Proof. Let $a\in \text {Fix}(\hat \varphi )$ and define $r,\,s,\,t\in \{0,\,\ldots,\, m\}$, and $i_l,\,j_l,\,k_l$ as above. We will prove that, for $b\in \text {Fix}(\hat \varphi )$, we have that $b\in (\text {Fix} \varphi )a$ if and only if $b_{i_l}=a_{i_l}$ for every $l\in [r]$, $b_{j_l}=a_{j_l}$, for every $l\in [s]$ and $b_{k_l}\not \in \{+\infty,\,-\infty \}$, for every $l\in [t]$, i.e., if their infinite entries coincide. If that is the case, then every orbit is defined by the position and the signal of their infinite entries. Obviously, for $i\in \{0,\,\ldots,\,m\}$, there are $m\choose i$ choices for $i$ infinite entries and each of them can be $+\infty$ or $-\infty$, hence the $2^{i}$ factor.

Start by supposing that $b\in \text {Fix}(\hat \varphi )$ is such that $b\in (\text {Fix}\varphi )a$. Then, there is some $c\in \text {Fix}\varphi$ such that $b=c+a$. This means that for every $l\in [r]$, we have that $b_{i_l}=c_{i_l}+a_{i_l}=c_{i_l}+ (+\infty )=+\infty$, since $c\in \mathbb {Z}^{m}$. Similarly, we have that $b_{j_l}=a_{j_l}$, for every $j\in [s]$ and $b_{k_l}\not \in \{+\infty,\,-\infty \}.$ It is clear that $|b_{k_l}|<\infty$ for $l\in [t].$

Now, suppose $b\in \text {Fix}(\hat \varphi )$ is such that $b_{i_l}=+\infty$, for $l\in [r]$, $b_{j_l}=-\infty$, for $l\in [s]$ and $b_{k_l}\not \in \{+\infty,\,-\infty \}$, for $l\in [t]$. Consider $c\in \mathbb {Z}^{m}$ defined by $c_{k_l}=b_{k_l}-a_{k_l}$ and all other entries are $0$. Clearly $b=c+a$. We only have to check that $c\in \text {Fix}(\varphi ),$ i.e., $\lambda _ic_{\alpha _i}=c_i$ for every $i\in [m]$. For $i$ such that $a_i=\pm \infty$, we have that $c_i=0$ and $a_{\alpha _i}=sgn(\lambda _i)a_i=\pm \infty$, which implies that $c_{\alpha _i}=0$. If not, $c_{i}=b_i-a_i=\lambda _i(b_{\alpha _i}-a_{\alpha _i})=\lambda _ic_{\alpha _i}$ and we are done.

Proof. We start by showing that $\text {Sing}(\hat \varphi )\subseteq \{(a,\,a\hat {\varphi _2})\mid a\in \text {Sing}(\hat {\varphi _1})\}$. Take some $(a,\,a\hat \varphi _2)\in (\text {Fix}(\varphi ))^{c}$ with $a\in \text {Fix}(\hat \varphi _1)$. Then, for every $\varepsilon >0$, the open ball of radius $\varepsilon$ centred in $(a,\,a\hat \varphi _2)$ contains an element $(b_\varepsilon,\,b_\varepsilon \varphi _2)\in \text {Fix}(\varphi )$, with $b_\varepsilon \in \text {Fix}(\varphi _1).$ Notice that $d(a,\,b_\varepsilon )\leq d((a,\,a\hat \varphi _2),\,(b_\varepsilon,\,b_\varepsilon \varphi _2))<\varepsilon$, thus, $a\in (\text {Fix}(\varphi _1))^{c}.$

For the reverse inclusion, take some $a\in \text {Sing}(\hat \varphi _1)$. As above, we know that for every $\varepsilon >0$, there is some $b_\varepsilon \in B(a;\varepsilon )\cap \text {Fix}(\varphi _1)$. Notice that, since $\hat \varphi _2$ is uniformly continuous, for every $\varepsilon >0$, there is some $\delta _\varepsilon$ such that, for all $a,\,b\in \widehat {\mathbb {Z}^{m}}$ such that $d(a,\,b)<\delta _\varepsilon$, we have that $d(a\hat \varphi _2,\,b\hat \varphi _2)<\varepsilon$. We want to prove that $(a,\,a\hat \varphi _2)\in (\text {Fix}(\varphi ))^{c}$, by showing that, for every $\varepsilon >0$, the ball centred in $(a,\,a\hat \varphi _2)$ contains a fixed point of $\varphi$. So, let $\varepsilon >0$ and consider $\delta =\min \{\delta _{\varepsilon },\,\varepsilon \}$. We have that $(b_\delta,\,b_\delta \hat \varphi _2)\in B((a,\,a\hat \varphi _2);\varepsilon )$ since, by definition of $b_\delta$, we have that $d(a,\,b_\delta )<\delta \leq \varepsilon$ and also, $d(a,\,b_\delta )<\delta _{\varepsilon }$ means that $d(a\hat \varphi _2,\,b_\delta \hat \varphi _2)<\varepsilon$.

Proof. For $(a,\,a\hat \varphi _2),\,(b,\,b\hat \varphi _2)\in \text {Fix}(\hat \varphi )$, we have that $(a,\,a\hat \varphi _2)$ belongs to $(\text {Fix}\varphi )(b,\,b\hat \varphi _2)$ if and only if there is some $(c,\,c\varphi _2) \in \text {Fix}\varphi$ such that $(a,\,a\hat \varphi _2)=(c,\,c\varphi _2)(b,\,b\hat \varphi _2)$, i.e., $a$ and $b$ belong to the same orbit of $\text {Fix}(\varphi _1)$.

Proof. Let $a\in \text {Per} (\hat \varphi )$. As done in the fixed point case, we will prove that, for $b\in \text {Per}(\hat \varphi )$, we have that $b\in (\text {Per}\varphi )a$ if and only if their infinite entries coincide. and that suffices.

Clearly, if $b\in \text {Per}(\hat \varphi )$ is such that $b=c+a$ for some $c\in \text {Per}(\varphi )$, then the infinite entries of $a$ and $b$ coincide.

Now, suppose $a$ and $b$ are two infinite periodic points whose infinite entries coincide. Then, there are $p,\,q\in \mathbb {N}$ such that $a\hat \varphi ^{p}=a$ and $b\hat \varphi ^{q}=b$, so $a\hat \varphi ^{pq}=a$ and $b\hat \varphi ^{pq}=b$. Consider $c\in \mathbb {Z}^{m}$ defined by $c_i=0$ if $a_i,\,b_i\in \{+ \infty,\,-\infty \}$ and $c_i=b_i-a_i$ otherwise. Clearly, $b=c+a$. We only have to check that $c\in \text {Per}(\varphi )$ and for that, we will show that $c\varphi ^{pq}=c.$ We have that, for $r>0$, the mapping $\varphi ^{r}$ is defined by

\[ [c_i]_{i\in [m]}\mapsto \left[\pi_i^{(r)}c_{i\psi^{r}}\right]_{i\in [m]}. \]

Now, we only have to see that, for every $i\in [m]$, we have that $c_i=\pi _i^{(pq)}c_{i\psi ^{pq}}$. Let $i\in [m]$ such that $a_i,\,b_i\in \{+\infty,\,-\infty \}$ and so $c_i=0$. Since $|a_i|=\infty$, $a_i=(a\hat \varphi ^{pq})_i=\pi _i^{(pq)}a_{i\psi ^{pq}}$ and all $\lambda _k$'s are finite, then we have that $a_{i\psi ^{pq}} \in \{+\infty,\,-\infty \}$ (and the same holds for $b_{i\psi ^{pq}}$), thus, $c_{i\psi ^{pq}}=0$. Then, we have that $\pi _i^{(pq)}c_{i\psi ^{pq}}=0=c_i$. Now, take $i\in [m]$ such that $a_i,\,b_i\not \in \{+\infty,\,-\infty \}$. Then $c_i=b_i-a_i=(b\hat \varphi ^{pq})_i-(a\hat \varphi ^{pq})_i=\pi _i^{(pq)}b_{i\psi ^{pq}}- \pi _i^{(pq)}a_{i\psi ^{pq}}=\pi _i^{(pq)}(b_{i\psi ^{pq}}-a_{i\psi ^{pq}})=\pi _i^{(pq)}c_{i\psi ^{pq}}$, since $a_{i\psi ^{pq}}$ and $b_{i\psi ^{pq}}$ are both finite.

Proof. It is easy to see, by induction on $r$ that, for every $r>0$, we have that $(a,\,u)\hat \varphi ^{r}=(a\hat \varphi _1^{r},\,a\hat \varphi _1^{r-1}\hat \varphi _2)$. Indeed, it is true for $r=1$ and if we have that $(a,\,u)\hat \varphi ^{r}=(a\hat \varphi _1^{r},\,a\hat \varphi _1^{r-1}\hat \varphi _2)$, then $(a,\,u)\hat \varphi ^{r+1}=(a\hat \varphi _1^{r},\,a\hat \varphi _1^{r-1}\hat \varphi _2)\hat \varphi =(a\hat \varphi _1^{r+1},\,a\hat \varphi _1^{r}\hat \varphi _2).$ So we have that

(4.2)\begin{equation} (a,u)\hat\varphi^{r}=([\pi_i^{(r)}a_{i\psi^{r}}]_{i\in[m]},z^{\lambda\pi_k^{(r-1)}a_{k\psi^{r-1}}}). \end{equation}

Let $(a,\,u),\,(b,\,v)\in \text {Per}(\hat \varphi )$. We have that $(b,\,v)\in (\text {Per} \varphi )(a,\,u)$ if and only if $b\in (\text {Per}(\varphi _1))a$. Indeed, if $b\in (Per(\varphi _1))a$, let $c\in \text {Per}(\varphi _1)$, defined as in the proof of Proposition 4.4, be  such that $b=c+a$ and denote by $p,\,q$ the periods of $a$ and $b$, respectively.

Then $(c,\,c\hat \varphi _1^{pq-1}\hat \varphi _2)\varphi ^{pq}=(c,\,c\hat \varphi _1^{pq-1}\hat \varphi _2)$, so $(c,\,c\hat \varphi _1^{pq-1}\hat \varphi _2)\in \text {Per}(\varphi )$ and

\begin{align*} (b,v)& =(b\hat\varphi_1^{pq},b\hat\varphi_1^{pq-1}\hat\varphi_2)=((c+a)\hat\varphi_1^{pq},(c+a)\hat\varphi_1^{pq-1}\hat\varphi_2)\\ & =(c\hat\varphi_1^{pq}+a\hat\varphi_1^{pq},c\hat\varphi_1^{pq-1}\hat\varphi_2a\hat\varphi_1^{pq-1}\hat\varphi_2)\\ & =(c\hat\varphi_1^{pq},c\hat\varphi_1^{pq-1}\hat\varphi_2)(a\hat\varphi_1^{pq},a\hat\varphi_1^{pq-1}\hat\varphi_2)\\ & =(c,c\hat\varphi_1^{pq-1}\hat\varphi_2)(a,u). \end{align*}

Hence, we have that $\text {Per} (\hat \varphi )$ has $\sum \nolimits \limits _{i=0}^{m} 2^{i} {m\choose i}$ $\text {Per}(\varphi )$-orbits.

Proof. By definition, $\text {Sing}(\hat \varphi )\subseteq \text {Fix}(\hat \varphi ).$

Let $\pi \in S_m$ be a permutation such that $\varphi$ maps $[a_i]_{i\in [m]}$ to $[\lambda _i a_{\pi (i)}]_{i\in [m]}$, and $\lambda _i=\pm 1$. Then $\text {Fix}(\varphi )=\{a\in \mathbb {Z}^{m} \mid \forall i\in [m],\, a_i=\lambda _i a_{\pi (i)}\}$ and $\text {Fix}(\hat \varphi )=\{a\in \widehat {\mathbb {Z}^{m}} \mid \forall i\in [m],\, a_i=\lambda _i a_{\pi (i)}\}$. Given $a\in \text {Fix}(\hat \varphi )$ and $\varepsilon >0$, choosing some

\[ n>\max\limits_{a_i\in\mathbb{Z}}\left\{\left|a_i\right|,\left\lceil \log_2\bigg(\frac 1 \varepsilon\bigg)\right\rceil\right\}, \]

consider $b\in \mathbb {Z}^{m}$ such that $b_i=n$ if $a_i=+\infty$; $b_i=-n$ if $a_i=-\infty$ and $b_i=a_i$, otherwise. Then $b_i\in \text {Fix}(\varphi )$ and $d(a,\,b)<\varepsilon$. Thus, $b$ is a point of closure of $\text {Fix}(\varphi )$ and we are done.

Proof. Let $a\in \text {Fix}\hat \varphi \setminus \text {Fix}\varphi$. Let $\varepsilon >0$ and define $q=\lceil \log _2(\tfrac {1}{\varepsilon })\rceil$. Take $p=\max \limits _{a_i\in \mathbb {Z}}\{q,\,|a_i|\}$. Take $b\in \mathbb {Z}^{m}$ such that $b_i=p$ for every $i$ such that $a_i=+\infty$; $b_i=-p$ for every $i$ such that $a_i=-\infty$ and $b_i=a_i$ otherwise. Then $d(a,\,b)<\varepsilon$ but $b\hat \varphi ^{n}\not \to a$ since $\max \limits _{i\in [m]} |b_i|=p$ and applying $\hat \varphi$ simply changes order and signal of the entries, so for every $n\in \mathbb {N}$, we have that $\max \limits _{i\in [m]} \{|({b\hat \varphi ^{n}})_i|\}=n$. Hence $d(a,\,b\hat \varphi ^{n})\geq 2^{-p}$ since there is some $k$ such that $a_k\in \{+\infty,\,-\infty \}$ and $d(a_k,\,b_k)\geq d(a_k,\, sgn(a_k)|p|)=2^{-p}$. The repeller case is analogous, since the inverse of a uniform matrix is uniform.

Proof. Let $\alpha =(a,\,u)$ be an infinite fixed point, where $a\in \text {Fix}(\hat \varphi _1)$ and $u\in \text {Fix}(\hat \phi )$. Clearly, if $a\in \text {Fix}(\hat \varphi _1)$ and $u\in \text {Fix}(\hat \phi )$ are attractors, then, $(a,\,u)\in \text {Fix}(\hat \varphi )$ is an attractor. Indeed, in that case, there are $\varepsilon _1>0$ and $\varepsilon _2>0$ such that

\[ \forall b\in\widehat{\mathbb{Z}^{m}},\, \left(d(a,b)<\varepsilon_1\Rightarrow \lim\limits_{n\to +\infty} b\hat\varphi_1^{n}=a\right) \]

and

\[ \forall v\in\widehat{ F_n},\, \left(d(u,v)<\varepsilon_2\Rightarrow \lim\limits_{n\to +\infty} v\hat\phi^{n}=u\right). \]

Thus, taking $\varepsilon =\min \{\varepsilon _1,\,\varepsilon _2\}$, we have that

\begin{align*} \forall (b,v)\in \widehat{\mathbb{Z}^{m}\times F_n} (d((a,u),(b,v))<\varepsilon & \Rightarrow d(a,b)<\varepsilon \wedge d(u,v)<\varepsilon \\ & \Rightarrow \lim\limits_{n\to+\infty} b\hat\varphi_1^{n}=a \wedge \lim\limits_{n\to+\infty} v\hat\phi^{n}=u\\ & \Rightarrow \lim\limits_{n\to +\infty}(b,v)\hat\varphi^{n}=(a,u). \end{align*}

Conversely, suppose w.l.o.g that $a$ is not an attractor for $\hat \varphi _1$. Then, for every $\varepsilon >0$, there is some $b_\varepsilon \in \widehat {\mathbb {Z}^{m}}$ such that $d(a,\,b)<\varepsilon$ but $b_\varepsilon \hat \varphi _1^n\not \to a$. In this case, we have that, for every $\varepsilon >0$, $d((a,\,u),\,(b_\varepsilon,\,u))<\varepsilon$ and $(b_\varepsilon,\, u)\hat \varphi ^{n}=(b_\varepsilon \hat \varphi _1^{n},\,u)\not \to (a,\,u)$.

Proof. There are only $2^{m}m!$ distinct uniform $m\times m$ matrices, so there are $0< p< q\leq 2^{m}m!+1$ such that $Q^{p}=Q^{q}$, thus $I_m=Q^{q-p}$.

Proof.

1. (1) This is obvious. Observe that if $Q$ is uniform, which is the case when we deal with automorphisms then the reverse implication holds as well, by Proposition 5.1.

2. (2) Suppose $u$ is $\hat \phi$-wandering. Then, take $\varepsilon >0$ and $N\in \mathbb {N}$ such that for all $n>N$, we have that $B(u;\varepsilon )\hat \phi ^{n}\cap B(u;\varepsilon )=\emptyset$. Let $n>N$ and take $V=B((a,\,u);\varepsilon )$ and $(b,\,v)\in V$. Then $v\in B(u;\varepsilon )$, thus $v\hat \phi ^{n}\not \in B(u;\varepsilon )$ and $(b,\,v)\hat \varphi ^{n}=(b\hat \varphi _1^{n},\,v\hat \phi ^{n})\not \in V$. Since $(b,\,v)$ is an arbitrary point of $V$, we have that $(a,\,u)$ is $\hat \phi$-wandering.

3. (3) Suppose $(a,\,u)$ is $\hat \varphi$-recurrent. Take $\varepsilon >0$. There is some $n\in \mathbb {N}$ such that $(a,\,u)\hat \varphi ^{n}\in B((a,\,u);\varepsilon )$ and so $u\hat \phi ^{n}\in B(u;\varepsilon ).$

Proof. We prove this by induction on $n$. By hypothesis, we have that $\hat d(x\widehat \phi,\,y\widehat \phi )\leq Kd(x,\,y)^{r}$, for all $x,\,y\in \hat F_n$. Now suppose that the result holds for all $n\leq k$, for some $k\in \mathbb {N}$. Let $x,\,y\in \hat F_n$. We have that

\begin{align*} \hat d(x\widehat\phi^{k+1},y\widehat\phi^{k+1}) & \leq K\hat d(x\widehat\phi^{k},y\widehat\phi^{k})^{r}\\ & \leq K (K^{\frac{1-r^{k}}{1-r}}d(x,y)^{r^{k}})^{r}\\ & \leq K^{1+r\frac{1-r^{k}}{1-r}}d(x,y)^{r^{k+1}}\\ & \leq K^{\frac{1-r}{1-r}+\frac{r-r^{k+1}}{1-r}}d(x,y)^{r^{k+1}}\\ & \leq K^{\frac{1-r^{k+1}}{1-r}}d(x,y)^{r^{k+1}}\end{align*}

□

Proof. By [Reference Carvalho6, Theorem 5.6] and [Reference Araújo and Silva1, Proposition 4.2], we know that $\widehat \phi$ satisfies a Hölder condition. Let $(K,\,r)$ be the constants and $x\in \widehat F_n$ be a non-wandering point. Then

\[ \forall \varepsilon >0 \forall\,N\in \mathbb{N}\exists n>\mathbb{N}: B(x,\varepsilon)\phi^{n}\cap B(x,\varepsilon)\neq\emptyset. \]

Let $\varepsilon \leq K^{\frac {1}{1-r}}$. We then have that

\[ \forall\,N\in \mathbb{N}\;\;\exists n>\mathbb{N} \;\;\exists y_n\in \hat F_n : \left(d(y_n,x)\leq \varepsilon \;\; \wedge \;\;d(y_n\phi^{n},x)\leq \varepsilon \right). \]

Let $N\in \mathbb {N}$ and take $n$ given by the condition. By the previous lemma, we have that $d(x\phi ^{n},\,y_n\phi ^{n})\leq K^{\frac {1-r^{n}}{1-r}}d(x,\,y_n)^{r^{n}}\leq K^{\frac {1-r^{n}}{1-r}}\varepsilon ^{r^{n}}.$ Thus,

\begin{align*} d(x\phi^{n},x)& \leq d(x\phi^{n},y_n\phi^{n})+d(y_n\phi^{n},x)\\ & \leq K^{\frac{1-r^{n}}{1-r}}\varepsilon^{r^{n}}+\varepsilon\\ & \leq \varepsilon(1+K^{\frac{1-r^{n}}{1-r}}\varepsilon^{r^{n}-1})\\ & \leq \varepsilon (1+K^{\frac{1-r^{n}}{1-r}}(K^{\frac{1}{1-r}})^{r^{n}-1})\\ & =2\varepsilon. \end{align*}

So, fixing $\varepsilon \leq K^{\frac {1}{1-r}}$, we have arbitrarily large integers $n$ such that $d(x\phi ^{n},\,x)\leq 2\varepsilon$. Thus, let $\varepsilon >0$ and take $\varepsilon '=\min \{\frac {\varepsilon }{2},\, \frac {K^{\frac {1}{1-r}}}{2} \}$. Then

\[ \forall\, N\in \mathbb{N}\;\;\exists\, n>N: x\phi^{n}\in B(x,2\varepsilon')\subseteq B(x,\varepsilon), \]

and so $x$ is recurrent.

Proof. By Proposition 5.10, non-wandering points are recurrent. Since recurrent points belong to its own $\omega$-limit, by Theorem 5.8, they must be periodic.

Proof. Let $(a,\,u)\in \widehat {\mathbb {Z}^{m} \times F_n}$ be a non-wandering point. Then, by Proposition 5.6, $u$ is a non-wandering point for $\phi$, which means that $u$ is periodic, by Corollary 5.11. Since $a$ is also periodic by Proposition 5.1, then so is $(a,\,u)$.

Proof. Since $\widehat {\mathbb {Z}^{m}\times F_n}$ is compact then $\omega$-limits are non-empty. Moreover, if a point $(a,\,u)$ belongs to an $\omega$-limit set, then it must be non-wandering, which, by Corollary 5.12, means that $(a,\,u)$ is periodic.

Proof. It is obvious that (ii) $\Rightarrow$ (i), (ii) $\Rightarrow$ (iii) and (iii) $\Rightarrow$ (iv). We will prove that (i) $\Rightarrow$ (ii) and that (iv)$\Rightarrow$ (i).

(i)$\Rightarrow$ (ii): Suppose that there is some $N>m$ such that for every $p>N$ we have that $|\lambda _{i\psi ^{p}}|=1$. We have that $\psi$ maps $[m]$ to a subset of $[m]$, and so, for every $i\in [m]$, there is some $k_i\leq m$ such that $i\psi ^{m+1}=i\psi ^{k_i}.$ This way, we have a periodic orbit (can be fixed) of $\psi$ given by $\{i\psi ^{k_i},\,\ldots i\psi ^{m}\}$. So, for every $p\geq N$, we define $j_p\in \{k_i,\,\ldots,\, m\}$ to be such that $i\psi ^{j_p}=i\psi ^{p}$. Also, if for some $j>j_N$, we had $|\lambda _{i\psi ^{j}}|>1$, we could obtain $p$ arbitrarily large such that $|\lambda _{i\psi ^{p}}|>1$, which is absurd. So, we have that,

\[ \forall p>j_N \;|\lambda_{i\psi^{p}}|=1 \]

and $j_N\leq m.$

(iv)$\Rightarrow$ (i) If, for some $i\in [m]$, we have that $|\pi _{i\psi ^{tr}}^{(r)}|=1$, for some $t\in \mathbb {N}$, then for every $j\in \{tr,\,\ldots,\,(t+1)r-1\}$, we have that $|\lambda _{i\psi ^{j}}|=1$. In this case, for some $s\geq (t+1)r$, we have that $i\psi ^{s}=i\psi ^{j}$ for some $j\in \{tr,\,\ldots,\,(t+1)r-1\}$ and so $|\lambda _{i\psi ^{s}}|=1$. Thus, (i) holds for $N=r$.

Proof. Consider a non-wandering point $(a,\,u)\in \widehat {\mathbb {Z}^{m} \times F_n}$ such that $a$ has finite entries, let $\delta =\max \limits _{a_i\in \mathbb {Z}}\{|a_i|\}$ and $U=B((a,\,u);\frac 1 {2^{\delta }})$. Consider a point $(b,\,v)\in U$ and a positive integer $r>m$ such that $(b,\,v)\hat \varphi ^{r}\in U$. So, for $i\in [m]$, we have that:

(5.1)\begin{equation} \text{if }|a_i|<\infty,\text{ then }a_i=b_i=\pi_i^{(r)}b_{i\psi^{r}} \end{equation}

and

(5.2)\begin{equation} \text{if }|a_i|=\infty,\text{ then }{\rm sgn}(a_i)={\rm sgn}(b_i)={\rm sgn}(\pi_i^{(r)}b_{i\psi^{r}}) \text{ and } |b_i|,|\pi_i^{(r)}b_{i\psi^{r}}|>\delta. \end{equation}

If $a\hat \varphi _1^{r}=a$, then $a_i= \pi _i^{(r)}a_{i\psi ^{r}}$ for every $i\in [m]$ and the first part of the lemma trivially holds.

If not, take $q\in [m]$ such that $a_q\neq \pi _q^{(r)}a_{q\psi ^{r}}$. We start by observing that $a_q$ must be infinite since if that is not the case, then, by (5.1), we have that $a_q=b_q=\pi _q^{(r)}b_{q\psi ^{r}}$. If $\pi _q^{(r)}=0$, then $a_q=\pi _q^{(r)}b_{q\psi ^{r}}=0=\pi _q^{(r)}a_{q\psi ^{r}}$. If not, then $b_{q\psi ^{r}}\leq b_q$, which means that $b_{q\psi ^{r}}\leq \delta$ and by (5.2), we have that $a_{q\psi ^{r}}$ is finite and by (5.1), it follows that $a_{q\psi ^{r}}=b_{q\psi ^{r}}$, so $a_q=\pi _q^{(r)}b_{q\psi ^{r}}=\pi _q^{(r)}a_{q\psi ^{r}}$. Also, $a_{q\psi ^{r}}$ must be finite, since, if it is infinite, then by (5.2) we have that $sgn(a_{q\psi ^{r}})=sgn(b_{q\psi ^{r}})$ and $sgn(a_q)=sgn(\pi _q^{(r)}b_{q\psi ^{r}})$, thus, $sgn(a_q)=sgn(\pi _q^{(r)}a_{q\psi ^{r}})$ and that implies that $a_q=\pi _q^{(r)}a_{q\psi ^{r}}$, since $\pi _q^{(r)}\neq 0$.

So, using (5.1) with $i=q\psi ^{r}$, we have that

(5.3)\begin{equation} a_{q\psi^{r}}=b_{q\psi^{r}}=\pi_{q\psi^{r}}^{(r)}b_{q\psi^{r}\psi^{r}}=\pi_{q\psi^{r}}^{(r)}b_{q\psi^{2r}}. \end{equation}

We have that $a_q$ is infinite, so, by (5.2), we know that $\pi _q^{(r)}b_{q\psi ^{r}}>\delta$. This means in particular that $a_{q\psi ^{r}}\neq 0$, because otherwise we would have $b_{q\psi ^{r}}=0$, by (5.1).

Suppose now that for every positive integer $t$, we have that $|\pi _{q\psi ^{tr}}^{(r)}|\neq 1$. If $\pi _{q\psi ^{r}}^{(r)}=0$, then $a_{q\psi ^{r}}=0$, which is absurd. Then, we have that $|\pi _{q\psi ^{r}}^{(r)}|>1$, and $|b_{q\psi ^{2r}}|<|b_{q\psi ^{r}}|<\infty$ and again, using (5.1) with $i=q\psi ^{2r}$, we get that $a_{q\psi ^{2r}}=b_{q\psi ^{2r}}=\pi _{q\psi ^{2r}}^{(r)}b_{q\psi ^{3r}}$. Again, if $\pi _{q\psi ^{2r}}^{(r)}=0,$ then $a_{q\psi ^{2r}}=0$ and by (5.1), it follows that $b_{q\psi ^{2r}}=0$. From (5.3), we reach a contradiction. So, we must have $|\pi _{q\psi ^{r}}^{(r)}|>1$ and $|b_{q\psi ^{3r}}|<|b_{q\psi ^{2r}}|<\infty.$ Proceeding like this, since the value of $|b_{q\psi ^{pr}}|$, for $p\in \mathbb {N}$ cannot decrease indefinitely, then we must have that $b_{q\psi ^{r}}=0$ and so $a_{q\psi ^{r}}=0$, which is absurd.