Proof. We have

$$ \begin{align*} h\not\in\partial W &\Leftrightarrow h\in\bigcup_{i\in\{0,1\}}V^i \Leftrightarrow h\in\bigcup_{i\in\{0,1\}}\bigcup_{n\in{\mathbb N}}\bigcup_{k\in{\mathcal P}_n^i\cap[0,p_n)}U_n(\Delta(k))\\ &\Leftrightarrow \text{there exists } i\in\{0,1\}\text{ there exists } n\in{\mathbb N}\text{ there exists }\\ &\qquad k\in{\mathcal P}_n^i\cap[0,p_n): h_n=k\\ &\Leftrightarrow \text{there exists } i\in\{0,1\}\text{ there exists } n\in{\mathbb N}: h_n\in{\mathcal P}_n^i\\ &\Leftrightarrow \text{there exists } n\in{\mathbb N}: h_n\not\in {\mathcal H}_n\\[-2.8pc] \end{align*} $$

Proof. (a) Let $h\in U_N(\Delta (k))\cap \partial W$ . Then $k\in h_N+p_N{\mathbb Z}\subseteq {\mathcal H}_N+p_N{\mathbb Z}={\mathcal H}_N$ by Lemma 2.1, and for all $n\ge N$ , we have in view of Lemma 2.1: $h_n\in {\mathcal H}_n\cap (h_N+p_N{\mathbb Z})={\mathcal H}_n\cap (k+p_N{\mathbb Z})$ . For the reverse implication, let $k\in {\widetilde {\mathcal H}}_N$ . We construct $h\in \partial W$ with $h_N=k\,\mod p_N$ : for $j>N$ , there is $r_j\in {\mathcal H}_j\cap (k+p_N{\mathbb Z})$ . Let $h^{(j)}=\Delta (r_j)$ and fix any accumulation point h of the sequence $(h^{(j)})_j$ . Consider any $n\ge N$ . For some sufficiently large $j_n\ge n$ , we have $h_n=h_n^{(j_n)}=r_{j_n}\,\mod p_n$ . Hence, $h_N\in r_{j_N}+p_N{\mathbb Z}=k+p_N{\mathbb Z}$ , so that $h\in U_N(\Delta (k))$ , and $h_n\in r_{j_n}+p_n{\mathbb Z}\subseteq {\mathcal H}_{j_n}+p_n{\mathbb Z}\subseteq {\mathcal H}_n+p_n{\mathbb Z}={\mathcal H}_n$ , so that $h\in \partial W$ by Lemma 2.1.

(b) We have

$$ \begin{align*} h\in\partial W &\Leftrightarrow\text{for all } N\in{\mathbb N}:U_N(h)\cap\partial W\neq\emptyset\\&\Leftrightarrow\text{for all } N\in{\mathbb N}:U_N(\Delta(h_N))\cap\partial W\neq\emptyset\\&\Leftrightarrow\text{for all } N\in{\mathbb N}:h_N\in{\widetilde{\mathcal H}}_N\quad\text{by part (a).}\\[-2.9pc] \end{align*} $$

Proof. (a) $h_n+\beta _n{\mathbb Z}\subseteq {\widetilde {\mathcal H}}_n$ for all n by Lemma 2.5(b). As ${\widetilde {\mathcal H}}_n$ is $\tilde {\tau }_n$ -periodic, this implies $h_n+\gcd (\beta _n,\tilde {\tau }_n){\mathbb Z}\subseteq {\widetilde {\mathcal H}}_n$ .

(b) Let $j\in {\widetilde {\mathcal H}}_n$ . Then there is some $h\in U_n(\Delta (j))\cap \partial W$ by Lemma 2.5(a). Now, $h+\beta {\mathbb Z}\subseteq \partial W$ by assumption. As $h_n=j$ , this implies, in view of part (a) of the lemma, $j+\gcd (\beta _n,\tilde {\tau }_n){\mathbb Z}\subseteq {\widetilde {\mathcal H}}_n$ . As this holds for each $j\in {\widetilde {\mathcal H}}_n$ , we have ${\widetilde {\mathcal H}}_n+\gcd (\beta _n,\tilde {\tau }_n){\mathbb Z}={\widetilde {\mathcal H}}_n$ , and as $\tilde {\tau }_n$ is the minimal period of ${\widetilde {\mathcal H}}_n$ , we conclude that $\tilde {\tau }_n\mid \beta _n$ .

Proof. (a) The first claim follows from Lemma 2.6(b), the second one is just a special case of this.

(b) In each residue class of $\operatorname {Aut}_\sigma (X_\eta )/\langle \sigma \rangle $ , there is exactly one element F for which the associated integer $k_F$ satisfying equation (4) equals $0$ . These elements F form a subgroup J of $\operatorname {Aut}_\sigma (X_\eta )$ . Suppose for a contradiction that there are $M+1$ different automorphisms $F_1,\ldots ,F_{M+1}\in J$ . In view of Lemma 2.6(b), they all satisfy

(5) $$ \begin{align} \tilde{\tau}_n\mid (y_{F_{i}})_{S_n} \quad\text{for all }n\in{\mathbb N}. \end{align} $$

Hence, there exists arbitrarily large $n\in {\mathbb N}$ such that $M=p_n/\tilde {\tau }_n$ and

$$ \begin{align*} (y_{F_i})_{S_n}/\tilde{\tau}_n\in\{0,\ldots,M-1\} \quad\text{for all }i=1,\ldots,M+1. \end{align*} $$

It follows that there exist two different $i,j\in \{1,\ldots ,M+1\}$ for which $(y_{F_i})_{S_n}=(y_{F_j})_{S_n}$ for infinitely many n, which is only possible if $y_{F_i}=y_{F_j}$ . In view of [Reference Donoso, Durand, Maass and Petite6, Lemma 2.4], this implies $F_i=F_j\,\mod \langle \sigma \rangle $ , so that $i=j$ , which is a contradiction. Hence, J is a finite group of order $m\leqslant M$ , say $J=\{F_1,\ldots ,F_m\}$ . In particular, $J\subseteq \operatorname {Tor}$ . However, if $F\in \operatorname {Tor}$ , then $ry_F=\Delta (0)\in G$ for some positive integer r. Hence, with the integer $k_F$ satisfying equation (4), we have

$$ \begin{align*} \partial W+y_{Id}-\Delta(rk_F)=\partial W-\Delta(rk_F)=\partial W+ry_F-\Delta(rk_F)\subseteq \partial W, \end{align*} $$

but $k_{Id}=0$ , so equation (4) implies $k_F=0$ . It follows that $F\in J$ , and we proved that $J=\operatorname {Tor}$ . Then [Reference Donoso, Durand, Maass and Petite7, Theorem 3.2(2)] implies that $\operatorname {Tor}$ is cyclic, and $\operatorname {Aut}_\sigma (X_\eta )=J\oplus \langle \sigma \rangle =\operatorname {Tor}\oplus \ \langle \sigma \rangle $ .

It remains to determine the order m of $\operatorname {Tor}$ : fix some $F\in \operatorname {Tor}$ . In view of equation (5),

$$ \begin{align*} p_n=M\cdot\tilde{\tau}_n\mid M\cdot(y_F)_{S_n} \equiv (y_{F^M})_{S_n}\,\mod\! p_n \end{align*} $$

for all $n\in {\mathbb N}$ . Hence, $F^M=\operatorname {id}_{X_\eta }$ , so that the order of F is a divisor of M.

Proof. Suppose that equation (6) holds. Then, for all $n\ge N$ , there is $r_n\in r+p_N{\mathbb Z}$ such that $U_n(\Delta (r_n))\cap \partial W\neq \emptyset $ , whence $r_n\in (r+p_N{\mathbb Z})\cap {\widetilde {\mathcal H}}_n$ in view of Lemma 2.5(a).

Now consider any $r_n\in (r+p_N{\mathbb Z})\cap {\widetilde {\mathcal H}}_n$ . By the inclusion in equation (6),

$$ \begin{align*} U_n(\Delta(r_n))&\cap\partial W\subseteq U_N(\Delta(r_n))\cap\partial W\\ &=U_N(\Delta(r))\cap\partial W\subseteq(\partial W-\beta)\cap(\partial W-2\beta-\Delta(k)), \end{align*} $$

so that, by Lemma 2.5(a) again, $r_n+\beta _n\in {\widetilde {\mathcal H}}_n$ and $r_n+2\beta _n+k\in {\widetilde {\mathcal H}}_n$ , which proves equation (7).

Conversely, suppose that equation (7) holds. For each $n\ge N$ , there is some $r_n\in (r+p_N{\mathbb Z})\cap {\widetilde {\mathcal H}}_n$ , and we find a subsequence $\Delta (r_{n_i})$ that converges to some $h\in G$ . Hence, for each $m>0$ , there is $n_i\ge m$ such that $h_m\in r_{n_i}+p_m{\mathbb Z}\subseteq {\widetilde {\mathcal H}}_{n_i}+p_m{\mathbb Z}\subseteq {\widetilde {\mathcal H}}_m+p_m{\mathbb Z}={\widetilde {\mathcal H}}_m$ , so that $h\in \partial W$ in view of Lemma 2.5(b). As $h_N\in r_{n_i}+p_N{\mathbb Z}=r+p_N{\mathbb Z}$ for some $n_i$ , this shows that $h\in U_N(\Delta (r))\cap \partial W$ .

Now consider any $h\in U_N(\Delta (r))\cap \partial W$ . Then, for all $n\ge N$ , $U_n(h)\cap \partial W\neq \emptyset $ , so that $h_n\in (r+p_N{\mathbb Z})\cap {\widetilde {\mathcal H}}_n$ , where we used Lemma 2.5(a) once more. The inclusion in equation (7) then implies $h_n\in ({\widetilde {\mathcal H}}_n-\beta _n)\cap ({\widetilde {\mathcal H}}_n-2\beta _n-k)$ for all $n\ge N$ . Now Lemma 2.5(b) shows that $h+\beta \in \partial W$ and $h+2\beta +\Delta (k)\in \partial W$ , which proves equation (6).

Proof. Recall from equation (3) that $\partial W+y_F \subseteq \bigcup _{k\in {\mathbb Z}}\partial W+\Delta (k)$ .

Let $y\in G$ and recall that $\pi \colon {X_\eta \to G}$ denotes the factor map onto the MEF. At the end of the proof, we show

(13) $$ \begin{align} |\pi^{-1}\{y\}|=2 \Leftrightarrow \text{there exists } j\in{\mathbb Z}:y+\Delta(j)\in R:=\partial W\setminus\bigcup_{k\in{\mathbb Z}\setminus\{0\}}\partial W+\Delta(k). \end{align} $$

As $F\colon X_\eta \to X_\eta $ is a bijection that maps $\pi $ -fibres to $\pi $ -fibres and as $\pi \circ F=f\circ \pi $ , it follows that when $\pi ^{-1}\{y\}=\{x_1,x_2\}$ with $x_1\neq x_2$ , then $\pi ^{-1}\{f(y)\}=\{x_1'=F(x_1),x_2'=F(x_2)\}$ , and there are exactly one index $j\in {\mathbb Z}$ such that $y+\Delta (j)\in \partial W$ and $(x_1)_j\neq (x_2)_j$ , and exactly one index $k\in {\mathbb Z}$ such that $f(y)+\Delta (k)\in \partial W$ and $(x_1')_{k}\neq (x_2')_{k}$ . As F is described by a block code $\{0,1\}^{[-m:m]}\to \{0,1\}$ , it follows that $|j-k|\leqslant m$ . (This argument is taken from the proof of [Reference Bułatek and Kwiatkowski2, Corollary 1].)

Consider any $y\in R$ . Then the index j in equation (13) equals $0$ and $y+y_F\in \partial W-\Delta (k)$ for some k with $|k|\leqslant m$ . In other words: $f(R)$ is contained in the closed set $\bigcup _{k=-m}^mT^k(\partial W)$ . Because of condition (D′), the set R defined in equation (13) is residual with respect to the subspace topology of $\partial W$ . Hence,

$$ \begin{align*} \partial W+y_F=f(\partial W)=f(\overline{R})\subseteq\overline{f(R)}\subseteq \bigcup_{k=-m}^mT^k(\partial W) =\bigcup_{k=-m}^m\partial W+\Delta(k). \end{align*} $$

It remains to prove equation (13). Recall first that $|\pi ^{-1}\{y\}|=1$ if and only if $y\in C_\phi $ , so that $|\pi ^{-1}\{y\}|>1$ if and only if $y\in \bigcup _{k\in {\mathbb Z}}\partial W+\Delta (k)$ . So all we must show is that (A) $|\pi ^{-1}\{y\}|>2$ if and only if (B) $y+\Delta (j)\in \partial W\cap (\partial W+\Delta (k))$ for some $j\in {\mathbb Z}$ and $k\in {\mathbb Z}\setminus \{0\}$ .

Suppose first that part (A) holds, i.e. that there are at least three different points in $\pi ^{-1}\{y\}$ . As points in the same $\pi $ -fibre can differ only at positions k where $y+\Delta (k)\in \partial W$ , there must be at least two such positions, and that is part (B).

Conversely, if part (B) holds and if $x=\phi (y)$ , then $x_j=x_{j-k}=1$ .

1. (i) As $\partial W\cup (\partial W+\Delta (k))$ is nowhere dense in G, there are arbitrarily small perturbations $y'$ of y such that $y'+\Delta (j)\not \in \partial W\cup (\partial W+\Delta (k))$ , resulting in points $x'=\phi (y')$ with $x^{\prime }_j=x^{\prime }_{j-k}=0$ .

2. (ii) As $\partial W\cap (\partial W+\Delta (k))$ is nowhere dense in $\partial W$ with respect to the subspace topology on $\partial W$ (because of condition (D′)), there are arbitrarily small perturbations $y'$ of y such that $y'+\Delta (j)\in \partial W\setminus (\partial W+\Delta (k))$ , resulting in $x'=\phi (y')$ with $x^{\prime }_j=1$ and $x^{\prime }_{j-k}=0$ .

Hence, $|\pi ^{-1}\{y\}|\geqslant 3$ , and that is part (A).

Proof. We start from the observation of equation (3), namely $\partial W+y_F\subseteq \bigcup _{k\in {\mathbb Z}}\partial W+\Delta (k)$ . This implies

$$ \begin{align*} \partial W= \bigcup_{k\in{\mathbb Z}}\partial W\cap(\partial W+\Delta(k)-y_F) =\bigcup_{k\in {\mathbb Z}}V_k. \end{align*} $$

Hence, $M:=\bigcup _{k\in {\mathbb Z}}V_k\setminus \operatorname {int}_{\partial W}(V_k)$ is a meagre subset of the compact space $\partial W$ and $\partial W=M\cup \bigcup _{k\in K}\operatorname {int}_{\partial W}(V_k)$ . Now Baire’s category theorem implies that $K\neq \emptyset $ . As $\partial W$ is separable, there is a countable collection $U_{{N_j}}(\Delta (r_j))$ , ${j\in {\mathbb N}}$ , of cylinder sets in G, for each of which there exists $k_j\in K$ such that $\emptyset \neq \partial W\cap U_{{N_j}}(\Delta (r_j))\subseteq \operatorname {int}_{\partial W}(V_{k_j})$ and such that

$$ \begin{align*} \partial W=M\cup\bigcup_{j\in{\mathbb N}}(\partial W\cap U_{{N_j}}(\Delta(r_j))). \end{align*} $$

As M is meagre in $\partial W$ , these cylinder sets satisfy equation (15), and as

$$ \begin{align*} (U_{{N_j}}(\Delta(r_j))\cap\partial W) \subseteq \operatorname{int}_{\partial W}(V_{k_j}) \subseteq \partial W+\Delta(k_j)-y_F, \end{align*} $$

also equation (14) holds.

Proof. Let $K':=K\cap [-m,m]$ , m as in Proposition 2.12. Because of that proposition and Proposition 2.13, $\partial W=\bigcup _{k\in K'}\overline {\operatorname {int}_{\partial W}(V_k)}$ . Suppose there are $k_i,k_j\in K$ such that $\operatorname {int}_{\partial W}(V_{k_i})\cap \operatorname {int}_{\partial W}(V_{k_j})\neq \emptyset $ . Then there is some cylinder set $U_{N}(\Delta (r))\cap \partial W$ contained in this intersection. Let $\tilde {U}:=(U_{N}(\Delta (r))\cap \partial W)+y_F-\Delta (k_i)$ . Then, $\tilde {U}\subseteq \partial W$ and $\tilde {U}+\Delta (k_i-k_j)\subseteq \partial W$ , so that $\tilde {U}\subseteq \partial W\cap (\partial W+\Delta (k_j-k_i))$ . As $\tilde {U}$ is non-empty and open in the relative topology on $\partial W$ , the weak disjointness assumption implies $k_i=k_j$ . This also proves that $K=K'\subseteq [-m,m]$ .

Proof. Fix some cylinder set $U_{{N'}}(\Delta (r'))$ such that

$$ \begin{align*} \emptyset\neq U_{{N'}}(\Delta(r'))\cap\partial W\subseteq \operatorname{int}_{\partial W}(V_{k_i})\cap (\operatorname{int}_{\partial W}(V_{k_j})-(y_F-\Delta(k_i))). \end{align*} $$

In view of Proposition 2.13, there are $N_i',N_j'\in {\mathbb N}$ and $r_i',r_j'\in {\mathbb Z}$ such that

(17) $$ \begin{align} U_{{N'}}(\Delta(r'))\cap U_{{N_i'}}(\Delta(r_i'))\cap (U_{{N_j'}}(\Delta(r_j'))-(y_F-\Delta(k_i))) \cap\partial W\neq\emptyset \end{align} $$

and

(18) $$ \begin{align} (U_{{N_i'}}(\Delta(r_i'))\cap\partial W)+(y_F-\Delta(k_i))&\subseteq\partial W,\nonumber\\ (U_{{N_j'}}(\Delta(r_j'))\cap\partial W)+(y_F-\Delta(k_j))&\subseteq\partial W. \end{align} $$

Because of equations (17) and (18), the set $U_{{N'}}(\Delta (r'))\cap U_{{N_i'}}(\Delta (r_i'))\cap (U_{{N_j'}}(\Delta (r_j'))-(y_F-\Delta (k_i)))$ contains a cylinder set $U_{N}(\Delta (r))$ for which $U_{N}(\Delta (r))\cap \partial W\neq \emptyset $ and

$$ \begin{align*} \begin{aligned} (U_{{N}}(\Delta(r))\cap\partial W)+(y_F-\Delta(k_i))\subseteq&\ \partial W\quad\text{and}\\ (U_{{N}}(\Delta(r))\cap\partial W)+(y_F-\Delta(k_i))+(y_F-\Delta(k_j))\subseteq&\ \partial W. \end{aligned} \end{align*} $$

Clearly, N can be chosen arbitrarily large.

Proof. It suffices to prove that $\operatorname {int}_{\partial W}(V_k)+(y_F-\Delta (k))\subseteq V_k'$ for all $k\in K$ . Suppose for a contradiction that there is $k_i\in K$ such that $\operatorname {int}_{\partial W}(V_{k_i})+(y_F-\Delta (k_i))\not \subseteq V_{k_i}'$ . By Proposition 2.13, there exists $k_j\in K\setminus \{k_i\}$ such that $(\operatorname {int}_{\partial W}(V_{k_i})+(y_F-\Delta (k_i)))\cap \operatorname {int}_{\partial W}(V_{k_j})\neq \emptyset $ . So we can apply Corollary 2.15. Hence, there are $N\in {\mathbb N}$ and $r\in {\mathbb Z}$ such that, setting $\beta =y_F-\Delta (k_i)$ ,

$$ \begin{align*} (U_N(\Delta(r))\cap \partial W)\subseteq(\partial W-\beta)\cap(\partial W-2\beta-\Delta(k_i-k_j)). \end{align*} $$

In view of condition (DD′), this implies $k_i=k_j$ . The remaining assertions follow from Corollary 2.14.

Proof. (a) If there is some $y\in \partial W\cap (\partial W-\Delta (k))$ , then $U_{N}(\Delta (y_N))\cap \partial W=U_{N}(y)\cap \partial W\neq \emptyset $ and $U_{N}(\Delta (k+y_N))\cap \partial W=U_{N}(\Delta (k)+y)\cap \partial W\neq \emptyset $ , so that $k=(k+y_N)-y_N\in {\widetilde {\mathcal H}}_N-{\widetilde {\mathcal H}}_N\subseteq {\mathcal M}_{A_N}-{\mathcal M}_{A_N}\subseteq \langle {\mathcal M}_{A_N}\rangle =\langle A_{N}\rangle $ for all $N>0$ by Lemma 2.5a). Hence, $k=0$ in view of property (TI). If there are $y_1,y_2\in \partial W$ and $k_1,k_2\in {\mathbb Z}$ such that $y_i+y_F\in \partial W-\Delta (k_i)$ ( $i=1,2$ ), then $U_N(\Delta (k_i+(y_i+y_F)_N))\cap \partial W\neq \emptyset $ ( $i=1,2$ ), so that $k_2-k_1\in \langle A_N\rangle $ for all $N>0$ as before. Hence, $k_2=k_1$ because of property (TI).

(b) This follows from part (a) of the lemma and from Theorem 2.8a).

Proof. As in [Reference Hall15, Lemma 1.17], we have

$$ \begin{align*} \boldsymbol{\delta}((\tilde{r}+\tilde{\ell}{\mathbb Z})\cap{\mathcal M}_{(A^{\div g})^{\perp{\tilde{\ell}}}}) =\frac{1}{\tilde{\ell}}\cdot\boldsymbol{\delta}({\mathcal M}_{(A^{\div g})^{\perp{\tilde{\ell}}}}). \end{align*} $$

As $g\cdot \tilde {\ell }=\ell \vee m$ , this together with equation (23) proves the identity in equation (28). As $\boldsymbol {\delta }((r+\ell {\mathbb Z})\cap (s+m{\mathbb Z}))={1}/{\ell \vee m}$ , equation (29) follows at once. For the inequality in equation (28), observe that ${\mathcal M}_{(A^{\div g})^{\perp {\tilde {\ell }}}}\subseteq {\mathcal M}_{A^{\div g}}\subseteq {\mathcal M}_{A^{\div \ell \vee m}}$ by equation (25).

Proof. Let $A:=(C^{\div g})^{\perp \tilde \ell }$ . In view of equation (27), our assumption implies $(\tilde {r}+\tilde {\ell }{\mathbb Z})\cap [N,\infty )\subseteq {\mathcal M}_{A}={\mathcal M}_{A^{\textrm {prim}}}$ . As $A^{\textrm {prim}}$ is taut, [Reference Dymek, Kasjan, Kułaga-Przymus and Lemańczyk12, Proposition 4.31] shows that there is $a\in A^{\textrm {prim}}$ such that $a\mid \gcd (\tilde {r},\tilde {\ell })=1$ , that is, $1\in A$ . Hence, there is $c\in C$ such that $c\mid g=\gcd (r,\ell )\vee \gcd (s,m)$ .

Proof. By Lemma 2.24, condition (a) implies condition (b). Conversely, condition (a) follows from condition (b), because $(r+\ell {\mathbb Z})\cap a{\mathbb Z}\subseteq (\gcd (r,\ell )\vee a){\mathbb Z}$ .

Suppose that conditions (a) and (b) hold. By condition (b), we have $c\mid \gcd (r,\ell )\vee a\mid \ell \vee a$ . Since $\gcd (a,\ell )\mid r$ , we have $(r+\ell {\mathbb Z})\cap a{\mathbb Z}\neq \emptyset $ . So by condition (a), we get $r\in c{\mathbb Z}+\ell {\mathbb Z}=\gcd (c,\ell ){\mathbb Z}$ . Hence, condition (c) holds.

Finally suppose that condition (c) holds. Then,

$$ \begin{align*} \gcd(c,\gcd(r,\ell)\vee a)&=\gcd(c,r,\ell)\vee\gcd(c,a)\\ &=\gcd(c,\ell)\vee\gcd(c,a)=\gcd(c,\ell\vee a)=c, \end{align*} $$

and condition (b) follows at once.

Proof. (a) Let $a\in {\mathbb N}\setminus {\mathcal M}_C$ . Then, $\emptyset \neq a{\mathbb Z}\setminus {\mathcal M}_C=a\cdot ({\mathbb Z}\setminus {\mathcal M}_{C^{\div a}})$ , and the claim follows from the observation that $\operatorname {lcm}((C^{\div a})^{\textrm {prim}})$ is the minimal period of ${\mathcal M}_{C^{\div a}}$ , see [Reference Kasjan, Keller and Lemańczyk18, Lemma 5.1b)].

(b) Let T be the minimal period of ${\mathcal M}_A\setminus {\mathcal M}_C$ . Clearly, $T|\operatorname {lcm}(A\cup C)$ . Now let $a\in A$ . Since $1\not \in C^{\div a}$ , $a\in {\mathcal M}_A\setminus {\mathcal M}_C$ , so $a+T{\mathbb Z}\subset {\mathcal M}_A$ . By Lemma 2.24 and primitivity of A, we get $a|T$ , so that $T+(a{\mathbb Z}\setminus {\mathcal M}_C)\subset a{\mathbb Z}\setminus {\mathcal M}_C$ for every $a\in A$ and therefore $a\cdot \operatorname {lcm}((C^{\div a})^{\textrm {prim}})\mid T$ in view of part (a). Hence, $\operatorname {lcm}(\{a\}\cup C)=a\operatorname {lcm}(C^{\div a})=a\operatorname {lcm}((C^{\div a})^{\textrm {prim}})\mid T$ for every $a\in A$ , so that $\operatorname {lcm}(A\cup C)|T$ .

Proof. In view of the assumption in equation (31), we have

(34) $$ \begin{align} \begin{aligned} ((r+p_N{\mathbb Z})\cap (a{\mathbb Z})\setminus{\mathcal M}_{S_n})\subseteq (-{b}+{\mathcal M}_{E_{n,a}})\cup(-{b}+ {\mathcal M}_{D_{n,a}}), \end{aligned} \end{align} $$

where $D_{n,a}=A_n\setminus E_{n,a}$ . We prove below that equation (32) implies

(35) $$ \begin{align} ((r+p_N{\mathbb Z})\cap(a{\mathbb Z})\setminus{\mathcal M}_{S_n})+{b}\not\subseteq{\mathcal M}_{E_{n,a}}. \end{align} $$

Hence and by equation (34), there is $a'\in D_{n,a}$ such that equation (33) holds. Hence, ${b}=0$ or $|{b}|\geqslant \gcd (a',\gcd (r,p_N)\vee a)\geqslant \gcd (a',a)$ . It remains to show that equation (32) implies the assumption in equation (35).

Consider any $a\in A_n$ for which $ ((r+p_N{\mathbb Z})\cap (a{\mathbb Z})\setminus {\mathcal M}_{S_n})\neq \emptyset $ . Then $\gcd (a,p_N)\mid r$ , and Lemma 2.23 implies

(36) $$ \begin{align} \begin{aligned} 0<\boldsymbol{\delta}((r+p_N{\mathbb Z})\cap(a{\mathbb Z})\setminus{\mathcal M}_{S_n}) &= \frac1{p_N\vee a}(1-\boldsymbol{\delta}({\mathcal M}_{(S_n^{\div g_n})^{\perp(p_N^{\div g_n})}})) , \end{aligned} \end{align} $$

where

$$ \begin{align*} g_n:=\gcd(r,p_N)\vee a\quad\text{and}\quad p_N^{\div g_n}=\frac{p_N\vee g_n}{g_n}=\frac{p_N\vee a}{g_n} \mid p_N^{\div a}. \end{align*} $$

Suppose for a contradiction that there is inclusion in equation (35). Then,

$$ \begin{align*} (r+p_N{\mathbb Z})\cap(a{\mathbb Z})\setminus{\mathcal M}_{S_n} \subseteq\bigcup_{a'\in E_{n,a}}(-{b}+a'{\mathbb Z}), \end{align*} $$

so that

(37) $$ \begin{align} &\boldsymbol{\delta}((r+p_N{\mathbb Z})\cap(a{\mathbb Z})\setminus{\mathcal M}_{S_n})\nonumber\\ &\quad\leqslant \sum_{a'\in E_{n,a}}\boldsymbol{\delta}((r+p_N{\mathbb Z})\cap(-{b}+a'{\mathbb Z})\cap(a{\mathbb Z})\setminus{\mathcal M}_{S_n})\nonumber\\ &\quad\leqslant \sum_{a'\in E_{n,a}} \boldsymbol{\delta}((g_n{\mathbb Z})\cap(-{b}+a'{\mathbb Z})\setminus{\mathcal M}_{S_n})\nonumber \\ &\quad= \sum_{a'\in E_{n,a},\ \gcd(g_n,a')\mid{b}} \boldsymbol{\delta}((g_n{\mathbb Z})\cap(-{b}+a'{\mathbb Z})\setminus{\mathcal M}_{S_n})\nonumber \\ &\quad= \sum_{a'\in E_{n,a},\ \gcd(g_n,a')\mid{b}} \frac{1}{g_n\vee a'}(1- \boldsymbol{\delta}({\mathcal M}_{(S_n^{\div (g_n\vee\gcd({b},a'))})^{\perp{g_n\vee a'}/{g_n\vee\gcd({b},a')}}})). \end{align} $$

The last equality follows from Lemma 2.23. Notice that $\gcd (g_n,a')\mid {b}$ implies

(38) $$ \begin{align} \frac{g_n\vee a'}{g_n\vee\gcd({b},a')} = \frac{(g_n\vee a')\cdot\gcd(g_n,a')}{g_n\cdot\gcd({b},a')}=\frac{a'}{\gcd({b},a')}=a^{{\prime}\div b}. \end{align} $$

As

(39) $$ \begin{align} \frac{p_N\vee a}{a'\vee g_n} \leqslant \frac{p_N\vee a}{a'\vee a} = \frac{p_N^{\div a}}{a^{\prime\div a}}, \end{align} $$

equations (36), (37) and (38) together yield

(40) $$ \begin{align} \begin{aligned} 1-\boldsymbol{\delta}({\mathcal M}_{(S_n^{\div g_n})^{\perp(p_N^{\div g_n})}}) &\leqslant \sum_{a'\in E_{n,a}} \frac{p_N^{\div a}}{a^{\prime\div a}}(1- \boldsymbol{\delta}({\mathcal M}_{(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}})). \end{aligned} \end{align} $$

Denote

$$ \begin{align*} \begin{aligned} R_n(a') :=& \{b^{\div g_n}: b\in S_n, b^{\div g_n}\perp p_N^{\div g_n}, b^{\div g_n}\not\perp a^{\prime\div g_n}\}. \end{aligned} \end{align*} $$

Then $R_n(a')\subseteq S_n^{\div g_n}$ trivially and we claim that

$$ \begin{align*} (S_n^{\div g_n})^{\perp(p_N^{\div g_n})}\setminus R_n(a') \subseteq {\mathcal M}_{(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}}. \end{align*} $$

Indeed, each $b^{\div g_n}\in S_n^{\div g_n}\setminus R_n(a')$ , which is also coprime to $p_N^{\div g_n}$ , is coprime to $a^{\prime \div g_n}$ and, a fortiori, to $a^{\prime \div b}$ because $\gcd (g_n,a')\mid b$ . Moreover, $b^{\div g_n}={b\vee g_n}/{g_n}$ is a multiple of ${b\vee g_n\vee \gcd ({b},a')}/{g_n\vee \gcd ({b},a')}$ , so that the latter is also coprime to ${a^{\prime \div b}}$ . Therefore,

$$ \begin{align*} (S_n^{\div g_n})^{\perp(p_N^{\div g_n})} \subseteq R_n(a')\cup {\mathcal M}_{(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}}, \end{align*} $$

so that Behrend’s inequality (see [Reference Hall15, Theorem 0.12] for a reference) yields

$$ \begin{align*} \begin{aligned} 1-\boldsymbol{\delta}({\mathcal M}_{(S_n^{\div g_n})^{\perp(p_N^{\div g_n})}}) &\geqslant 1-\boldsymbol{\delta}({\mathcal M}_{R_n(a')\cup {\mathcal M}_{(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}}})\\ &= 1-\boldsymbol{\delta}({\mathcal M}_{R_n(a')\cup {(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}}})\\ &\geqslant (1-\boldsymbol{\delta}({\mathcal M}_{R_n(a')}))\cdot (1- \boldsymbol{\delta}({\mathcal M}_{(S_n^{\div(g_n\vee\gcd({b},a'))})^{\perp{a^{\prime\div b}}}})). \end{aligned} \end{align*} $$

Therefore, equation (40) leads to

$$ \begin{align*} 1-\boldsymbol{\delta}({\mathcal M}_{(S_n^{\div g_n})^{\perp(p_N^{\div g_n})}}) \leqslant \sum_{a'\in E_{n,a}} \frac{{p_N}^{\div a}}{a^{\prime\div a}}\cdot \frac{1-\boldsymbol{\delta}({\mathcal M}_{(S_n^{\div g_n})^{\perp(p_N^{\div g_n})}})}{1-\boldsymbol{\delta}({\mathcal M}_{R_n(a')})}. \end{align*} $$

As $1-\boldsymbol {\delta }({\mathcal M}_{(S_n^{\div g_n})^{\perp (p_N^{\div g_n})}})>0$ , in view of equation (36), we can divide the last inequality by this expression, so that

(41) $$ \begin{align} \begin{aligned} 1 &\leqslant \sum_{a'\in E_{n,a}} \frac{p_N^{\div a}}{a^{\prime\div a}}\cdot \frac{1}{1-\boldsymbol{\delta}({\mathcal M}_{R_n(a')})} \leqslant \sum_{a'\in E_{n,a}} \frac{p_N^{\div a}}{a^{\prime\div a}}\cdot \frac{1}{1-\boldsymbol{\delta}({\mathcal M}_{\operatorname{Spec}(a^{\prime\div a})})}\\ &= \sum_{a'\in E_{n,a}} \frac{{p_N}^{\div a}}{a^{\prime\div a}}\cdot \prod_{p\mid a^{\prime\div a}}\frac{1}{1-\frac{1}{p}} \leqslant {p_N}\cdot \sum_{a'\in E_{n,a}}\frac{1}{\varphi(a^{\prime\div a})}, \end{aligned} \end{align} $$

where we used the fact that $R_n(a')\subseteq {\mathcal M}_{\operatorname {Spec}(a^{\prime \div g_n})}\subseteq {\mathcal M}_{\operatorname {Spec}(a^{\prime \div a})}$ . However, the last estimate contradicts the assumption in equation (32).

Proof. Suppose for a contradiction that condition (D′) does not hold, equivalently that condition (Seh′) does not hold. Then there are $k\in {\mathbb Z}\setminus \{0\}$ and an arithmetic progression $r+p_N{\mathbb Z}$ such that for all $n\ge N$ ,

$$ \begin{align*} \emptyset\neq(r+p_N{\mathbb Z})\cap{\widetilde{\mathcal H}}_n\subseteq {\widetilde{\mathcal H}}_n-k. \end{align*} $$

Let $n\ge N$ . In view of property (AS), there is $a\in A_n$ such that

(43) $$ \begin{align} \emptyset\neq(r+p_N{\mathbb Z})\cap a{\mathbb Z}\setminus{\mathcal M}_{S_n}+k\subseteq {\mathcal M}_{A_n}. \end{align} $$

Let $E_{n,a}=\{a'\in A_n: \gcd (a',a)\leqslant |k|\}$ . Since

$$ \begin{align*} \varphi(a^{\prime\div a})=a^{\prime\div a}\prod_{p\mid a^{\prime\div a}}\bigg(1-\frac{1}{p}\bigg)\geqslant\frac{1}{|k|}a'\prod_{p\mid a'}\bigg(1-\frac{1}{p}\bigg)=\frac{1}{|k|}\varphi(a') \end{align*} $$

for any $a'\in E_{n,a}$ , the assumption in equation (42) above implies the assumption in equation (32) of Proposition 2.27. So this proposition applies to the inclusion in equation (43), and there is $a'\in A_n\setminus E_{n,a}$ such that $\gcd (a',a)\mid k$ . As $|k|<\gcd (a',a)$ for all $a'\in A_n\setminus E_{n,a}$ , this contradicts the assumption $k\in {\mathbb Z}\setminus \{0\}$ .

The remaining conclusions follow from Corollary 2.14.

Proof. Suppose for a contradiction that condition (DD′) does not hold, equivalently that condition (DSeh′) does not hold. Then there are $k\in {\mathbb Z}\setminus \{0\}$ , $\beta \in G$ , an arithmetic progression $r+p_N{\mathbb Z}$ , such that for all $n\ge N$ ,

$$ \begin{align*} \emptyset\neq(r+p_N{\mathbb Z})\cap{\widetilde{\mathcal H}}_n\subseteq ({\widetilde{\mathcal H}}_n-\beta_n)\cap({\widetilde{\mathcal H}}_n-2\beta_n-k). \end{align*} $$

Let $n\ge N$ . In view of the property in equation (AS), there is $a\in A_n$ such that

(45) $$ \begin{align} &\emptyset\neq(r+p_N{\mathbb Z})\cap a{\mathbb Z}\setminus{\mathcal M}_{S_n}+\beta_n\subseteq {\mathcal M}_{A_n}\quad\text{and }\nonumber\\ &\emptyset\neq(r+p_N{\mathbb Z})\cap a{\mathbb Z}\setminus{\mathcal M}_{S_n}+2\beta_n+k\subseteq {\mathcal M}_{A_n}. \end{align} $$

Let $E_{n,a}= A_n\setminus \{a\}$ . In view of the assumption in equation (44), Proposition 2.27 applies to both inclusions in equation (45), and as $A_n\setminus E_{n,a}=\{a\}$ , we can conclude that $a\mid \beta _n$ and $a\mid 2\beta _n+k$ , so that $a\mid k$ . As $a\in A_n$ and $\min A_n\to \infty $ , this contradicts the assumption $k\in {\mathbb Z}\setminus \{0\}$ .

The final conclusion follows from Proposition 2.16.

Proof. (a) Let $k\in K$ . Then $V_k'+(y_F-\Delta (k)){\mathbb Z}\subseteq V_k'\subseteq \partial W$ by Proposition 2.30, and because of Lemma 2.5, $h_n+(y_F-\Delta (k))_n{\mathbb Z}\subseteq {\widetilde {\mathcal H}}_n\subseteq {\mathcal M}_{A_n}$ for each $h\in V_k'$ . It follows that there exists $a'\in A_n$ such that $a'\mid \gcd (h_{n},(y_F-\Delta (k))_{n})$ .

(b) Let $a\in A_n$ . Notice that $a\not \in {\mathcal M}_{S_n(a)}$ . Otherwise, $a{\mathbb Z}\subseteq {\mathcal M}_{S_n(a)}$ which contradicts equation (AS). So, $a\in {\widetilde {\mathcal H}}_n$ and, by Lemma 2.5, there exists some $h\in U_{n}(\Delta (a))\cap \partial W$ . In particular, $h_{n}=a\in {\widetilde {\mathcal H}}_n$ . Because of Corollary 2.14, there exists $k\in K$ such that $h\in V_k'$ . As in the proof of part (a), it follows that there exists $a'\in A_n$ such that $a'\mid \gcd (h_{n},(y_F-\Delta (k))_{n})$ . As $a'$ and $a=h_n$ belong to the same primitive set $A_n$ , this implies $a=a'\mid (y_F-\Delta (k))_{n}$ .

Suppose there is another $k'\in K$ such that $a\mid (y_F-\Delta (k'))_{n}$ . Then $a\mid k-k'$ , so that $k=k'$ or $a\leqslant |k-k'|\leqslant 2m$ .

(c) It suffices to prove that $\kappa _n(a)=\kappa _n(a')$ for every edge $(a,a')$ of ${\mathcal G}_n$ , i.e. whenever $\gcd (a,a')>2m$ . However, as part (b) implies

$$ \begin{align*} \gcd(a,a')\mid((y_F)_{n}-\kappa_n(a))-((y_F)_{n}-\kappa_n(a'))=\kappa_n(a')-\kappa_n(a), \end{align*} $$

it follows that $\kappa _n(a)=\kappa _n(a')$ or $2m<\gcd (a,a')\leqslant |\kappa _n(a)-\kappa _n(a')|\leqslant 2m$ .

(d) If $K=\{k\}$ , then $\partial W=V_k'$ , that is, $\partial W+(y_F-\Delta (k))\subseteq \partial W$ , and the claim follows from Lemma 2.6(b).

(e) It follows from part (d) that $y_F=\Delta (k)$ for some $k\in {\mathbb Z}$ .

Proof. Let $n_1=1$ . If $n_1<n_2<\cdots <n_k$ are chosen, let $n_{k+1}=\min \{j\in {\mathbb N}: S_{n_k}^{\textrm {sat}}\subseteq S_j\}$ .

Proof. (i) Let $s\in {\mathcal M}_S$ . Then $b \mid s$ for some $b \in S$ . Since $b \mid \ell _{S}$ , $s+\ell _{S} \mathbb {Z} \subseteq b \mathbb {Z} \subseteq \mathcal {M}_{S}$ .

(ii) Let $s+\ell _S{\mathbb Z}\subseteq {\mathcal M}_{\mathcal B}$ . By [Reference Dymek, Kasjan, Kułaga-Przymus and Lemańczyk12, Proposition 4.31], the tautness of ${\mathcal B}$ implies $b \mid \gcd (s,\ell _S)$ for some $b \in {\mathcal B}$ . Since S is saturated, $b\in S$ . So $s \in \mathcal {M}_{S}$ .

(iii) Let $s\in {\mathcal F}_{{\mathcal A}_S}$ . Assume for a contradiction that $\gcd (b,\ell _S) \mid s+\ell _{S} k$ for some $k \in {\mathbb Z}$ and some $b \in {\mathcal B}$ . Since $\gcd (b,\ell _S) \mid \ell _{S}$ , $\gcd (b, \ell _{S}) \mid s$ , which contradicts $s\in {\mathcal F}_{{\mathcal A}_S}$ . So $s+\ell _S{\mathbb Z}\subseteq {\mathcal F}_{{\mathcal A}_S}$ .

(iv) Let $s+\ell _S{\mathbb Z}\subseteq {\mathcal F}_{\mathcal B}$ . Assume for a contradiction that $\gcd (b,\ell _S)\mid s$ for some $b\in {\mathcal B}$ . Then there is $x\in {\mathbb Z}$ such that $x\equiv 0\,\mod b$ and $x\equiv s\,\mod \ell _S$ , i.e. $x\in b{\mathbb Z}\cap (s+\ell _S{\mathbb Z})$ , in contradiction to $s+\ell _S{\mathbb Z}\subseteq {\mathcal F}_{\mathcal B}$ . Hence, $s\in {\mathcal F}_{{\mathcal A}_S}$ .

Now part (a) follows from items (i) and (ii), while part (b) follows from items (iii) and (iv).

Proof. (a) Let $a\in {\mathcal A}_S^\infty $ . If $a=\gcd (b,\ell _S)\in {\mathcal M}_S$ for some $b\in {\mathcal B}$ , then there is $b'\in S$ such that $b'\mid a\mid b$ , and the primitivity of ${\mathcal B}$ implies $b=a=b'\in S$ . This contradicts to a having infinitely many sources.

(b) Let $a\in {\mathcal A}_S^\infty $ . There are infinitely many $b\in {\mathcal B}\setminus S'$ such that $a=\gcd (b,\ell _S)$ . Let $a_b':=\gcd (b,\ell _{S'})$ for these b. Then, $\gcd (a^{\prime }_b,\ell _S)=\gcd (b,\ell _S)=a$ for all these b, and as ${\mathcal A}_{S'}$ is finite, there exists some $a'\in {\mathcal A}_{S'}$ such that $a'=a^{\prime }_b$ for infinitely many of them. Hence, $a'\in {\mathcal A}_{S'}^\infty $ .

(c) Clear.

(d) In the situation of item (b), suppose that $a\in {\mathcal A}_S^{\infty ,p}$ and that there is $a^{\prime }_0\in {\mathcal A}_{S'}^{\infty ,p}$ such that $a^{\prime }_0\mid a'$ . Then, $a_0:=\gcd (a^{\prime }_0,\ell _S)\in {\mathcal A}_S^\infty $ and $a_0\mid \gcd (a',\ell _S)=a$ . As $a\in {\mathcal A}_S^{\infty ,p}$ , this implies $a_0=a$ , so that $\gcd (a_0',\ell _S)=a$ .

(e) This follows from item (d).

Proof. Suppose there exists some $h\in U_{S_N}(\Delta (r))\cap \partial W$ . Then, $h_{S_n}\in (r+\ell _{S_N}{\mathbb Z})\cap ({\mathcal M}_{{\mathcal A}_{S_n}}\setminus {\mathcal M}_{S_n})$ for all $n\geqslant N$ , see Lemma 2.1 and Proposition 3.7. Hence, $r\not \in {\mathcal M}_{S_N}$ , and there are numbers $k_n\in {\mathbb Z}$ and $b_n\in {\mathcal B}\setminus S_n (n\geqslant N)$ such that $\gcd (b_n,\ell _{S_N})\mid \gcd (b_n,\ell _{S_n})\mid h_{S_n}=r+k_n\ell _{S_N}$ . In particular, $\gcd (b_n,\ell _{S_N})\mid r$ for all $n\geqslant N$ , and as ${\mathcal A}_{S_N}$ is a finite set, there exist $a\in {\mathcal A}_{S_N}$ and infinitely many $b_{n_i}$ such that $\gcd (b_{n_i},\ell _{S_N})=a$ . It follows that $a\in {\mathcal A}_{S_N}^\infty $ and $a\mid r$ . This proves equation (47).

Observe that, trivially, $h\in U_{S_n}(\Delta (h_{S_n}))\cap \partial W$ for each $n\geqslant N$ . Hence, we can apply equation (47) to n and $h_{S_n}$ instead of N and r, respectively. It follows that $h_{S_n}\in {\mathcal M}_{{\mathcal A}_{S_n}^\infty }\setminus {\mathcal M}_{S_n}$ . As $h_{S_n}=r+k_n\ell _{S_N}$ , this proves the ‘ $\Rightarrow $ ’-direction of equation (48) and also the final claim.

The ‘ $\Leftarrow $ ’-implication of equation (48) follows from Lemma 2.5 and Proposition 3.7, because ${\mathcal A}_{S_n}^\infty \subseteq {\mathcal A}_{S_n}$ for all $n\in {\mathbb N}$ .

Proof. For $n\in {\mathbb N}$ , denote by ${\mathcal U}_n$ the family of all sets $U_{S_n}(\Delta (r))$ that have non-empty intersection with $\partial W$ and by $\bigcup {\mathcal U}_n$ the union of these sets. Then, equation (47) implies $m_G(\partial W) \le m_G(\bigcup {\mathcal U}_n) = {|{\mathcal U}_n|}/{\ell _{S_n}} \le d({\mathcal M}_{{\mathcal A}_{S_n}^\infty }\setminus {\mathcal M}_{S_n}) \le d({\mathcal M}_{{\mathcal A}_{S_n}^{\infty ,p}})$ for all n, and the second inequality is the Heilbronn–Rohrbach inequality [Reference Hall15, Theorem 0.9].

Proof. Notice that in view of Remark 3.12,

$$ \begin{align*} r\in{\widetilde{\mathcal H}}_N\Leftrightarrow \text{for all } n\geqslant N: (r+\ell_{S_N}{\mathbb Z})\cap1({\mathcal M}_{{\mathcal A}_{S_n}^{\infty}}\setminus{\mathcal M}_{S_n})\neq\emptyset. \end{align*} $$

Hence,

$$ \begin{align*} \begin{aligned} r\not\in{\widetilde{\mathcal H}}_N \Leftrightarrow\, &\text{there exists } n\ge N: (r+\ell_{S_N}{\mathbb Z})\cap({\mathcal M}_{{\mathcal A}_{S_n}^{\infty}}\setminus{\mathcal M}_{S_n})=\emptyset\\\Leftrightarrow\, &\text{there exists } n\ge N\ \text{ for all } a_n\in{\mathcal A}_{S_n}^{\infty}: (r+\ell_{S_N}{\mathbb Z})\cap a_n{\mathbb Z}\subseteq{\mathcal M}_{S_n}\\\Leftrightarrow\, &(\text{there exists } n\ge N\ \text{ for all } a_n\in{\mathcal A}_{S_n}^{\infty}: (r+\ell_{S_N}{\mathbb Z})\cap a_n{\mathbb Z}\neq\emptyset\\&\Rightarrow \text{there exists } b\in S_n: (r+\ell_{S_N}{\mathbb Z})\cap a_n{\mathbb Z}\subseteq b{\mathbb Z})\\\Leftrightarrow\, &(\text{there exists } n\ge N\ \text{ for all } a_n\in{\mathcal A}_{S_n}^{\infty}: \gcd(a_n,\ell_{S_N})\mid r\\&\Rightarrow\text{there exists } b\in S_n: b\mid a_n\vee\ell_{S_N}\text{ and }\gcd(b,\ell_{S_N})\mid r). \end{aligned} \end{align*} $$

The third equivalence follows from Lemma 2.24 and the last one from Lemma 2.25. Now, the first of the following two equivalences is immediate:

(50) $$ \begin{align} &\hspace{-5pt}r\in{\widetilde{\mathcal H}}_N\nonumber\\ &\Leftrightarrow \text{for all } n\ge N\quad\text{there exists } a_n\in{\mathcal A}_{S_n}^{\infty}: \gcd(a_n,\ell_{S_N})\mid r \quad\text{and}\nonumber\\ &\quad\qquad[\text{for all } b\in S_n: b\mid a_n\vee\ell_{S_N}\Rightarrow\gcd(b,\ell_{S_N})\nmid r]\notag\\ &\Leftrightarrow \text{there exists } a\in{\mathcal A}_{S_N}^{\infty} \quad\text{there exists }\text{ an }(a,{\mathcal A})\text{-sequence }(a_n)_{n\ge N}: a\mid r\quad\text{and }\\ &\quad\qquad \text{for all } b\in{\mathcal B}\quad \text{for all } n\ge N: b\mid a_n\vee\ell_{S_N}\Rightarrow\gcd(b,\ell_{S_N})\nmid r.\notag \end{align} $$

The ‘ $\Leftarrow $ ’-direction of the second equivalence is obvious—just a matter of notation. For the ‘ $\Rightarrow $ ’-direction, we construct a suitable $(a,{\mathcal A})$ -sequence $(a_m')_{m\ge N}$ from the given numbers $a_n$ : there is $a\in {\mathcal A}_{S_N}^\infty $ such that $a=\gcd (a_n,\ell _{S_N})$ for infinitely many indices n. Obviously $a\mid r$ , and we choose $a_N'=a$ . Suppose inductively that suitable $a_N',\ldots ,a_m'$ are constructed in such a way that $a_m'=\gcd (a_{n},\ell _{S_m})$ for infinitely many different indices $n\ge m$ . Then there is an increasing subsequence $(a_{n_i})_i$ such that $\gcd (a_{n_i},\ell _{S_{m+1}})$ is the same value for all $n_i\ge m+1$ . This common value is denoted by $a_{m+1}'$ . It satisfies $\gcd (a_{m+1}',\ell _{S_m})=\gcd (a_{n_i},\ell _{S_{m+1}},\ell _{S_m})=\gcd (a_{n_i},\ell _{S_m})=a_m'$ for all $n_i$ . Suppose now that $b\in {\mathcal B}$ , $n\ge N$ and $b\mid a_n\vee \ell _{S_N}$ . Fix $n'\ge n$ such that $b\in S_{n'}$ . Then $b\mid a_{n'}\vee \ell _{S_n}$ , because $a_n\mid a_{n'}$ , and we conclude that $\gcd (b,\ell _{S_N})\nmid r$ . The claim follows.

Proof. Consider any fixed $a\in {\mathcal A}_{S_N}^\infty $ . In view of Proposition 3.16, the set $S_N(a)$ must be constructed in such a way that ${\mathcal M}_{S_N(a)}=\bigcap {\mathcal M}_{S_N((a_n)_{n\ge N})}$ , where the intersection runs over all $(a,{\mathcal A})$ -sequences with $a_N=a$ . As all sets $S_N((a_n)_{n\ge N})$ consist of divisors of $\ell _{S_N}$ , this set is only a finite intersection, say of r sets of multiples ${\mathcal M}_{R_1},\ldots ,{\mathcal M}_{R_r}$ . Hence, we may choose as $S_N(a)$ the primitivization of the set of all $c_1\vee \cdots \vee c_r$ , where $c_i\in R_i$ for $i=1,\ldots ,r$ .

Suppose for a contradiction that $a{\mathbb Z}\subseteq {\mathcal M}_{S_N(a)}$ . Then, $a{\mathbb Z}\subseteq {\mathcal M}_{S_N((a_n)_{n\ge N})}$ for each $(a,{\mathcal A})$ -sequence $(a_n)_{n\ge N}$ with $a_N=a$ . Hence, for each such sequence, there is $b\in {\mathcal B}$ such that $\gcd (b,\ell _{S_N})\mid a=\gcd (a_n,\ell _{S_N})$ and $b\mid a_n\vee \ell _{S_N}$ for some $n\ge N$ . It follows that $b^{\div \ell _{S_N}}\mid (a_n\vee \ell _{S_N})^{\div \ell _{S_N}}=a_n^{\div \ell _{S_N}}$ , so that $b=b^{\div \ell _{S_N}}\cdot \gcd (b,\ell _{S_N})\mid a_n^{\div \ell _{S_N}}\cdot \gcd (a_n,\ell _{S_N})=a_n$ . However, this is impossible, because $a_n\in {\mathcal A}_{S_n}^\infty $ and ${\mathcal B}$ is primitive.