^i,z^i)$ of $x^i$ in $X_{j+1}^{(1)}$ . Then, equation $\triangle _1$ allows us to compute:

$$ \begin{align*} T^{k^0}\sigma_{j,0}\phi_{j,1}(z^0) = T^{k^0}\pi_j(\sigma_{j+1,0}(z^0)) = \pi_j(x^0) = \pi_j(x^1) = T^{k^1}\sigma_{j,0}\phi_{j,1}(z^1). \end{align*} $$

This implies that $(k^i,z^i)$ is a $\sigma _{j,0}\phi _{j,1}$ -factorization of y in $X_{j+1}^{(1)}$ for $i=0,1$ . Moreover, these are centered factorizations as, by $\bullet $ , $|\sigma _{j,0}\phi _{j,1}(a)| = |\sigma _{j+1,0}(a)|$ for all $a \in {\mathcal A}_{j+1,1}$ . Now, being $(X_j^{(1)},\sigma _{0,j})$ and $(X_{j+1}^{(1)}, \phi _{j,1})$ recognizable, Lemma 2.3 gives that $(X_{j+1}^{(1)},\sigma _{j,1}\phi _{j,1})$ is recognizable, and thus we have that $(k^0,z^0) = (k^1,z^1)$ . Therefore, $x^0 = x^1$ and $\pi $ is a conjugacy.

Now we can finish the proof. We assume, by contradiction, that $\pi _j$ is not a conjugacy for all j. Then, by the claim,

(27) $$ \begin{align} (X_j^{(1)},\phi_{1,j})\ \text{is not recognizable for every}\ j \in \{0,\ldots,L-1\}. \end{align} $$

Let

$$ \begin{align*} \boldsymbol{\nu} = (\phi_{0,1}, \phi_{1,1}, \phi_{2,1}, \ldots,\phi_{L-1,1},\sigma_{L,1}, \sigma_{L,2}, \sigma_{L,3}, \ldots). \end{align*} $$

The idea is to use Theorem 2.7 with $\boldsymbol {\nu }$ to obtain a contradiction. To do so, we first note that, since $\boldsymbol {\nu }$ and $\boldsymbol {\sigma }^{(L)}$ have the same ‘tail’, $X_{\boldsymbol {\nu }}^{(m+L)} = X_{L}^{(m+1)}$ for all $m \in {\mathbb N}$ . Moreover, $\triangle _2$ and the previous relation imply that

$$ \begin{align*} X_{\boldsymbol{\nu}}^{(j)} &= \bigcup_{k\in{\mathbb Z}} T^k\phi_{j,1}(X_{\boldsymbol{\nu}}^{(j+1)}) = \cdots = \bigcup_{k\in{\mathbb Z}} T^k\phi_{j,1}\ldots\phi_{L-1,1}(X_{\boldsymbol{\nu}}^{(L)}) \\[3pt] &= \bigcup_{k\in{\mathbb Z}} T^k\phi_{j,1}\ldots\phi_{L-1,1}(X_{L}^{(1)}) = \bigcup_{k\in{\mathbb Z}} T^k\phi_{j,1}\ldots\phi_{L-2,1}(X_{L-1}^{(1)}) = \cdots = X_j^{(1)}. \end{align*} $$

This and (27) imply that for every $j \in \{1,\ldots ,L-1\}$ , the level $(X_{\boldsymbol {\nu }}^{(j)}, \phi _{j,1})$ of $\boldsymbol {\nu }$ is not recognizable. Being $\boldsymbol {\nu }$ everywhere growing as $\boldsymbol {\sigma _L}$ has this property, we conclude that Theorem 2.7 can be applied and, therefore, that $X_0^{(1)} = X_{\boldsymbol {\nu }}$ is periodic. However, then $X_0 = \bigcup _{k\in {\mathbb Z}}T^k\sigma _{0,0}(X_0^{(1)})$ is periodic, contrary to our assumptions.

Proof. We sketch the proof from [Reference Durand, Host and SkauDHS99] that we mentioned above.

Let $({\mathcal P}_n)_{n\in {\mathbb N}}$ be a sequence of clopen partitions of Y such that ${\mathcal P}_{n+1}$ is finer than ${\mathcal P}_n$ and their union generates the topology of Y. Also, let $Y_n$ be the subshift obtained by codifying the orbits of $(Y,T)$ by using the atoms of ${\mathcal P}_n$ . Then, the fact that ${\mathcal P}_n$ is a clopen partition induces a factor map $\pi _n\colon (Y,T) \to (Y_n,T)$ . Moreover, since ${\mathcal P}_{n+1}$ is finer than ${\mathcal P}_n$ , there exists a factor map $\xi _n\colon (Y_{n+1},T)\to (Y_n,T)$ such that $\xi _n\pi _{n+1} = \pi _n$ . Hence, we have the following chain of factors:

$$ \begin{align*} (X,T) \overset{\pi}{\longrightarrow} (Y,T) \overset{\pi_n}{\longrightarrow} (Y_n,T) \overset{\xi_{n-1}}{\longrightarrow} (Y_{n-1},T) \overset{\xi_{n-2}}{\longrightarrow} \cdots \overset{\xi_{1}}{\longrightarrow} (Y_0,T). \end{align*} $$

We conclude, by also using the fact that the partitions ${\mathcal P}_n$ generate the topology of Y, that $(Y,T)$ is conjugate to the inverse limit $\overleftarrow {\lim }_{n\to \infty }(Y_n;\xi _n)$ .

Now we consider two cases. If $Y_n$ is periodic for every $n \in {\mathbb N}$ , then Y is the inverse limit of periodic system, and hence an odometer. In the other case, we have, by Corollary 1.5, that $\xi _n$ is a conjugacy for all big enough $n \in {\mathbb N}$ , and thus that $(Y,T)$ is conjugate to one of the subshifts $Y_n$ .

Proof. We start by introducing the necessary objects for the proof and doing some general observations about them. First, thanks to Theorem 1.6, we know that $\pi _j$ is almost $k_j$ -to-1 for some $k_j \leq K$ , so, by the pigeonhole principle, there exist $J_1 \subseteq J$ and $k \leq K$ such that $\#J_1 \geq \#J/K> (144K^7)^K$ and $k_j = k$ for every $j \in J_1$ . For $j \in J_1$ , we fix $z^j = (z_1^j, \ldots , z_k^j) \in R_{\pi _j}^k$ with $z_n^j \not = z_m^j$ for all $n\not =m$ . Let $Z_j = \overline {\mathrm {orb}}_{T^{[k]}}(z^j)$ and $\rho \colon X^k\to X$ be the factor map that projects onto the first coordinate. By Lemma 6.1, $\pi _j$ is k-to-1 and $Z_j$ minimal. This implies that if $x = (x_1,\ldots ,x_k) \in Z_j$ , then

(30) $$ \begin{align} \{x_1,\ldots,x_k\} &= \pi_j^{-1}(\pi_j(x_n)) \quad\text{for all}\ n\in \{1,\ldots,k\}, \end{align} $$

(31) $$ \begin{align} x_n &\not= x_m \quad\text{for all}\ n,m \in \{1,\ldots,k\}. \end{align} $$

Indeed, since $Z_j$ is minimal, $(T^{[k]})^{n_\ell }z \to x$ for some sequence $(n_\ell )_\ell $ , so,

$$\begin{align*}\inf_{n\not=m}\mathrm{dist}(x_n,x_m) \geq \inf_{n\not=m, l\in{\mathbb Z}}\mathrm{dist}(T^lz_n,T^lz_m)> 0,\end{align*}$$

where in the last step is due to the fact that $\pi _j$ is distal. This gives (31). For (30), we first note that $\{x_1,\ldots ,x_k\} \subseteq \pi _j^{-1}(\pi _j(x_n))$ as $x \in R_{\pi _j}$ , and then that the equality must hold since $\#\pi _j^{-1}(\pi _j(x_n)) = k = \#\{x_1,\ldots ,x_k\}$ by (31).

The next step is to prove that asymptotic pairs in $Z_j$ are well-behaved.

Claim 6.3.1. Let $j \in J_1$ and $(x^j = (x_1^j,\ldots ,x_k^j)$ , $\tilde {x}^j = (\tilde {x}_1^j,\ldots ,\tilde {x}_k^j))$ be a right asymptotic pair in $Z_j$ , this is,

(32) $$ \begin{align} \lim_{n\to-\infty}\mathrm{dist}((T^{[k]})^nx^j, T^{[k]}\tilde{x}^j) = 0\quad\text{and}\quad x^j \not= \tilde{x}^j. \end{align} $$

Then, $(x_n^j,\tilde {x}_n^j)$ is right asymptotic for every $n \in \{1,\ldots ,k\}$ .

Proof. Suppose, with the aim to obtain a contradiction, that $(x_n^j, \tilde {x}_n^j)$ is not right asymptotic for some $n \in \{1,\ldots ,k\}$ . Observe that (32) implies that

(33) $$ \begin{align} \text{for every}\ m \in \{1,\ldots,k\}, \text{either}\ (x_m^j,\tilde{x}_m^j)\ \text{is right asymptotic or}\ x_n^j = \tilde{x}_n^j. \end{align} $$

Therefore, $x_n^j = \tilde {x}_n^j$ . Using this and that $x^j,\tilde {x}^j \in R_{\pi _j}^k$ , we can compute:

$$ \begin{align*} \pi_j(x_m^j) = \pi_j(x_n^j) = \pi_j(\tilde{x}_n^j) = \pi_j(\tilde{x}_l^j) \quad\text{for all } m,l\in\{1,\ldots,k\}, \end{align*} $$

and thus, by (30),

$$ \begin{align*} \{x_1^j,\ldots,x_k^j\} = \pi_j^{-1}(\pi_j(x_n^j)) = \pi_j^{-1}(\pi_j(\tilde{x}_n^j)) = \{\tilde{x}_1^j,\ldots,\tilde{x}_k^j\}. \end{align*} $$

The last equation, (31), and that $x^j\not =\tilde {x}^j$ imply that there exist $m\not =l \in \{1,\ldots ,k\}$ such that $\tilde {x}_{l}^j = x_m^j$ . This last equality and (33) tell us that $x^j_m$ and $x^j_l$ are either asymptotic or equal. However, in both cases a contradiction occurs: in the first one with the distality of $\pi $ and in the second one with (31).

Let $j \in J_1$ . Since $Y_j$ is infinite, $Z_j$ is a infinite subshift. It is a well-known fact from symbolic dynamics that this implies that there exists a right asymptotic pair $(x^j = (x_1^j,\ldots ,x_k^j)$ , $\tilde {x}^j = (\tilde {x}_1^j,\ldots ,\tilde {x}_k^j))$ in $Z_j$ . We are now going to use Theorem 2.8 to prove the following.

Proof. On one hand, by the previous claim, $(x_n^j, \tilde {x}_n^j) \in X^2$ is right asymptotic for every $n \in \{1,\ldots , k\}$ and $j \in J_1$ . Let $p_n^j \in {\mathbb Z}$ be such that $(T^{p_n^j}x_n^j, T^{p_n^j}\tilde {x}_n^j)$ is centered right asymptotic. On the other hand, Theorem 2.8 asserts that the set

$$ \begin{align*} \{ x_{(0,\infty)} : (x,\tilde{x}) \text{ is centered right asymptotic in }X\} \end{align*} $$

has at most $144K^7$ elements. Since $\#J_1> (144K^7)^K$ , we conclude, by the pigeonhole principle, that there exist $i,j \in J_1$ , $i\not =j$ , such that

(34) $$ \begin{align} T^{p_n^i}x_n^i \text{ and } T^{p_n^j}x_n^j \text{ agree on }(0,\infty) \text{ for every }n \in \{1,\ldots,k\}. \end{align} $$

We are going to show that $Z_i = Z_j$ .

Using (34), we can find $u \in E(X,T)$ such that $uT^{p_n^i}x_n^i = uT^{p_n^j}x_n^j$ for every n. Then, by putting $y_n^i = ux_n^i$ , $y_n^j = ux_n^j$ and $q_n = p_n^j-p_n^i$ , we have

$$ \begin{align*} y^i := (y_1^i,\ldots,y_k^i) \in Z_i, y^j := (y_1^j,\ldots,y_k^j) \in Z_j\text{ and } y_n^i = T^{q_n}y_n^j. \end{align*} $$

Hence, $\pi (y_n^i) = T^{q_n}\pi (y_n^j)$ and Lemma 6.2 can be applied to deduce that $q := q_n$ has the same value for every n. We conclude that $y^i = T^qy^j \in T^qZ_j = Z_j$ , that $Z_i \cap Z_j$ is not empty, and, therefore, that $Z_i = Z_j$ as these are minimal systems.