Dynamics of $\mathcal B$-free sets: a view through the window

TL;DR

This paper explores the properties of $$-free sets using a window approach, linking their arithmetic and dynamical features to the topological and measure-theoretic characteristics of an associated set.

Contribution

It introduces a novel perspective by interpreting $$-free sets as weak model sets and characterizes their properties through the geometry of the window.

Findings

Tautness of $$ corresponds to Haar regularity of the window.

The associated dynamical system is a Toeplitz system if the window is topologically regular.

Proximality of the system occurs when the window has empty interior.

Abstract

Let $\mathcal B$ be an infinite subset of $\{1,2,\dots\}$. We characterize arithmetic and dynamical properties of the $\mathcal B$-free set $\mathcal F_{\mathcal B}$ through group theoretical, topological and measure theoretic properties of a set $W$ (called the window) associated with $\mathcal B$. This point of view stems from the interpretation of the set $\mathcal F_{\mathcal B}$ as a weak model set. Our main results are: $\mathcal B$ is taut if and only if the window is Haar regular; the dynamical system associated to $\mathcal F_{\mathcal B}$ is a Toeplitz system if and only if the window is topologically regular; the dynamical system associated to $\mathcal F_{\mathcal B}$ is proximal if and only if the window has empty interior; and the dynamical system associated to $\mathcal F_{\mathcal B}$ has the "na\"ively expected" maximal equicontinuous factor if and only if the interior of the window is aperiodic.

Equations

$${\mathcal{M}}_{\mathcal{B}}:=\bigcup _{b\in \mathcal{B}}b\text{mathbbm}Z$$

$${\mathcal{F}}_{\mathcal{B}}:=\text{mathbbm}Z\setminus {\mathcal{M}}_{\mathcal{B}}.$$

$${\mathcal{B}}^{prim}:=\mathcal{B}\setminus \bigcup _{b\in \mathcal{B}}b\cdot (\text{mathbbm}N\setminus \{1\}).$$

$$W:=\{h\in H:h_b\ne 0(\forall b\in \mathcal{B})\}.$$

$$U_S(h):=\{h^{\prime }\in H:\forall b\in S:h_b=h_b^{\prime }\},\text{ defined for finite }S\subset \mathcal{B}\text{ and }h\in H.$$

$$X_{\eta }=\overline{\phi (\Delta (\text{mathbbm}Z))}.$$

$$h_b=n\text{ mod }b\text{ for each }b\in S.$$

$$h\text{ satisfies the CRT iff}h\in H.$$

$$h\text{ satisfies the B-free CRT iff}h\in \overline{\Delta (\text{mathbbm}Z)\cap W}.$$

$${\mathcal{A}}_S:=\{\gcd (b,\mathrm{lcm}(S)):b\in \mathcal{B}\},$$

$${\mathcal{A}}_{\infty }:=\{n\in \text{mathbbm}N:{\forall }_{S\subset \mathcal{B}}{\exists }_{S^{\prime }:S\subseteq S^{\prime }}:n\in {\mathcal{A}}_{S^{\prime }}\setminus S^{\prime }\}.$$

$$\underset{k\to \infty }{\mathrm{limsup}}\left({\mathcal{A}}_{S_k}\setminus S_k\right)={\mathcal{A}}_{\infty }.$$

$$H_A:=\left\{h\in H:A+h=A\right\}$$

$$\Delta (1)+H_{\mathrm{int}(W)}$$

$$d_k:=\underset{j\to \infty }{\lim }\gcd (s_k,c_{k+j}),\text{ where }{s}_k:=\mathrm{lcm}(S_k)\text{ and }{c}_l:=\text{minimal period of }{\mathcal{M}}_{{\mathcal{A}}_{S_l}}.$$

$$0\to H_{\mathrm{int}(W)}\to H\cong \underleftarrow{\lim }\text{mathbbm}Z/s_k\text{mathbbm}Z\to \underleftarrow{\lim }\text{mathbbm}Z/d_k\text{mathbbm}Z\to 0$$

$$0\to M^{\prime }\stackrel{f}{⟶}M\stackrel{g}{⟶}{M}^{\prime \prime }\to 0$$

$$Y:=\left\{x\in \{0,1{\}}^{\text{mathbbm}Z}:\mathrm{card}(\mathrm{supp}(x)\text{ mod }b)=b-1\forall b\in \mathcal{B}\right\}.$$

$$W=\overline{\mathrm{int}W}\Rightarrow X_{\eta }\subseteq Y\Rightarrow H_W=H_{\overline{\mathrm{int}W}}.$$

$$\overline{d}(\mathcal{M})=\underset{N\to \infty }{\mathrm{limsup}}\frac{1}{N}\mathrm{card}\left(\mathcal{M}\cap \{1,\dots ,N\}\right)\text{ resp. }\underline{d}(\mathcal{M})=\underset{N\to \infty }{\mathrm{liminf}}\frac{1}{N}\mathrm{card}\left(\mathcal{M}\cap \{1,\dots ,N\}\right)$$

$$\mathbf{\delta }(\mathcal{M})=\underset{N\to \infty }{\lim }\frac{1}{\log N}\sum _{n\in \mathcal{M}\cap \{1,\dots ,N\}}\frac{1}{n}$$

$$\begin{array}{rl}\mathcal{B}\text{ is taut}& \iff \forall b\in \mathcal{B}:\underline{d}({\mathcal{M}}_{\mathcal{B}})>\underline{d}({\mathcal{M}}_{\mathcal{B}\setminus \{b\}})\iff \forall b\in \mathcal{B}:c^{-1}\underline{d}({\mathcal{M}}_{\mathcal{B}})>c^{-1}\underline{d}({\mathcal{M}}_{\mathcal{B}\setminus \{b\}})\\ & \iff \forall b\in \mathcal{B}:\underline{d}({\mathcal{M}}_{c\mathcal{B}})>\underline{d}({\mathcal{M}}_{c(\mathcal{B}\setminus \{b\})})=\underline{d}({\mathcal{M}}_{c\mathcal{B}\setminus \{cb\}})\\ & \iff \forall b^{\prime }\in c\mathcal{B}:\underline{d}({\mathcal{M}}_{c\mathcal{B}})>\underline{d}({\mathcal{M}}_{c\mathcal{B}\setminus \{b^{\prime }\}})\\ & \iff c\mathcal{B}\text{ is taut.}\end{array}$$

$${\mathcal{B}}^{\prime }(q)=\left\{\frac{b}{\gcd (b,q)}:b\in \mathcal{B}\right\},$$

$$c=\frac{\ell b}{q}=k\cdot \frac{b}{\gcd (b,q)}\in {\mathcal{M}}_{{\mathcal{B}}^{\prime }(q)}.$$

$$U_S(h):=\{h^{\prime }\in H:\forall b\in S:h_b=h_b^{\prime }\},\text{ defined for finite }S\subset \mathcal{B}\text{ and }h\in H,$$

$${\mathcal{Z}}_k:=\left\{U_{S_k}(\Delta (n)):n=1,\dots ,s_k\right\}.$$

$$\#{\mathcal{H}}_W^{\ell }/s_{\ell }⩽\#\left(c(\text{mathbbm}Z\setminus {\mathcal{M}}_{{\mathcal{A}}^{\ell }})\cap [0,s_l)\right)/s_{\ell }\le \#\left((\text{mathbbm}Z\setminus {\mathcal{M}}_{{\mathcal{A}}^{\ell }})\cap [0,s_l)\right)/s_{\ell }=d(\text{mathbbm}Z\setminus {\mathcal{M}}_{{\mathcal{A}}^{\ell }})\to 0.$$

$$U\cap W=\varnothing \text{or}{m}_H(U\cap W)>0.$$

$${\mathcal{G}}_{\ell }:=(s_k\text{mathbbm}Z+n)\cap \left\{m\in \text{mathbbm}Z:U_{S_{\ell }}(\Delta (m))\cap W=\varnothing \right\}=(s_k\text{mathbbm}Z+n)\cap {\mathcal{M}}_{S_{\ell }},$$

$${\mathcal{G}}_{\ell }={\mathcal{G}}_{\ell }+s_{\ell }\text{mathbbm}Z=\left({\mathcal{G}}_{\ell }\cap [0,s_{\ell })\right)+s_{\ell }\text{mathbbm}Z.$$

Full text

MSC 2010 clasification: 37A35, 37A45, 37B05.Keywords: ${\mathcal{B}}$-free dynamics, sets of multiples, maximal equicontinuous factor.

1 Introduction and main results

For any given set ${\mathcal{B}}\subseteq{\mathbbm{N}}=\{1,2,\dots\}$ one can define its set of multiples

[TABLE]

and the set of ${\mathcal{B}}$-free numbers

[TABLE]

The investigation of structural properties of ${\mathcal{M}}_{\mathcal{B}}$ or, equivalently, of ${\mathcal{F}}_{\mathcal{B}}$ has a long history (see the monograph [10] and the recent paper [4] for references), and dynamical systems theory provides some useful tools for this. Namely, denote by $\eta\in\{0,1\}^{\mathbbm{Z}}$ the characteristic function of ${\mathcal{F}}_{\mathcal{B}}$, i.e. $\eta(n)=1$ if and only if $n\in{\mathcal{F}}_{\mathcal{B}}$, and consider the orbit closure $X_{\eta}$ of $\eta$ in the shift dynamical system $(\{0,1\}^{\mathbbm{Z}},\sigma)$, where $\sigma$ stands for the left shift. Then topological dynamics and ergodic theory provide a wealth of concepts to describe various aspects of the structure of $\eta$, see [16] which originated this point of view by studying the set of square-free numbers, and also [1], [4] which continued this line of research.

In this paper we continue to provide a dictionary that characterizes arithmetic properties of ${\mathcal{B}}$ in terms of dynamical properties of $X_{\eta}$, and, as an intermediate step, also in terms of topological and measure theoretic properties of a pair $(H,W)$ associated with the passage from ${\mathcal{B}}$ to $X_{\eta}$, where $H$ is a compact abelian group and $W$ a compact subset of $H$. This latter point of view is borrowed from the theory of weak model sets, which applies here, because ${\mathcal{F}}_{\mathcal{B}}$ is a particular example of such a set, see e.g. [3, 13]. Finally the Chinese Remainder Theorem allows us to interpret our dynamical results combinatorially.

In order to formulate our main results, we need to recall some notions from the theory of sets of multiples [10] and also to introduce some further notation. Let ${\mathcal{B}}$ be a non-empty subset of ${\mathbbm{N}}$.

• •

${\mathcal{B}}$ is primitive, if there are no $b,b^{\prime}\in{\mathcal{B}}$ with $b\mid b^{\prime}$. From any set ${\mathcal{B}}\subseteq{\mathbbm{N}}$ one can remove all multiples of other numbers in ${\mathcal{B}}$, which results in the set

[TABLE]

${\mathcal{B}}^{prim}$ is primitive by construction, and ${\mathcal{M}}_{\mathcal{B}}={\mathcal{M}}_{{\mathcal{B}}^{prim}}$.

• •

${\mathcal{B}}$ is taut, if $\boldsymbol{\delta}({\mathcal{M}}_{{\mathcal{B}}\setminus\{b\}})<\boldsymbol{\delta}({\mathcal{M}}_{\mathcal{B}})$ for each $b\in{\mathcal{B}}$, where $\boldsymbol{\delta}({\mathcal{M}}_{\mathcal{B}}):=\lim_{n\to\infty}\frac{1}{\log n}\sum_{k\leqslant n,k\in{\mathcal{M}}_{\mathcal{B}}}k^{-1}$ denotes the logarithmic density of this set, which is known to exist by the Theorem of Davenport and Erdös [6, 7]. So a set is taut, if removing any single point from it changes its set of multiples drastically and not only by “a few points”.

• •

$\tilde{H}:=\prod_{b\in{\mathcal{B}}}{\mathbbm{Z}}/b{\mathbbm{Z}}$ and $\Delta:{\mathbbm{Z}}\to\tilde{H}$, $\Delta(n)=(n,n,\dots)$ – the canonical diagonal embedding.

• •

$H:=\overline{\Delta({\mathbbm{Z}})}$ is a compact abelian group, and we denote by $m_{H}$ its normalised Haar measure.

• •

$R_{\Delta(1)}:H\to H$ denotes the rotation by $\Delta(1)$, i.e. $(R_{\Delta(1)}h)_{b}=(h_{b}+1)$ mod $b$ for all $b\in{\mathcal{B}}$.

• •

The window is defined as

[TABLE]

• •

$\varphi:H\to\{0,1\}^{\mathbbm{Z}}$ is the coding function: $\varphi(h)(n)=1$, if and only if $R_{\Delta(1)}^{n}h\in W$, equivalently, if and only if $h_{b}+n\neq 0$ mod $b$ for all $b\in{\mathcal{B}}$.

• •

By $S,S^{\prime}\subset{\mathcal{B}}$ we always mean finite subsets.

• •

The topology on $H$ is generated by the (open and closed) cylinder sets

[TABLE]

A recurring theme of the main results in this paper is to characterize arithmetic and dynamical properties of a ${\mathcal{B}}$-free set ${\mathcal{F}}_{\mathcal{B}}$ through group theoretical, topological and measure theoretic properties of the window $W$ defined above.

Remark 1.1**.**

With the notation introduced above, we can write

[TABLE]

This is certainly a subset of $X_{\varphi}:=\overline{\varphi(H)}$, the set studied in [13] under the name ${\mathcal{M}}^{\scriptscriptstyle G}_{\scriptscriptstyle W}$. In Proposition 2.2 we show that $X_{\eta}=X_{\varphi}$ when ${\mathcal{B}}$ has light tails (see Subsection 2.5 for a definition), but we do not know whether also tautness of ${\mathcal{B}}$ suffices (see also Subsection 2.5).

1.1 Tautness as a measure theoretic property

Theorem A**.**

111The authors are indebted to J. Kułaga-Przymus for pointing out the relevance of [10, Lemma 1.17] for the proof of this theorem.

Suppose that the set ${\mathcal{B}}$ is primitive. Then the following are equivalent:

1. (i)

${\mathcal{B}}$* is taut.* 2. (ii)

The window $W$ is Haar regular, i.e. $\operatorname{supp}(m_{H}|_{W})=W$.

Moreover, these properties imply

1. (iii)

$\overline{\Delta({\mathbbm{Z}})\cap W}=W$.

The proof of the theorem is provided in Section 2. The concept of a Haar regular window was introduced in [14] in the context of general weak model sets.

Given a set ${\mathcal{B}}\subset{\mathbbm{N}}$, one says that $h:=(h_{b})_{b\in{\mathcal{B}}}\in{\mathbbm{Z}}^{{\mathcal{B}}}$ satisfies the CRT (Chinese Remainder Theorem) if for each finite $S\subset{\mathcal{B}}$ there exists $n\in{\mathbbm{Z}}$ such that

[TABLE]

Clearly,

[TABLE]

We are looking for solutions of (3) with $n\in{\mathcal{F}}_{\mathcal{B}}$. If for $h$ as above we can solve (3) with $n=n_{S}\in{\mathcal{F}}_{\mathcal{B}}$ for all finite $S\subset{\mathcal{B}}$, then we say that $h$ satisfies the ${\mathcal{B}}$-free CRT. A necessary condition for $h=(h_{b})_{b\in{\mathcal{B}}}$ to satisfy the ${\mathcal{B}}$-free CRT is, of course, that $h_{b}\neq 0\text{ mod }b$ for each $b\in{\mathcal{B}}$, and a moment’s reflection shows that

[TABLE]

Therefore the implication $(i)$ $\Rightarrow$ $(iii)$ of Theorem A is an immediate consequence of the following proposition.

Proposition 1.1**.**

Assume that ${\mathcal{B}}$ is taut. Let $h\in W$ and $S\subset{\mathcal{B}}$ finite. Then the set of ${\mathcal{B}}$-free integers $n$ that solve $n=h_{b}$ mod $b$ for $b\in S$ has asymptotic density $m_{H}(U_{S}(h)\cap W)>0$.

In Subsection 2.4 we provide a sequence ${\mathcal{B}}$, which is not taut, but for which $\overline{\Delta({\mathbbm{Z}})\cap W}=W$ (Example 2.2). Hence $(iii)$ of Theorem A is not equivalent to $(i)$ and $(ii)$. Here we provide two simpler examples which throw some light on property $(iii)$. Denote by ${\mathcal{P}}\subseteq{\mathbbm{N}}$ the set of all prime numbers.

Example 1.1**.**

If ${\mathcal{B}}={\mathcal{P}}$ then $H=\prod_{p\in{\mathcal{P}}}{\mathbbm{Z}}/p{\mathbbm{Z}}$, $W$ is uncountable (although of Haar measure zero) and $\overline{\Delta({\mathbbm{Z}})\cap W}\neq W$, since for each $n$ we find $p\in{\mathcal{P}}$ such that $p\mid n$, so $n=0$ mod $p$.

Example 1.2**.**

If ${\mathcal{B}}\subset{{\mathcal{P}}}$ is thin, i.e. if $\sum_{p\in{\mathcal{B}}}1/p<+\infty$, then $\overline{\Delta({\mathbbm{Z}})\cap W}=W$ in view of (4), because each $h\in H$ satisfies the ${\mathcal{B}}$-free CRT. Indeed, if $S\subset{\mathcal{B}}$ is finite and $n=h_{b}\text{ mod }b$ for $b\in S$, then $n+\operatorname{lcm}(S){\mathbbm{Z}}$ is the set of all solutions to this system of congruences. Moreover, if $h\in W$, then $\gcd(n,b)=1$ for all $b\in S$. We only need to find $r\in{\mathbbm{Z}}$ so that $n+r\operatorname{lcm}(S)$ is a prime number which is not in ${\mathcal{B}}$. The latter follows from Dirichlet’s theorem: The set of prime numbers contained in $n+\operatorname{lcm}(S){\mathbbm{Z}}$ is not thin. Of course this is a special case of Theorem A.

Remark 1.2**.**

Denote by $\nu_{\eta}:=m_{H}\circ\varphi^{-1}$ the Mirsky measure on $X_{\eta}$. There are two independent proofs of the fact that the two equivalent conditions from Theorem A imply that the measure preserving dynamical system $(X_{\eta},\sigma,\nu_{\eta})$ is isomorphic to the group rotation $(H,R_{\Delta(1)},m_{H})$: In [4, Theorem F] it is proved that this is implied by $(i)$. That it is also a direct consequence of $(ii)$ follows - in the more general context of model sets - from [14]. The proof uses our observation that $W$ is aperiodic (see Proposition 5.1). To see this, denote by $H_{W}:=\{h\in H:W+h=W\}$ the period group of $W$ and by $H_{W}^{Haar}:=\{h\in H:m_{H}((W+h)\triangle W)=0\}$ its group of Haar periods. It is easily seen that $H_{W}=H_{W}^{Haar}$ for Haar regular $W$, in particular whenever the sequence ${\mathcal{B}}$ is taut. Hence, if $W$ is aperiodic, it is also Haar aperiodic, and this is what is needed to apply the general theorem from [14] to the present context.

A word of caution is in order at this point: Althoug, in the ${\mathcal{B}}$-free context, the window $W$ is always aperiodic (Proposition 5.1), this is not necessarily true for its Haar regularization $W_{reg}:=\operatorname{supp}(m_{H}|_{W})$, because that window is not of the same arithmetic type as $W$. On the other hand, as proved in [4, Theorem C], each non-taut set ${\mathcal{B}}$ can be modified into a taut set ${\mathcal{B}}^{\prime}$ whose corresponding Mirsky measure $\nu_{\eta^{\prime}}$ coincides with $\nu_{\eta}$ (as a measure on $\{0,1\}^{\mathbbm{Z}}$). The (arithmetic!) window $W^{\prime}\subseteq H^{\prime}$ defined by ${\mathcal{B}}^{\prime}$ is then aperiodic and Haar regular, and we suspect that it to be closely related to $W_{reg}\subseteq H$.

1.2 The proximal and the Toeplitz case

From [4, Theorem A] we know that $X_{\eta}$ has a unique minimal subset $M$. In Lemma 3.10, we prove that $M=\overline{\varphi(C_{\varphi})}$, where $C_{\varphi}$ denotes the set of continuity points of $\varphi:H\to\{0,1\}^{\mathbbm{Z}}$, see also [13, Lemma 6.3]. $M$ is degenerate to a singleton, namely to $M=\{(\dots,0,0,0,\dots)\}$, if and only if $\operatorname{int}(W)=\emptyset$ [13], and we collect a number of equivalent characterizations of this extreme case in Theorem C below. Assuming primitivity of ${\mathcal{B}}$ and property $(iii)$ of Theorem A, we prove the following equivalent characterizations of minimality of $(X_{\eta},\sigma)$, i.e. of $M=X_{\eta}$, in Subsection 3.2. For $S\subset{\mathcal{B}}$ let

[TABLE]

and note that ${\mathcal{F}}_{{\mathcal{A}}_{S}}\subseteq{\mathcal{F}}_{\mathcal{B}}$, because $b\mid m$ for some $b\in{\mathcal{B}}$ implies $\gcd(b,\operatorname{lcm}(S))\mid m$ for any $S\subset{\mathcal{B}}$. Let

[TABLE]

In Lemma 3.2 we prove: If $(S_{k})_{k}$ is a filtration of ${\mathcal{B}}$ with finite sets, then

[TABLE]

Theorem B**.**

Suppose that ${\mathcal{B}}$ is primitive. Consider the following list of properties:

1. (B)

The window $W$ is topologically regular, i.e. $\overline{\operatorname{int}(W)}=W$. 2. (B)

${\mathcal{F}}_{\mathcal{B}}=\bigcup_{S\subset{\mathcal{B}}\text{ finite}}{\mathcal{F}}_{{\mathcal{A}}_{S}}$. 3. (B)

${\mathcal{A}}_{\infty}=\emptyset$. 4. (B)

There are no $d\in{\mathbbm{N}}$ and no infinite pairwise coprime set ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$ such that $d\,{\mathcal{A}}\subseteq{\mathcal{B}}$. 5. (B)

$\eta=\varphi(0)$* is a Toeplitz sequence (see [8], [12] for the definition) different from $(\dots,0,0,0,\dots)$.* 6. (B)

$0\in C_{\varphi}$* and $\varphi(0)\neq(\dots,0,0,0,\dots)$.* 7. (B)

$\eta\in M$* and $\eta\neq(\dots,0,0,0,\dots)$.* 8. (B)

$X_{\eta}$* is minimal., i.e. $X_{\eta}=M$, and $\operatorname{card}(X_{\eta})>1$.* 9. (B)

The dynamics on $X_{\eta}$ is a minimal almost 1-1 extension of $(H,R_{\Delta(1)})$, the rotation by $\Delta(1)$ on $H$.

1. a)

(B1) - (B6) are all equivalent, and each of these conditions implies that ${\mathcal{B}}$ is taut. 2. b)

(B7) and (B8) are equivalent. 3. c)

Each of (B1) - (B6) implies (B9). 4. d)

(B9) implies (B7) and (B8). 5. e)

If $\overline{\Delta({\mathbbm{Z}})\cap W}=W$ (in particular if ${\mathcal{B}}$ is taut), then (B1) - (B9) are all equivalent.

Remark 1.3**.**

One ingredient of the proof of Theorem B is the observation that the set ${\mathcal{B}}$ is taut whenever $\eta$ is a Toeplitz sequence. This was pointed out to us by A. Bartnicka who also gave a proof of it, which we recall in Lemma 3.7 below.

Moreover, we can interpret the result purely arithmetically as follows: If ${\mathcal{B}}$ is primitive and satisfies (B4) then the set of elements for which the ${\mathcal{B}}$-free CRT holds is topologically regular, i.e. it contains a dense subset of points for which all sufficiently close points satisfying the CRT satisfy also the ${\mathcal{B}}$-free CRT.

The following characterization of regular Toeplitz sequences is included in Proposition 4.1 in Subsection 4.2, where also the precise definition of regularity of a Toeplitz sequence is recalled.

Proposition 1.2**.**

Assume that $\overline{\operatorname{int}(W)}=W$. Then the Toeplitz sequence $\eta$ is regular, if and only if $m_{H}(\partial W)=0$.

In Subsection 4.2 we also provide examples of sets ${\mathcal{B}}$ that give rise to regular Toeplitz sequences and others giving rise to irregular Toeplitz sequences. Note also that $m_{H}(\partial W)=0$ if and only if $\inf_{S\subset{\mathcal{B}}}\bar{d}({\mathcal{M}}_{{\mathcal{A}}_{S}}\setminus{\mathcal{M}}_{\mathcal{B}})=0$, see Lemma 4.3, and observe that $m_{H}(\partial W)=0$ implies unique ergodicity of the dynamics on $X_{\eta}$ [13, Theorem 2c].

The next theorem is complementary to Theorem B. Most of its equivalences follow from results in [4] and [13] and are proved in Subsection 3.3. They do not rely on the more advanced arithmetic concept of tautness.

Theorem C**.**

The following are equivalent:

1. (C)

$\operatorname{int}(W)=\emptyset$** 2. (C)

$\bigcup_{S\subset{\mathcal{B}}\text{ finite}}{\mathcal{F}}_{{\mathcal{A}}_{S}}=\emptyset$, i.e. ${\mathcal{F}}_{{\mathcal{A}}_{S}}=\emptyset$ for all finite $S\subset{\mathcal{B}}$. 3. (C)

$\forall S\subset{\mathcal{B}}:1\in{\mathcal{A}}_{S}$. 4. (C)

${\mathcal{B}}$* contains an infinite pairwise coprime subset.* 5. (C)

If ${\mathcal{C}}\subseteq{\mathbbm{N}}$ is finite and if ${\mathcal{B}}\subseteq{\mathcal{M}}_{\mathcal{C}}$, then $1\in{\mathcal{C}}$. 6. (C)

$M=\{(\dots,0,0,0,\dots)\}$. 7. (C)

The dynamics on $X_{\eta}$ are proximal.

Remark 1.4**.**

Under the conditions of Theorem C no element of $W$ is stable, that is, for each $h$ staisfying the ${\mathcal{B}}$-free CRT there is an element $h^{\prime}\in{\mathbbm{Z}}^{{\mathcal{B}}}$ arbitrarily close to $h$ which satisfies the CRT but not the ${\mathcal{B}}$-free CRT.

1.3 The maximal equicontinuous factor

We finish with a result that identifies the maximal equicontinuous factor of the dynamics on $X_{\eta}$ and answers Question 3.14 in [4]. Given a subset $A\subseteq H$, denote by

[TABLE]

the period group of $A$. The set $A\subseteq H$ is topologically aperiodic, if $H_{A}=\{0\}$. Observe also that $H_{\operatorname{int}(A)}$ is a closed subgroup of $H$, whenever $A$ is closed [14, Lemma 6.1].

In Proposition 5.1 we prove that $H_{W}=\{0\}$ whenever ${\mathcal{B}}$ is primitive. If $\operatorname{int}(W)=\emptyset$, then of course $H_{\operatorname{int}(W)}=H$. If $\operatorname{int}(W)\neq\emptyset$, the situation is more complicated: $H_{\operatorname{int}(W)}$ is obviously always a strict subgroup of $H$, and very often $H_{\operatorname{int}(W)}=\{0\}$, but there are examples where $H_{\operatorname{int}(W)}$ is a non-trivial strict subgroup of $H$, see Subsection 5.3. In any case, however, $H_{\operatorname{int}(W)}$ determines the maximal equicontinuous factor. The following is proved in [14, Theorem A2]:

[TABLE]

Let $S_{1}\subset S_{2}\subset\dots$ be any filtration of ${\mathcal{B}}$ by finite sets. In Subsection 5.1 we define divisors $d_{k}$ of $\operatorname{lcm}(S_{k})$:

[TABLE]

By Remark 5.1 we have $\frac{s_{k}}{d_{k}}\mid\frac{s_{k+1}}{d_{k+1}}$ for any $k$. The sequences $(s_{k})$, $(d_{k})$ and $(c_{k})$ determine $H_{\operatorname{int}(W)}$ in the following way:

Proposition 1.3**.**

1. a)

[TABLE]

*is an exact sequence.*222A sequence of abelian groups and homomorphisms $...\smash{\mathop{\longrightarrow}\limits}M_{k-1}\smash{\mathop{\longrightarrow}\limits^{f_{k-1}}}M_{k}\smash{\mathop{\longrightarrow}\limits^{f_{k}}}M_{k+1}\rightarrow...$ is called exact if the kernel of $f_{k}$ is equal to the image of $f_{k-1}$ for any $k$. In particular, a sequence

$0\rightarrow M^{\prime}\smash{\mathop{\longrightarrow}\limits^{f}}M\smash{\mathop{\longrightarrow}\limits^{g}}M^{\prime\prime}\rightarrow 0$

is exact, when $f$ is injective, the kernel of $g$ equals the image of $f$ and $g$ is surjective. We say that it is a ”short exact sequence”. In particular, the homomorphism $g$ induces an isomorphism $M^{\prime\prime}\cong M/f(M^{\prime})$ in this case. ** 2. b)

$H_{\operatorname{int}(W)}\cong\lim\limits_{\leftarrow}{\mathbbm{Z}}/\frac{s_{k}}{d_{k}}{\mathbbm{Z}}$. 3. c)

$H/H_{\operatorname{int}(W)}\cong\lim\limits_{\leftarrow}{\mathbbm{Z}}/{d_{k}}{\mathbbm{Z}}$. 4. d)

$H_{\operatorname{int}(W)}=\{0\}$* if and only if $s_{k}=d_{k}$ for each $k\in{\mathbbm{N}}$, equivalently if for each $b\in{\mathcal{B}}$ there is $n>0$ such that $b$ divides $c_{n}$.*

Theorem D**.**

1. a)

The translation by $(1,1,\dots)$ on $H/H_{\operatorname{int}(W)}\cong\lim\limits_{\leftarrow}{\mathbbm{Z}}/{d_{k}}{\mathbbm{Z}}$ is the maximal equicontinuous factor of the dynamics on $X_{\eta}$. 2. b)

In case d) of Proposition 1.3, the translation by $\Delta(1)$ on $H\cong\lim\limits_{\leftarrow}{\mathbbm{Z}}/{s_{k}}{\mathbbm{Z}}$ is the maximal equicontinuous factor of the dynamics on $X_{\eta}$.

In Subsection 5.3 we provide a number of examples illustrating this theorem.

Remark 1.5**.**

In [4], the following set $Y$ is defined: 333Versions of this set occur also in [15] and [2].

[TABLE]

Observe that $\operatorname{card}(\operatorname{supp}(x)\text{ mod }b)\leqslant b-1$ for all $x\in X_{\eta}$ and $b\in{\mathcal{B}}$. 444Indeed, if $\operatorname{card}(\operatorname{supp}(x)\text{ mod }b)=b$ for some $x\in X_{\eta}$ and $b\in{\mathcal{B}}$, then this happens on some integer interval $[-M,M]$, and hence $\operatorname{card}(\operatorname{supp}(\eta)\text{ mod }b)=b$, which contradicts the fact that $\operatorname{supp}(\eta)\subseteq{\mathcal{F}}_{\mathcal{B}}$. Proposition 3.27 of [4] asserts that $(H,R_{\Delta(1)})$ is the maximal equicontinuous factor of $(X_{\eta},S)$, whenever $X_{\eta}\subseteq Y$. Hence, in that case, $H_{\overline{\operatorname{int}W}}=H_{\operatorname{int}W}=\{0\}=H_{W}$ by Theorem D and Proposition 5.1. This is the second one of the following two implications:

[TABLE]

The first one is proved in Proposition 3.3.

2 Tautness of ${\mathcal{B}}$ and Haar regularity of $W$

2.1 Arithmetic of ${\mathcal{B}}$ and topology of $W$, part I

Definition 2.1**.**

Let ${\mathcal{M}}\subseteq{\mathbbm{N}}$.

1. a)

The upper resp. lower density of ${\mathcal{M}}$ is

[TABLE]

If the limit exists, we write $d({\mathcal{M}})$. 2. b)

The logarithmic density of ${\mathcal{M}}$ is

[TABLE]

whenever the limit exists.

The theorem of Davenport and Erdös [6, 7] asserts that $\boldsymbol{\delta}({\mathcal{M}}_{\mathcal{B}})=\underline{d}({\mathcal{M}}_{\mathcal{B}})$ exists for any subset ${\mathcal{B}}\subseteq{\mathbbm{N}}$.

Definition 2.2**.**

${\mathcal{B}}\subseteq{\mathbbm{N}}\setminus\{1\}$* is a Behrend sequence, if $\boldsymbol{\delta}({\mathcal{M}}_{\mathcal{B}})=1$.*

Recall that ${\mathcal{B}}$ is taut, if $\boldsymbol{\delta}({\mathcal{M}}_{{\mathcal{B}}\setminus\{b\}})<\boldsymbol{\delta}({\mathcal{M}}_{\mathcal{B}})$ for each $b\in{\mathcal{B}}$. The following is a corollary to a theorem of Behrend [5]:

Proposition 2.1**.**

A set ${\mathcal{B}}\subseteq{\mathbbm{N}}$ is taut, if and only if it is primitive and there are no $q\in{\mathbbm{N}}$ and no Behrend set ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$ such that $q\,{\mathcal{A}}\subseteq{\mathcal{B}}$ [10, Corollary 0.19].

This motivates the next definition:

Definition 2.3**.**

A set ${\mathcal{B}}\subseteq{\mathbbm{N}}$ is pre-taut, if there are no $q\in{\mathbbm{N}}$ and Behrend set ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$ such that $q\,{\mathcal{A}}\subseteq{\mathcal{B}}$.

Lemma 2.1**.**

Let ${\mathcal{B}}\subseteq{\mathbbm{N}}$ and $c\in{\mathbbm{N}}$.

1. a)

If $c\,{\mathcal{B}}$ is pre-taut, then also ${\mathcal{B}}$ is pre-taut. Moreover, ${\mathcal{B}}$ is taut if and only if $c\,{\mathcal{B}}$ is taut. 2. b)

Each subset of a (pre-)taut set is (pre-)taut. 3. c)

A finite union of pre-taut sets is pre-taut. 4. d)

If ${\mathcal{B}}$ is taut, then ${\mathcal{B}}=\{1\}$ or $d({\mathcal{M}}_{\mathcal{B}})\neq 1$ (possibly non-existing). Equivalently, if $d({\mathcal{M}}_{\mathcal{B}})=1$, then ${\mathcal{B}}=\{1\}$ or ${\mathcal{B}}$ is not taut. 5. e)

If ${\mathcal{B}}$ is pre-taut, then $1\in{\mathcal{B}}$ or $d({\mathcal{M}}_{\mathcal{B}})\neq 1$ (possibly non-existing). Equivalently, if $d({\mathcal{M}}_{\mathcal{B}})=1$, then $1\in{\mathcal{B}}$ or ${\mathcal{B}}$ is not pre-taut.

Proof.

a) The first implication is obvious. It is also clear that ${\mathcal{B}}$ is primitive if and only if $c{\mathcal{B}}$ is primitive. Moreover,

[TABLE]

b) is obvious (see Remark 2.1).

c) follows from [10, Corollary 0.14], see also [4, Proposition 2.33].

d) Suppose that ${\mathcal{B}}$ is taut. Then ${\mathcal{B}}$ is primitive, and $d({\mathcal{M}}_{\mathcal{B}})\neq 1$ unless $1\in{\mathcal{B}}$ by [10, Corollary 0.19]. Hence $d({\mathcal{M}}_{\mathcal{B}})\neq 1$ or ${\mathcal{B}}=\{1\}$.

e) follows directly from Definition 2.3 555Note that d) follows from e) and Remark 2.2.. ∎

Remark 2.1**.**

${\mathcal{B}}$ is taut if and only if it is pre-taut and primitive. If ${\mathcal{B}}$ is pre-taut, then ${\mathcal{B}}^{prim}$ is taut in view of Lemma 2.1b.

For $q\in{\mathbbm{N}}$ and ${\mathcal{B}}\subseteq{\mathbbm{Z}}$ let

[TABLE]

and note that $1\in{\mathcal{B}}^{\prime}(q)$ if and only if $q\in{\mathcal{M}}_{\mathcal{B}}$.

Lemma 2.2**.**

Let $q\in{\mathbbm{N}}$, ${\mathcal{B}},{\mathcal{C}}\subseteq{\mathbbm{Z}}$, and $q\,{\mathcal{C}}\subseteq{\mathcal{M}}_{\mathcal{B}}$. Then ${\mathcal{M}}_{\mathcal{C}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$.

Proof.

Let $c\in{\mathcal{C}}$. There are $\ell\in{\mathbbm{Z}}$ and $b\in{\mathcal{B}}$ such that $qc=\ell b$. Since $q\mid\ell b$, it follows that $q\mid\ell\gcd(b,q)$, thus $k=\frac{\ell\gcd(b,q)}{q}$ is an integer. We have

[TABLE]

This shows that ${\mathcal{C}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$ and hence also ${\mathcal{M}}_{\mathcal{C}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$. ∎

Lemma 2.3**.**

Let ${\mathcal{B}}\subseteq{\mathbbm{N}}$ and $q\in{\mathbbm{N}}$.

1. a)

If ${\mathcal{B}}$ is pre-taut, then ${\mathcal{B}}^{\prime}(q)$ is pre-taut. 2. b)

If ${\mathcal{B}}$ is taut, then ${\mathcal{B}}^{\prime}(q)$ is a finite disjoint union of taut sets ${\mathcal{B}}_{i}^{\prime}$ defined below in the proof of a). 3. c)

If $d({\mathcal{M}}_{\mathcal{B}})=1$, then $d({\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)})=1$. 4. d)

If ${\mathcal{B}}=\bigcup_{i=1}^{N}{\mathcal{C}}_{i}$ and if $d({\mathcal{M}}_{\mathcal{B}})=1$, then $d({\mathcal{M}}_{{\mathcal{C}}_{i}})=1$ for at least one $i\in\{1,\dots,N\}$.

Proof.

Let $I:=\left\{\frac{q}{\gcd(b,q)}:b\in{\mathcal{B}}\right\}$. For $i\in I$ denote ${\mathcal{B}}_{i}:=\left\{b\in{\mathcal{B}}:\frac{q}{\gcd(b,q)}=i\right\}$ and ${\mathcal{B}}_{i}^{\prime}:=\left\{\frac{b}{\gcd(b,q)}:b\in{\mathcal{B}}_{i}\right\}$. Then $I$ is finite, ${\mathcal{B}}=\bigcup_{i\in I}{\mathcal{B}}_{i}$, and ${\mathcal{B}}^{\prime}(q)=\bigcup_{i\in I}{\mathcal{B}}_{i}^{\prime}$. Moreover, ${\mathcal{B}}_{i}=\left\{\frac{q}{i}b^{\prime}:b^{\prime}\in{\mathcal{B}}_{i}^{\prime}\right\}=\frac{q}{i}{\mathcal{B}}_{i}^{\prime}$.

a)  If ${\mathcal{B}}$ is pre-taut, then all ${\mathcal{B}}_{i}$ are pre-taut (Lemma 2.1b), then all ${\mathcal{B}}_{i}^{\prime}$ are pre-taut (Lemma 2.1a), and then ${\mathcal{B}}^{\prime}(q)$ is pre-taut (Lemma 2.1c).

b) If ${\mathcal{B}}$ is taut, then all ${\mathcal{B}}_{i}$ are taut (Lemma 2.1b), and then all ${\mathcal{B}}_{i}^{\prime}$ are taut (Lemma 2.1a).

c) As ${\mathcal{B}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$, we have also ${\mathcal{M}}_{\mathcal{B}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$.

d) If ${\mathcal{B}}$ is Behrend, then at least one of the sets ${\mathcal{C}}_{i}$ is Behrend [10, Corollary 0.14], and so ${d({\mathcal{M}}_{{\mathcal{C}}_{i}})=1}$. Otherwise $1\in{\mathcal{B}}$, so that $1\in{\mathcal{C}}_{i}$ for some $i$, whence ${\mathcal{M}}_{{\mathcal{C}}_{i}}={\mathbbm{Z}}$. ∎

Lemma 2.4**.**

(compare [4, Proposition 4.25]) Assume that ${\mathcal{B}}\subseteq{\mathbbm{N}}$ is taut and $d({\mathcal{M}}_{\mathcal{C}})=1$ for some ${\mathcal{C}}\subseteq{\mathbbm{Z}}$. If $q\,{\mathcal{C}}\subseteq{\mathcal{M}}_{\mathcal{B}}$ for some $q\geqslant 1$, then $b\mid q$ for some $b\in{\mathcal{B}}$.

Proof.

By Lemma 2.2, ${\mathcal{M}}_{\mathcal{C}}\subseteq{\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)}$, so that $d({\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)})=1$. Then ${\mathcal{B}}^{\prime}(q)=\{1\}$ or ${\mathcal{B}}^{\prime}(q)$ is not taut (Lemma 2.1d). If ${\mathcal{B}}^{\prime}(q)=\{1\}$, then $\gcd(b,q)=b$ for all $b\in{\mathcal{B}}$, i.e $b\mid q$ for all $b\in{\mathcal{B}}$, which is impossible because ${\mathcal{B}}$ is infinite. Hence ${\mathcal{B}}^{\prime}(q)$ is not taut. On the other hand, as ${\mathcal{B}}$ is taut by assumption, ${\mathcal{B}}^{\prime}(q)$ is a finite union of taut sets ${\mathcal{B}}_{i}^{\prime}$ (Lemma 2.3b). As $d({\mathcal{M}}_{{\mathcal{B}}^{\prime}(q)})=1$, also $d({\mathcal{M}}_{{\mathcal{B}}_{i}^{\prime}})=1$ for at least one of the sets ${\mathcal{B}}_{i}^{\prime}$ (Lemma 2.3d), so that ${\mathcal{B}}_{i}^{\prime}=\{1\}$ for this set (Lemma 2.1d). This implies $q\in{\mathcal{M}}_{\mathcal{B}}$. ∎

Recall that the topology on $H$ is generated by the (open and closed) cylinder sets

[TABLE]

and recall also the definition of ${\mathcal{A}}_{S}:=\{\gcd(b,\operatorname{lcm}(S)):b\in{\mathcal{B}}\}$. Note that ${\mathcal{A}}_{S}$ is finite and $S\subseteq{\mathcal{A}}_{\mathcal{S}}$.

Lemma 2.5**.**

Let $U=U_{S}(\Delta(n))$ for some $S\subset{\mathcal{B}}$ and $n\in{\mathbbm{Z}}$.

1. a)

If $n\in{\mathcal{M}}_{S}$, then $U\cap W=\emptyset$. 2. b)

If $U\cap W=\emptyset$, then $n+\operatorname{lcm}(S)\cdot{\mathbbm{Z}}\subseteq{\mathcal{M}}_{{\mathcal{B}}\cap{\mathcal{A}}_{S}}$. 3. c)

There is a filtration of ${\mathcal{B}}$ by finite sets $S$ for which ${\mathcal{B}}\cap{\mathcal{A}}_{S}=S$. 4. d)

If ${\mathcal{B}}\cap{\mathcal{A}}_{S}=S$, then $n\in{\mathcal{M}}_{S}$ iff $U\cap W=\emptyset$ iff $n+\operatorname{lcm}(S)\cdot{\mathbbm{Z}}\subseteq{\mathcal{M}}_{S}$.

Proof.

a)  This follows immediately from the definitions of $U_{S}(\Delta(n))$ and $W$.

b)  For each $h\in U$ there is $b\in{\mathcal{B}}$ such that $h_{b}=0$. As $U$ is compact, the Heine-Borel argument produces a finite set $S^{\prime}\subset{\mathcal{B}}$ such that for each $h\in U$ there is $b\in S^{\prime}$ such that $h_{b}=0$. Let $s=\operatorname{lcm}(S)$. This observation applies in particular to all $h\in\Delta(n+s{\mathbbm{Z}})\subseteq U_{S}(\Delta(n))=U$. That means, for each $k\in{\mathbbm{Z}}$ there is $b_{k}\in S^{\prime}$ such that $b_{k}\mid n+sk$. In other words: $n+s{\mathbbm{Z}}\subset{\mathcal{M}}_{S^{\prime}}$. The set $S^{\prime}$ need not be primitive automatically, but we can replace it w.l.o.g. by a primitive subset without changing its set of multiples. Then, as $S^{\prime}$ is finite, it is taut. Denote $q=\gcd(n,s)$ and ${\mathcal{C}}=\frac{n}{q}+\frac{s}{q}\cdot{\mathbbm{Z}}$. Then $q\,{\mathcal{C}}=n+s{\mathbbm{Z}}\subseteq{\mathcal{M}}_{S^{\prime}}$, and as $\gcd(n/q,s/q)=1$, $d({\mathcal{M}}_{\mathcal{C}})=1$ (Dirichlet, see [4, Corollary 4.24]). Now Lemma 2.4 shows that $b\mid q=\gcd(n,s)$ for some $b\in S^{\prime}$. In particular, $n\in b{\mathbbm{Z}}$ and $b\mid s=\operatorname{lcm}(S)$ for that $b\in S^{\prime}$, so that $b=\gcd(b,s)\in{\mathcal{B}}\cap{\mathcal{A}}_{S}$ and $n+\operatorname{lcm}(S)\cdot{\mathbbm{Z}}\subseteq b{\mathbbm{Z}}\subseteq{\mathcal{M}}_{{\mathcal{B}}\cap{\mathcal{A}}_{S}}$.

c)  It suffices to prove that for any finite $S\subset{\mathcal{B}}$ there exists a finite $S^{\prime}\subset{\mathcal{B}}$ with ${\mathcal{B}}\cap{\mathcal{A}}_{S^{\prime}}=S^{\prime}$. So let $S\subset{\mathcal{B}}$ and $S^{\prime}:={\mathcal{B}}\cap{\mathcal{A}}_{S}$. $S^{\prime}$ is finite, because ${\mathcal{A}}_{S}$ is finite, and obviously $S\subseteq S^{\prime}\subseteq{\mathcal{B}}\cap{\mathcal{A}}_{S^{\prime}}$. As each $b^{\prime}\in S^{\prime}\subseteq{\mathcal{A}}_{S}$ divides $\operatorname{lcm}(S)$, also $\operatorname{lcm}(S^{\prime})$ divides $\operatorname{lcm}(S)$. Therefore $\operatorname{lcm}(S^{\prime})=\operatorname{lcm}(S)$, so that ${\mathcal{A}}_{S^{\prime}}={\mathcal{A}}_{S}$. Hence $S^{\prime}={\mathcal{B}}\cap{\mathcal{A}}_{S^{\prime}}$.

d)  This follows from a) and b). ∎

2.2 Proof of Theorem A

Let ${\mathcal{B}}=\{b_{1},b_{2},\dots\}$ be primitive, and denote $S_{1}\subset S_{2}\subset\dots\subset{\mathcal{B}}$ a filtration of ${\mathcal{B}}$ by finite sets $S_{k}$. Let $s_{k}=\operatorname{lcm}(S_{k})$. We can assume without loss of generality that $b\mid s_{k}\Rightarrow b\in S_{k}$ holds for all $b\in{\mathcal{B}}$ and all $k\in{\mathbbm{N}}$. For each $k\in{\mathbbm{N}}$, the collection of all cylinder sets $U_{S_{k}}(h)$, $h\in H$, can be written explicitly as

[TABLE]

Suppose first that ${\mathcal{B}}$ is not taut. Then it contains a scaled copy $c{\mathcal{A}}$ of a Behrend set ${\mathcal{A}}\subseteq\{2,3,\dots\}$. Enlarging ${\mathcal{A}}$, if necessary, we can assume that $c{\mathcal{A}}={\mathcal{B}}\cap c{\mathbbm{Z}}$. (As ${\mathcal{B}}$ is primitive, also the enlarged ${\mathcal{A}}$ does not contain the number $1$.) Let $a_{0}>1$ be the smallest element of ${\mathcal{A}}$ and denote $b_{0}=ca_{0}$. Let $H_{0}=\{h\in H:h_{b_{0}}\in c{\mathbbm{Z}}\}$. Then $H_{0}$ is open and closed, and we will show that $H_{0}\cap W\neq\emptyset$ but $m_{H}(H_{0}\cap W)=0$, so that $W$ is not Haar regular.

First observe that $(\Delta(c))_{b_{0}}=c\in c{\mathbbm{Z}}$, so that $\Delta(c)\in H_{0}$. Suppose for a contradiction that $H_{0}\cap W=\emptyset$. Then $\Delta(c)\not\in W$, i.e. there is $b\in{\mathcal{B}}$ such that $c\in b{\mathbbm{Z}}$. Hence $c{\mathcal{A}}\subseteq b{\mathbbm{Z}}$, so that $c{\mathcal{A}}=\{b\}$, because $b\in{\mathcal{B}}$ and ${\mathcal{B}}$ is primitive. Hence $b=ca_{0}=b_{0}$, so that ${\mathcal{A}}=\{a_{0}\}$, a contradiction, as ${\mathcal{A}}$ is Behrend.

We turn to the proof of $m_{H}(H_{0}\cap W)=0$. Let ${\mathcal{H}}_{W}^{\ell}=\{n\in\{0,\dots,s_{\ell}-1\}:U_{S_{\ell}}(\Delta(n))\cap H_{0}\cap W\neq\emptyset\}$. It suffices to show that $\sum_{n\in{\mathcal{H}}_{W}^{\ell}}m_{H}(U_{S_{\ell}}(\Delta(n))\to 0$ as $\ell\to\infty$. As all cylinder sets $U_{S_{\ell}}(\Delta(n))$ have identical Haar measure $s_{\ell}^{-1}$, this is equivalent to $\#{\mathcal{H}}_{W}^{\ell}/s_{\ell}\to 0$ as $\ell\to\infty$. So let $\ell$ be so large that $b_{0}\in S_{\ell}$. Denote ${\mathcal{A}}^{\ell}=\{a\in{\mathcal{A}}:ca\mid s_{\ell}\}$. As $c{\mathcal{A}}\subseteq{\mathcal{B}}$, the sequence $({\mathcal{A}}^{\ell})_{\ell}$ is increasing and exhausts the set ${\mathcal{A}}$.

If $n\in{\mathcal{H}}_{W}^{\ell}$, then $n\in c{\mathbbm{Z}}$ and, by Lemma 2.5a, $n\in{\mathcal{F}}_{S_{\ell}}$. Hence $n=cn^{\prime}\in{\mathcal{F}}_{S_{\ell}}$ for some $n^{\prime}\in{\mathbbm{Z}}$. Suppose for a contradiction that $n^{\prime}\in{\mathcal{M}}_{{\mathcal{A}}^{\ell}}$, i.e. there are $k\in{\mathbbm{Z}}$ and $a\in{\mathcal{A}}_{\ell}$ such that $n^{\prime}=ka$. Then $n=kca$, where $ca\in{\mathcal{B}}$ and $ca\mid s_{\ell}$, so that $ca\in S_{\ell}$, which contradicts $n\in{\mathcal{F}}_{S_{\ell}}$. Hence $n^{\prime}\in{\mathcal{F}}_{{\mathcal{A}}^{\ell}}$ so that $n\in c{\mathcal{F}}_{{\mathcal{A}}^{\ell}}=c\,({\mathbbm{Z}}\setminus{\mathcal{M}}_{{\mathcal{A}}^{\ell}})$. As ${\mathcal{A}}$ is Behrend, $\bar{d}({\mathbbm{Z}}\setminus{\mathcal{M}}_{{\mathcal{A}}^{\ell}})\to 0$ as $\ell\to\infty$. Hence

[TABLE]

Suppose now that ${\mathcal{B}}$ is taut. We must show that for any $k\in{\mathbbm{N}}$ and $U\in{\mathcal{Z}}_{k}$

[TABLE]

So fix some $U=U_{S_{k}}(\Delta(n))$ such that $m_{H}(U\cap W)=0$. We have to show that $U\cap W=\emptyset$. Observe first that $U_{S_{k}}(\Delta(m))=U$ if and only if $m\in s_{k}{\mathbbm{Z}}+n$. For $\ell>k$ let

[TABLE]

where we used Lemma 2.5c for the last equality. Observe that

[TABLE]

Hence, for each $\ell>k$,

[TABLE]

As all $U^{\prime}\in{\mathcal{Z}}_{\ell}$ have identical Haar measure $m_{H}(U^{\prime})=s_{\ell}^{-1}$ and as $m_{H}(U\setminus W)=m_{H}(U)$ by assumption, it follows that

[TABLE]

so that

[TABLE]

Let $q=\gcd(s_{k},n)$, $a^{\prime}=s_{k}/q$ and $r^{\prime}=n/q$. Then $\gcd(a^{\prime},r^{\prime})=1$ and $q{\mathbbm{Z}}\cap{\mathcal{M}}_{\mathcal{B}}=q{\mathbbm{Z}}\cap{\mathcal{M}}_{q\cdot{\mathcal{B}}^{\prime}(q)}$, in particular $(s_{k}{\mathbbm{Z}}+n)\cap{\mathcal{M}}_{\mathcal{B}}=(s_{k}{\mathbbm{Z}}+n)\cap{\mathcal{M}}_{q\cdot{\mathcal{B}}^{\prime}(q)}$. Hence

[TABLE]

In view of Lemma 1.17 in [10], this suffices to conclude that ${\mathcal{B}}^{\prime}(q)$ is Behrend.

On the other hand, as ${\mathcal{B}}$ is taut, ${\mathcal{B}}^{\prime}(q)$ is pre-taut (Lemma 2.3), so that $1\in{\mathcal{B}}^{\prime}(q)$ or ${\mathcal{B}}^{\prime}(q)$ is not Behrend (Lemma 2.1e). Hence $1\in{\mathcal{B}}^{\prime}(q)$. This implies $q\in{\mathcal{M}}_{\mathcal{B}}$, which in turn implies $U\cap W=U_{S_{k}}(\Delta(n))\cap W=\emptyset$ (the property to be proved): Indeed, if $q\in{\mathcal{M}}_{\mathcal{B}}$, then there is some $b\in{\mathcal{B}}$ with $b\mid q$, and as $q\mid s_{k}$, this implies $b\mid s_{k}$, so that $b\in S_{k}$. From $b\mid q\mid n$ we then conclude that $n\in{\mathcal{M}}_{S_{k}}$, and Lemma 2.5a implies $U_{S_{k}}(\Delta(n))\cap W=\emptyset$.

It remains to show that the implication $(i)$ $\Rightarrow$ $(iii)$ follows from Proposition 1.1, which will be proved in the next subsection. So let $h\in W$. By the proposition there exists $n\in{\mathcal{F}}_{\mathcal{B}}$ such that $\Delta(n)\in U_{S}(h)$, hence $\Delta(n)\in U_{S}(h)\cap(\Delta({\mathbbm{Z}})\cap W)$. As this holds for all finite $S\subset{\mathcal{B}}$, this proves the claim.

2.3 Tautification of the set ${\mathcal{B}}$ and regularization of the window $W$

In [4, Section 4.2] the authors provide a construction that associates to each (non-taut) set ${\mathcal{B}}$ a taut set ${\mathcal{B}}^{\prime}$ such that ${\mathcal{F}}_{{\mathcal{B}}^{\prime}}\subseteq{\mathcal{F}}_{\mathcal{B}}$ but $\overline{d}({\mathcal{F}}_{\mathcal{B}}\setminus{\mathcal{F}}_{{\mathcal{B}}^{\prime}})=0$, and such that the two Mirsky measures $\nu_{\eta}$ and $\nu_{\eta^{\prime}}$ determined by ${\mathcal{B}}$ and ${\mathcal{B}}^{\prime}$ coincide. ${\mathcal{B}}$ and ${\mathcal{B}}^{\prime}$ determine groups $H$ resp. $H^{\prime}$ with windows $W$ resp. $W^{\prime}$, and while the window $W$ is not Haar regular (if ${\mathcal{B}}$ is non-taut), the window $W^{\prime}$ is Haar regular because of Theorem A.

On the abstract level one can also pass from the window $W\subseteq H$ to its Haar regularization $W_{reg}:=\operatorname{supp}(m_{H}|_{W})$ (introduced in [14]), which also determines the same Mirsky measure on $\{0,1\}^{\mathbbm{Z}}$. However, $W_{reg}$ will not be a window of the particular arithmetic type defined in (2), in particular it need not be aperiodic. The construction of ${\mathcal{B}}^{\prime}$ given ${\mathcal{B}}$ in [4] suggests an obvious factor map $f:H\to H^{\prime}$, and we expect that also $f(W_{reg})=W^{\prime}$, so that in this sense the regularization of $W$ and the tautification of ${\mathcal{B}}$ are two sides of the same medal.

The following example illustrates this discussion.

Example 2.1**.**

Let ${\mathcal{P}}=\{p_{1},p_{2},\ldots\}$ denote the set of primes. Let ${\mathcal{B}}:=\bigcup_{i\geq 1}p_{i}^{2}({\mathcal{P}}\setminus\{p_{i}\})$. Note that ${\mathcal{B}}$ is primitive. It is not taut, because it contains rescalings of Behrend sets. The corresponding taut set is ${\mathcal{B}}^{\prime}=\{p_{i}^{2}:\>i\geq 1\}$, which generates the square-free system. 666Note that $\eta(n)=1$ at all square-free numbers and also at $p_{i}^{k}$ for $i\geq 1$ and $k\geq 2$.

2.4 The property $\overline{\Delta({\mathbbm{Z}})\cap W}=W$

Proof of Proposition 1.1.

Given $h\in W$, we need to show that for each finite $S\subset{\mathcal{B}}$ the set

[TABLE]

has asymptotic density $m_{H}(U_{S}(h)\cap W)>0$.

By Theorem A, the tautness assumption on ${\mathcal{B}}$ implies that $W$ is Haar regular, so that indeed

[TABLE]

Let ${\mathcal{B}}=\{b_{1},b_{2},\ldots\}$ and, for $K\geq 1$, $W_{K}:=\{g\in H:\>g_{i}\neq 0\text{ for }i=1,\ldots,K\}$. Then $W_{K}$ is clopen and $W\subseteq W_{K}$. Moreover, $W_{K}\supseteq W_{K+1}$ and $\bigcap_{K}W_{K}=W$. Fix $\varepsilon>0$. We now choose $K\geq 1$ so that

[TABLE]

Since $U_{S}(h)\cap W_{K}$ is clopen (and $T$ is strictly ergodic)

[TABLE]

for all $N\geq N_{0}$. Moreover, we can choose $N_{1}$ so that for $N\geq N_{1}$, we also have

[TABLE]

Indeed, if

[TABLE]

then (by setting ${\mathcal{B}}_{K}=\{b_{1},\ldots,b_{K}\}$), we have

[TABLE]

Therefore, by the Davenport-Erdös theorem [10, Eq. (0.67)], we can choose first $K\geq 1$ sufficiently large so that $\overline{d}({\mathcal{M}}_{{\mathcal{B}}}\setminus{\mathcal{M}}_{{\mathcal{B}}_{K}})<\varepsilon$ and then $N_{1}$ so that

[TABLE]

for all $N\geq N_{1}$, so in particular (13) holds. In view of (11), (12) and (13), it follows that

[TABLE]

As $T^{n}0=\Delta(n)\in U_{S}(h)\cap W$ if and only if $n\in{\mathcal{L}}_{S}(h)$, this finishes the proof. ∎

Example 2.2**.**

($\overline{\Delta({\mathbbm{Z}})\cap W}=W$ does not imply tautness)

Suppose that $(m_{k},r_{k})$, ${k\in{\mathbbm{N}}}$, is an enumeration of all coprime pairs of natural numbers. For any $k$ choose a prime $p_{k}\in r_{k}+m_{k}{\mathbbm{Z}}$ such that $p_{k}>2^{k+1}$. Let ${\mathcal{B}}={\mathcal{P}}\setminus\{p_{k}:k\in{\mathbbm{N}}\}$. Clearly ${\mathcal{B}}$ is primitive, and ${\mathcal{M}}_{\{p_{k}:k\in{\mathbbm{N}}\}}$ has upper density less than or equal to $\sum_{k=1}^{\infty}1/2^{k+1}=1/2$. Thus $d({\mathcal{M}}_{{\mathcal{B}}})=1$ and ${\mathcal{B}}$ is not taut [10, Corollary 0.14]. But $\overline{\Delta({\mathbbm{Z}})\cap W}=W$. Indeed, let $h=(h_{b})_{b\in{\mathcal{B}}}\in W$ and take any finite set $S\subset{\mathcal{B}}$. We are going to show that $U_{S}(h)\cap W\cap\Delta({\mathbbm{Z}})\neq\emptyset$. Let $n\in{\mathbbm{Z}}$ be such that $n=h_{b}$ mod $b$ for $b\in S$. Since $h\in W$, $b$ does not divide $n$ for any $b\in S$, i.e. $\operatorname{lcm}(S)$ and $n$ are coprime. Then $(\operatorname{lcm}(S),n)=(m_{k},r_{k})$ for some $k$, and the prime number $p_{k}$ belongs to arithmetic progression $r_{k}+m_{k}{\mathbbm{Z}}=n+\operatorname{lcm}(S){\mathbbm{Z}}$, in other words $\Delta(p_{k})\in U_{S}(\Delta(n))=U_{S}(h)$. Finally, $\Delta(p_{k})\in W$, because the prime number $p_{k}$ does not belong to ${\mathcal{B}}$ and hence also not to ${\mathcal{M}}_{\mathcal{B}}$.

2.5 $X_{\eta}$ and $X_{\varphi}$

The set ${\mathcal{B}}\subseteq{\mathbbm{N}}$ has light tails, if

[TABLE]

If ${\mathcal{B}}$ has light tails, then ${\mathcal{B}}$ is taut, but the converse doses not hold [4, Section 4.3]. Here we prove:

Proposition 2.2**.**

If ${\mathcal{B}}$ has light tails, then $X_{\eta}=X_{\varphi}$.

Proof.

Let $H=(h_{k})\in H$ and $n\in{\mathbbm{N}}$. We are going to show that $\varphi(h)[-n,n]=\eta[l+1,l+2n+1]$ for some $l\in{\mathbbm{Z}}$. We know that $\varphi(h)(i)=1$ if and only if $h_{j}+i$ is not a multiple of $b_{j}$ for any $j\in{\mathbbm{N}}$. For any $i\in[-n,n]$ such that $\varphi(h)(i)=0$ let $k_{i}$ be such that $b_{k_{i}}|h_{k_{i}}+i$.

Let $K\in{\mathbbm{N}}$ be such that the set $\mathscr{A}:=\{b_{1},\ldots,b_{K}\}$ contains $b_{k_{i}}$, for $i\in[-n,n]$ and any $b_{k}$ with $k>K$ has a prime factor $p>2n+1$. Since $h\in H$, there exists $m\in Z$ such that

[TABLE]

for all $k\leq K$. It follows that

[TABLE]

Indeed, if $i\in\operatorname{supp}\varphi(h)\cap[-n,n]$, then $h_{k}+i$ is not a multiple of $b_{k}$ for any $k\in{\mathbbm{N}}$. By (15) we get that $m+i$ is not a multiple of $b_{k}$ for any $k\leq K$, that is, $m+i\in{\mathcal{F}}_{{\mathcal{A}}}$. On the other hand, if $i\notin{\rm supp}\varphi(h)\cap[-n,n]$, then $b_{k_{i}}|h_{k_{i}}+i$. Since $k_{i}\leq K$, again by (15), we obtain $b_{k_{i}}|m+i$, that is $m+i\notin{\mathcal{F}}_{{\mathcal{A}}}$.

By [4, Proposition 5.11] 777Assume that ${\mathcal{B}}\subset{\mathbbm{N}}$ has light tails and ${\mathcal{B}}^{(n)}\subset{\mathcal{A}}\subset{\mathcal{B}}$. Suppose that

$\{k+1,\ldots,k+n\}\cap{\cal M}_{{\mathcal{A}}}=\{k+i_{0},k+i_{1},\ldots,k+i_{r}\}$

(16)

for some $1\leq i_{0},\ldots,i_{r}\leq n$, $r<n$. Then the density of $k^{\prime}\in{\mathbbm{N}}$ such that

$\{k^{\prime}+1,\ldots,k^{\prime}+n\}\cap{\cal M}_{{\mathcal{B}}}=\{k^{\prime}+i_{0},k^{\prime}+i_{1},\ldots,k^{\prime}+i_{r}\}$

is positive. (Here ${\mathcal{B}}^{(n)}:=\{b\in{\mathcal{B}}:p\leq n\text{ for any }p\in{\rm Spec}(b)\}$. If ${\mathcal{B}}$ is primitive, then ${\mathcal{B}}^{(n)}$ is finite.) there exists $l\in{\mathbbm{Z}}$ such that

[TABLE]

It follows that $\varphi(h)[-n,n]=\eta[l+1,l+2n+1]$. ∎

We now present a Behrend set (hence a non-taut set), for which $X_{\eta}$ is a strict subeset of $X_{\varphi}$.

Example 2.3**.**

Let ${\mathcal{B}}=\{p_{2},p_{3},\ldots\}=\{3,5,7,11,\ldots\}$ - the set of all odd prime numbers. Since we are in the coprime case,

[TABLE]

Now, $\eta=\varphi(\Delta(0))$ is the characteristic function of the ${{\mathcal{B}}}$-free set $\{\pm 2^{m}:\>m\geq 0\}$. We compute an initial block of $\varphi(h)$ for

[TABLE]

We have $\varphi(h)(0)=0$, $\varphi(h)(1)=1$, $\varphi(h)(2)=1$, $\varphi(h)(3)=0$, $\varphi(h)(4)=0$888If we add 4 to each coordinate of $h$, we obtain the sequence $(1,0,4,4,\ldots)$, whence $\varphi(h)(4)=0$., $\varphi(h)(5)=1$, $\varphi(h)(6)=0$, $\varphi(h)(7)=0$ and $\varphi(h)(8)=1$. It follows that the block $11001001$ appears on $\varphi(h)$. But there is no block $\underline{a}$ of length 8 appearing on $\eta$ and such that $11001001\leq\underline{a}$. Indeed, the two neighboring 1’s at the beginning of $\underline{a}$ could only appear at the positions 1,2 or -2,-1 in $\eta$. In the both cases this would force $\eta(5)=1$, which is not true. This shows that $\varphi(h)\not\in X_{\eta}$, although it belongs to $X_{\varphi}$.999Indeed, $\varphi(h)$ does not even belong to $\widetilde{X}_{\eta}$, the hereditary closure of $X_{\eta}$, see [4].

Question 2.1**.**

If ${\mathcal{B}}$ is taut, is then $X_{\eta}=X_{\varphi}$? 101010We recall that in case of ${\mathcal{B}}$ taut, the Mirsky measure is supported on $X_{\eta}$.

3 Minimality/proximality of $X_{\eta}$ and topological properties of $W$

Throughout this section we assume that ${\mathcal{B}}$ is primitive.

3.1 Arithmetic of ${\mathcal{B}}$ and topology of $W$, part II

Recall from (5) that ${\mathcal{A}}_{S}:=\{\gcd(b,\operatorname{lcm}(S)):b\in{\mathcal{B}}\}$ and ${\mathcal{F}}_{{\mathcal{A}}_{S}}\subseteq{\mathcal{F}}_{\mathcal{B}}$ for $S\subset{\mathcal{B}}$. If $S\subseteq S^{\prime}\subset{\mathcal{B}}$, then the following inclusions and implications are obvious:

[TABLE]

Let ${\mathcal{E}}:=\bigcup_{S\subset{\mathcal{B}}}{\mathcal{F}}_{{\mathcal{A}}_{S}}$ and observe that ${\mathcal{E}}\subseteq{\mathcal{F}}_{\mathcal{B}}$.

Lemma 3.1**.**

1. a)

For all $S\subset{\mathcal{B}}$ and $n\in{\mathbbm{Z}}$ we have: $U_{S}(\Delta(n))\subseteq W\Leftrightarrow n\in{\mathcal{F}}_{{\mathcal{A}}_{S}}$. 2. b)

If $(S_{k})_{k}$ is a filtration of ${\mathcal{B}}$ with finite sets and $\lim_{k}\Delta(n_{S_{k}})=h$ (see Remark 5.2), then $h\in\operatorname{int}(W)$ if and only if $n_{S_{k}}\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$ for some $k$. 3. c)

For all $n\in{\mathbbm{Z}}$ we have: $\Delta(n)\in{\operatorname{int}(W)}\Leftrightarrow n\in{\mathcal{E}}$ . 4. d)

$\operatorname{int}(W)=\emptyset\Leftrightarrow{\mathcal{E}}=\emptyset\Leftrightarrow\forall S\subset{\mathcal{B}}:{\mathcal{F}}_{{\mathcal{A}}_{S}}=\emptyset\Leftrightarrow\forall S\subset{\mathcal{B}}:1\in{\mathcal{A}}_{S}$**

Proof.

1. a)

As $U_{S}(\Delta(n))$ is clopen,

[TABLE]

That the only implication is also an equivalence is a consequence of the CRT. Indeed, if $\gcd(c,\operatorname{lcm}(S))\mid n$, then there exist $k,l\in{\mathbbm{Z}}$ such that $l\cdot c-k\cdot\operatorname{lcm}(S)=n$, thus $c\mid n+k\cdot\operatorname{lcm}(S)$. 2. b)

Assume that $h\in\operatorname{int}(W)$, that is $U_{S}(h)\subseteq W$ for some $S$. Then, for $k$ such that $S\subseteq S_{k}$, we have $U_{S_{k}}(\Delta(n_{S_{k}}))=U_{S_{k}}(h)\subseteq W$, which is equivalent to $n_{S_{k}}\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$ by a). Conversely, if $n_{S_{k}}\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$ then, again by a), $U_{S_{k}}(\Delta(n_{S_{k}}))=U_{S_{k}}(h)\subseteq W$ and $h\in\operatorname{int}(W)$. 3. c)

Follows from a). 4. d)

Follows from c).

∎

Recall from (6) that ${\mathcal{A}}_{\infty}:=\{n\in{\mathbbm{N}}:\forall_{S\subset{\mathcal{B}}}\ \exists_{S^{\prime}:S\subseteq S^{\prime}}:n\in{\mathcal{A}}_{S^{\prime}}\setminus S^{\prime}\}$.

Lemma 3.2**.**

1. a)

If $(S_{k})_{k}$ is a filtration of ${\mathcal{B}}$ with finite sets, then

[TABLE] 2. b)

For each $n\in{\mathcal{A}}_{\infty}$ there is a filtration $(S_{k})_{k}$ of ${\mathcal{B}}$ with finite sets such that

[TABLE]

Proof.

a) Assume that $n\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$ for infinitely many $k$, and let $S\subset{\mathcal{B}}$. Then there is $k$ such that $S\subseteq S_{k}$ and $n\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$. Hence $n\in{\mathcal{A}}_{\infty}$. Conversely, let $n\in{\mathcal{C}}_{\infty}$. There is a finite set $S_{1}$ such that $n\in{\mathcal{A}}_{S_{1}}\setminus S_{1}$. Assume that we have constructed sets $S_{1}\subset S_{2}\subset\ldots\subset S_{k}$ with the property that $n\in{\mathcal{A}}_{S_{i}}\setminus S_{i}$ for $i=1,\ldots,k$ and $\{1,\ldots,k\}\cap{\mathcal{B}}\subset S_{k}$. Then there is a set $S_{k+1}$ containing $S_{k}\cup\{k+1\}$ and such that $n\in{\mathcal{A}}_{S_{k+1}}\setminus S_{k+1}$. In this inductive way we construct a filtration $(S_{k})_{k}$ as required.

b) follows from a). ∎

Lemma 3.3**.**

The sets ${\mathcal{E}}$ and ${\mathcal{A}}_{\infty}$ are related by the identity

[TABLE]

Proof.

Let $n\in{\mathcal{E}}$ and chose $S$ such that $n\in{\mathcal{F}}_{{\mathcal{A}}_{S}}$. Take arbitrary $b\in{\mathcal{B}}$ and $c\in{\mathcal{A}}_{\infty}$. There exists a finite set $S^{\prime}$ such that $S\cup\{b\}\subseteq S^{\prime}$ and $c\in{\mathcal{A}}_{S^{\prime}}\setminus S^{\prime}$. Since ${\mathcal{F}}_{{\mathcal{A}}_{S}}\subseteq{\mathcal{F}}_{{\mathcal{A}}_{S^{\prime}}}$, $n\in{\mathcal{F}}_{{\mathcal{A}}_{S^{\prime}}}$, hence neither $b$ nor $c$ divides $n$. We have proved that ${\mathcal{E}}\subseteq{\mathcal{F}}_{{\mathcal{B}}\cup{\mathcal{A}}_{\infty}}$. In order to prove the other inclusion assume that $n\in{\mathbbm{N}}$ and that for any $S$ there exists $c_{S}\in{\mathcal{A}}_{S}$ dividing $n$. As $n$ has only finitely many divisors, it has a divisor $c$ such that there exists a filtration $(S_{k})_{k}$ of ${\mathcal{B}}$ such that $c\in{\mathcal{A}}_{S_{k}}$ for any $k\in{\mathbbm{N}}$. If $c\notin{\mathcal{B}}$, then $c\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$ for any $k\in{\mathbbm{N}}$. This proves $n\notin{\mathcal{F}}_{{\mathcal{B}}\cup{\mathcal{A}}_{\infty}}$. ∎

Lemma 3.4**.**

${\mathcal{A}}_{\infty}=\emptyset$* if and only if ${\mathcal{E}}={\mathcal{F}}_{\mathcal{B}}$.*

Proof.

If ${\mathcal{A}}_{\infty}=\emptyset$, then ${\mathcal{E}}={\mathcal{F}}_{\mathcal{B}}$ by Lemma 3.3. Conversely, assume that ${\mathcal{E}}={\mathcal{F}}_{\mathcal{B}}$. Then ${\mathcal{F}}_{\mathcal{B}}\subseteq{\mathcal{F}}_{{\mathcal{A}}_{\infty}}$ by Lemma 3.3, so that ${\mathcal{A}}_{\infty}\subseteq{\mathcal{M}}_{{\mathcal{A}}_{\infty}}\subseteq{\mathcal{M}}_{\mathcal{B}}$. Suppose for a contradiction that there exists some $n\in{\mathcal{A}}_{\infty}$. Then there is $b\in{\mathcal{B}}$ such that $n\in b{\mathbbm{Z}}$, i.e. $b\mid n$, and there is a finite set $S=S_{k}\subset{\mathcal{B}}$ such that $n\in{\mathcal{A}}_{S}\setminus S$, see Lemma 3.2b. Hence there exists $b^{\prime}\in{\mathcal{B}}$ such that $n=\gcd(\operatorname{lcm}(S),b^{\prime})$. It follows that $b\mid n\mid b^{\prime}$, which is impossible, because ${\mathcal{B}}$ is assumed to be primitive. ∎

Proposition 3.1**.**

The following conditions are equivalent:

1. (i)

$W\neq\overline{\operatorname{int}(W)}$** 2. (ii)

For any filtration $S_{0}\subset S_{1}\subset\ldots\subset{\mathcal{B}}$ of ${\mathcal{B}}$ with finite subsets $S_{k}$, there exists a number $d$ such that $d\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$, for infinitely many $k\in{\mathbbm{N}}$. 3. (iii)

There exists a filtration $S_{0}\subset S_{1}\subset\ldots\subset{\mathcal{B}}$ of ${\mathcal{B}}$ with finite subsets $S_{k}$ and there exists a number $d$ such that $d\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$, for every $k\in{\mathbbm{N}}$. 4. (iv)

There are $d\in{\mathbbm{N}}$ and an infinite pairwise coprime set ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$ such that $d\,{\mathcal{A}}\subseteq{\mathcal{B}}$.

Proof.

$(i)\Rightarrow(ii)$: Let $h=(h_{b})\in W\setminus\overline{\operatorname{int}(W)}$. There exists $S$ such that $U_{S}(h)\cap\operatorname{int}(W)=\emptyset$. We can assume that any $b\in{\mathcal{B}}$ such that $b|\operatorname{lcm}(S)$, belongs to $S$.111111Otherwise we can incorporate all such $b$’s into $S$, there are finitely many of them. Let $n$ be a number such that

[TABLE]

for $b\in S$.

Then $\Delta(n+k\operatorname{lcm}(S))\in U_{S}(h)$, hence $\Delta(n+k\operatorname{lcm}(S))\notin\operatorname{int}(W)$ for any $k\in{\mathbbm{Z}}$. This means (see Lemma 3.1) that for any finite set $T$, in particular for any $T=S_{k}$, the arithmetic progression $n+\operatorname{lcm}(S){\mathbbm{Z}}$ is contained in ${\mathcal{M}}_{{\mathcal{A}}_{T}}$. Since the set ${\mathcal{A}}_{T}$ is finite, it follows that ${\mathcal{A}}_{T}$ contains a divisor of $\gcd(n,\operatorname{lcm}(S))$121212Apply Dirichlet theorem on primes in arithmetic progressions.. There is only finitely many divisors of $\gcd(n,\operatorname{lcm}(S))$, hence one of them, denote it by $d$, appears in ${\mathcal{A}}_{S_{k}}$ for infinitely many $k$. To finish the proof it is enough to observe that $d\notin{\mathcal{B}}$ (consequently, $d\notin S_{k}$, for any $k$). Indeed, otherwise $d\in S$, by our assumption on $S$. Moreover, $d|n$ and then, by (18), $d|h_{b}$, where $b=d$, which leads to a contradiction with the assumption $h\in W$.

$(ii)\Rightarrow(iii)$: obvious

$(iii)\Rightarrow(i)$: Assume that $d\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$ for any $k$. Then $d\notin{\mathcal{M}}_{{\mathcal{B}}}$131313Otherwise $d$ is divisible by some $b\in{\mathcal{B}}$. On the other hand, $d$ divides some $b^{\prime}\in{\mathcal{B}}$ as a member of ${\mathcal{A}}_{S_{k}}$, which in view of the fact that ${\mathcal{B}}$ is primitive, leads to the conclusion that $d=b=b^{\prime}\in{\mathcal{B}}$. But it is not true, since $d\notin S_{k}$ for any $k$., hence $\Delta(d)\in W$. We prove that $\Delta(d)\notin\overline{\operatorname{int}(W)}$. It is enough to show that $U_{S_{0}}(\Delta(d))\cap\operatorname{int}(W)=\emptyset$.

Assume that $h=(h_{b})\in U_{S_{0}}(\Delta(d))\cap\operatorname{int}(W)$. It means that

[TABLE]

and there exists a finite set $T\subset{\mathcal{B}}$ such that $U_{T}(h)\subset W$. We can assume that $T=S_{k}$ for some $k$.

Let $m\in{\mathbbm{Z}}$ be such that

[TABLE]

Let $c\in{\mathcal{B}}$ be such that $\gcd(c,\operatorname{lcm}(S_{k}))=d$. Clearly, $c\notin S_{k}$, since $d\notin{\mathcal{B}}$. Since $U_{S_{k}}(h)\subset W$, it follows that there exists $b\in S_{k}$ such that

[TABLE]

Indeed, otherwise there would exist $l\in{\mathbbm{Z}}$ such that $l\equiv h_{b}$ mod $b$ for $b\in S_{k}$ and $l=0$ mod $c$, hence $\Delta(l)\in U_{S_{k}}(h)$, but $\Delta(l)\notin W$, a contradiction. Thus, in view of (20),

[TABLE]

On the other hand,

[TABLE]

Since $d\in{\mathcal{A}}_{S_{0}}\setminus S_{0}$ we get

[TABLE]

By (19) and (20),

[TABLE]

Now, (23), (24) and (25) imply $\gcd(c,b)|m$, a contradiction with (22).

$(iii)\Rightarrow(iv)$: Assume that $d\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$ for any $k$. Then

[TABLE]

As $d\not\in S_{k}$, we have $b_{k}\neq d$ for all $k$. We choose a subsequence $b_{k_{1}},b_{k_{2}},\dots$ of $(b_{k})_{k}$ in the following way: Let $k_{1}=1$, and given $k_{1},\dots,k_{j}$, let

[TABLE]

Let $a_{j}=b_{k_{j}}/d$ for all $j\in{\mathbbm{N}}$ and denote ${\mathcal{A}}=\{a_{j}:j\in{\mathbbm{N}}\}$. Then ${\mathcal{A}}\subseteq{\mathbbm{N}}$ and $d\,{\mathcal{A}}\subseteq{\mathcal{B}}$ by construction. Suppose that $1\in{\mathcal{A}}$. Then $d\in{\mathcal{B}}$, a contradiction to (26), as ${\mathcal{B}}$ is primitive. Hence ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$.

It remains to prove that ${\mathcal{A}}$ is pairwise coprime. Suppose for a contradiction that there is a prime number $p$ dividing some $a_{i}$ and $a_{j}$, $i<j$. Then $pd\mid b_{k_{i}}$ and $pd\mid b_{k_{j}}$. As $b_{k_{i}}\in S_{k_{j}}$, it follows that $pd\mid\operatorname{lcm}(S_{k_{j}})$, so that $pd\mid\gcd(b_{k_{j}},\operatorname{lcm}(S_{k_{j}}))=d$ (see (26)), which is impossible.

$(iv)\Rightarrow(iii)$: Let $d\in{\mathbbm{N}}$ and ${\mathcal{A}}=\{a_{1}<a_{2}<\dots\}$ be as in $(iv)$. Then $d\not\in{\mathcal{B}}$, because ${\mathcal{B}}$ is primitive. For $k\in{\mathbbm{N}}$ let $S_{k}={\mathcal{B}}\cap\{1,\dots,k\}\cup\{da_{k}\}$. As all $a_{j}$ are pairwise coprime, there are $j_{1}<j_{2}<\dots\in{\mathbbm{N}}$ such that $a_{j_{k}}$ is coprime to $\operatorname{lcm}(S_{k})$. On the other hand, $d\mid\operatorname{lcm}(S_{k})$. Hence $d=\gcd(da_{j_{k}},\operatorname{lcm}(S_{k}))\in{\mathcal{A}}_{S_{k}}$. As $d\not\in{\mathcal{B}}$, we see that $d\in{\mathcal{A}}_{S_{k}}\setminus S_{k}$ for all $k\in{\mathbbm{N}}$. ∎

Proposition 3.2**.**

The following conditions are equivalent:

1. (i)

$W$* is topologically regular, i.e. $W=\overline{\operatorname{int}(W)}$.* 2. (ii)

There are no $d\in{\mathbbm{N}}$ and no infinite pairwise coprime set ${\mathcal{A}}\subseteq{\mathbbm{N}}\setminus\{1\}$ such that $d\,{\mathcal{A}}\subseteq{\mathcal{B}}$. 3. (iii)

${\mathcal{A}}_{\infty}=\emptyset$. 4. (iv)

${\mathcal{E}}={\mathcal{F}}_{\mathcal{B}}$.

Proof.

The equivalence of $(i)$ and $(ii)$ follows from Proposition 3.1, that of $(iii)$ and $(iv)$ from Lemma 3.4.. In view of Lemma 3.2, Proposition 3.1 finally implies the equivalence of $(i)$ and $(iii)$, too. ∎

Lemma 3.5**.**

$\Delta({\mathbbm{Z}})\cap\left(\overline{\operatorname{int}(W)}\setminus\operatorname{int}(W)\right)=\emptyset$.

Proof.

Assume $\Delta(m)\in\overline{\operatorname{int}(W)}\setminus\operatorname{int}(W)$. Then for any $S\subset{\mathcal{B}}$ there exists $n_{S}\in{\mathbbm{Z}}$ such that $\Delta(n_{S})\in U_{S}(\Delta(m))\cap\operatorname{int}(W)$. It means that for any $S$ there exist: a finite set $T_{S}\subset{\mathcal{B}}$, (we can assume that $S\subset T_{S}$), $b_{S}\in{\mathcal{B}}$ and $n_{S}\in{\mathbbm{Z}}$ such that (see Lemma 3.1 c)):

1. $\bullet$

$\operatorname{lcm}(S)|m-n_{s}$ (that is, $\Delta(n_{s})\in U_{S}(\Delta(m))$) 2. $\bullet$

$\gcd(b_{S},\operatorname{lcm}(T_{S}))$ does not divide $n_{S}$ ($\Delta(n_{S})$ is chosen to be an element of $U_{S}(\Delta(m))\cap\operatorname{int}(W)$) 3. $\bullet$

$\gcd(b_{S},\operatorname{lcm}(T_{S}))|m$ (since $\Delta(m)\notin\operatorname{int}(W)$)

Then $\gcd(b_{S},\operatorname{lcm}(T_{S}))$ does not divide $\operatorname{lcm}(S)$. Let us iterate: $S_{0}$ is arbitrary and $S_{k+1}:=T_{S_{k}}$, $c_{k}:=\operatorname{lcm}(S_{k+1})$, $d_{k}:=\gcd(b_{S_{k}},\operatorname{lcm}(S_{k+1}))$. We have:

1. $\bullet$

$c_{k}|c_{k+1}$ 2. $\bullet$

$d_{k}|m$ 3. $\bullet$

$d_{k}|c_{k}$ 4. $\bullet$

$d_{k}$ does not divide $c_{k-1}$

Since $d_{k}|m$ for every $k$, the sequence $(\operatorname{lcm}(d_{1},\ldots d_{k}))_{k}$ stabilizes on $\operatorname{lcm}(d_{1},\ldots d_{k_{0}})$ for some $k_{0}$, which means $d_{l}$ divides $\operatorname{lcm}(d_{1},\ldots d_{k_{0}})$, and consequently $d_{l}$ divides $\operatorname{lcm}(c_{1},\ldots,c_{k_{0}})=c_{k_{0}}$, for any $l$, a contradiction. ∎

For $x\in\{0,1\}^{\mathbbm{Z}}$ denote $\operatorname{supp}x:=\{n\in{\mathbbm{Z}}:x(n)=1\}$. Following [4] we consider the set

[TABLE]

As $\operatorname{supp}\eta={\mathcal{F}}_{\mathcal{B}}$ is disjoint from $b{\mathbbm{Z}}$ for all $b\in{\mathcal{B}}$, we have

[TABLE]

Lemma 3.6**.**

If $\overline{\Delta({\mathbbm{Z}})\cap W}=W$, then $\eta\in Y$.

Proof.

For $b\in{\mathcal{B}}$ and $r\in\{0,\dots,b-1\}$ let $V_{b}(r):=\{h\in H:h_{b}=r\}$ and observe that these sets are open and closed in $H$. Hence $\overline{\Delta({\mathbbm{Z}})\cap V_{b}(r)\cap W}=V_{b}(r)\cap W$, because $\overline{\Delta({\mathbbm{Z}})\cap W}=W$.

Suppose for a contradiction that $\eta\not\in Y$. Then (27) implies that there are $b\in{\mathcal{B}}$ and $r\in\{1,\dots,b-1\}$ such that

[TABLE]

which implies that also $V_{b}(r)\cap W=\emptyset$. Hence

[TABLE]

and as $V_{b}(r)$ is compact and the $V_{b^{\prime}}(0)$ are open, there is a finite $S\subset{\mathcal{B}}$ such that

[TABLE]

In other words, whenever $h_{b}=r$ for some $h\in H$, then $h_{b^{\prime}}=0$ for some $b^{\prime}\in S$. Applied to any $h=\Delta(n)$ this yields:

[TABLE]

Since $r$ is not divisible by $b$, we can assume that $b\notin S$. Let $q=\gcd(b,r)$, $\tilde{b}=b/q$, $\tilde{r}=r/q$. Then $q\,(\tilde{b}{\mathbbm{Z}}+\tilde{r})=b{\mathbbm{Z}}+r\subseteq{\mathcal{M}}_{S}$, so that ${\mathcal{M}}_{\tilde{b}{\mathbbm{Z}}+\tilde{r}}\subseteq{\mathcal{M}}_{S^{\prime}(q)}$ by Lemma 2.2. But $d({\mathcal{M}}_{\tilde{b}{\mathbbm{Z}}+\tilde{r}})=1$ by Dirichlet’s theorem, whereas $d({\mathcal{M}}_{S^{\prime}(q)})<1$, because $S^{\prime}(q)\subseteq\{1,\dots,\max S\}$ is finite and $1\notin S^{\prime}(q)$ 141414As $q|b$ and ${\mathcal{B}}$ is primitive, $q\notin S$, thus $1\notin S^{\prime}(q)$.. This is a contradiction. ∎

Remark 3.1**.**

Together with Theorem A this shows that $\eta\in Y$ whenever ${\mathcal{B}}$ is taut. This implication was proved previously in [4, Corollary 4.27].

Lemma 3.6 provides the implication

[TABLE]

The reverse implication does not hold, as is shown by the next example.

Example 3.1**.**

Observe that for every $k\in{\mathbbm{Z}}$ there exists a prime divisor $p_{k}$ of $5+12k$ such that

[TABLE]

Let

[TABLE]

Let us enumerate the elements of ${\mathcal{B}}$ as $b_{0},b_{1},b_{2},\ldots$ and $b_{0}=4,b_{1}=6$. Observe that

[TABLE]

Since niether 2 nor 3 divides an element of the progression $5+12{\mathbbm{Z}}$, in view of (28) we see that $1,2,3,11,22\in{\mathcal{F}}_{{\mathcal{B}}}$. It follows that

[TABLE]

We claim that

[TABLE]

It is clear that $\gcd(12,b_{k})=1$ for any $k\geq 2$. Let $k\geq 2$ and take arbitrary $r\in\{1,\ldots,b_{k}-1\}$. There exists $r^{\prime}\in{\mathbbm{Z}}$ such that

[TABLE]

Then $\gcd(12b_{k},r^{\prime})=1$ and, by Dirichlet Theorem, there exists a prime number $q$ of the form $q=12b_{k}l+r^{\prime}$ for some $l\in{\mathbbm{Z}}$. Since, by (32), $q\equiv 1\text{ mod }12$, $q\in{\mathcal{F}}_{{\mathcal{B}}}$ by (28). Moreover, $q\equiv r\text{ mod }b_{k}$ by (32).

Thus the claim (31) follows. Clearly, (31) and (30) yield $\eta\in Y$.

We shall construct $h\in W$ such that $h\notin\overline{\Delta({\mathbbm{Z}})\cap W}$. We denote $S_{k}=\{b_{0},b_{1},\ldots b_{k}\}$. Inductively we construct a sequence $(n_{S_{k}})$ of integers satisfying:

1. a)

$n_{S_{1}}=5$ 2. b)

$\operatorname{lcm}(S_{k})|n_{S_{k+1}}-n_{S_{k}}$ for $k=1,2,\ldots$ 3. c)

$n_{S_{k}}\in{\mathcal{F}}_{S_{k}}$ for $k=1,2,\ldots$

Assume that $n_{S_{1}},\ldots,n_{S_{k}}$ have been constructed. If $b_{k+1}$ does not divide $n_{S_{k}}$, we set $n_{S_{k+1}}=n_{S_{k}}$. Otherwise we set $n_{S_{k+1}}=n_{S_{k}}+\operatorname{lcm}(S_{k})$. The conditions a), b), c) follow easily by induction. Let

[TABLE]

Thanks to c), $h\in W$.

But

[TABLE]

the last equality by (29). (Clearly, $d({\mathcal{M}}_{{\mathcal{B}}})=1$ and ${\mathcal{B}}$ is not taut.)

3.2 Proof of Theorem B

Lemma 3.7**.**

151515The authors are indebted to A. Bartnicka for pointing out and proving this lemma.

If ${\mathcal{B}}$ is primitive and $\eta$ is a Toeplitz sequence, then ${\mathcal{B}}$ is taut.

Proof.

Suppose that ${\mathcal{B}}$ is not taut. Then there are $c\in{\mathbbm{N}}$ and a Behrend set ${\mathcal{A}}$ such that $c{\mathcal{A}}\subseteq{\mathcal{B}}$. Hence

[TABLE]

because ${\mathcal{M}}_{\mathcal{A}}$ has density one. As ${\mathcal{B}}$ is primitive, $c$ must be ${\mathcal{B}}$-free. So $\eta(c)=1$, and (since $\eta$ is Toeplitz) there exists $m\in{\mathbbm{N}}$ such that $c+m{\mathbbm{Z}}\subseteq{\mathcal{F}}_{\mathcal{B}}$. But then

[TABLE]

which contradics (33). ∎

Lemma 3.8**.**

Assume that $\eta\in Y$. If $\eta={\mathbbm{1}}_{{\mathcal{F}}_{\mathcal{B}}}$ is almost periodic (i.e. if the orbit closure of $\eta$ is minimal), then $X_{\eta}\subseteq Y$.

Proof.

Fix $k\geq 1$. Since $\eta\in Y$, the support of $\eta$ taken mod $b_{k}$ misses exactly one residue class mod $b_{k}$ (that is, it misses zero). Let $B$ be a block on $\eta$ such that its support mod $b_{k}$ misses exactly one residue class mod $b_{k}$. Since $\eta$ is almost periodic, the block $B$ appears on $\eta$ with bounded gaps. It follows that if $C$ is any sufficiently long block that appears on $\eta$, its support misses exactly one residue class. Clearly this property passes to limits in the product topology, so each $y=\lim S^{m_{i}}\eta$ is also in $Y$. ∎

In general, we can define a map $\theta:Y\to\prod_{k\geq 1}{\mathbbm{Z}}/b_{k}{\mathbbm{Z}}$ by setting

[TABLE]

Remark 2.51 in [4] tells us that

[TABLE]

while Remark 2.52 says that $\theta$ is continuous.

Corollary 3.1**.**

By the definitions of $\varphi$ and $\theta$, we have $\theta\circ\varphi(h)=h$ provided $\varphi(h)\in Y$. In particular, $\theta(\eta)=0$ and $\theta$ is continuous at $\eta$. Moreover, $\theta$ is equivariant.

For any map $\psi:X\to Y$ denote by $C_{\psi}\subseteq X$ the set of continuity points of this map.

Lemma 3.9**.**

Let $(X,S)$ and $(Y,T)$ be compact dynamical systems and assume that $(X,S)$ is minimal. Let $\psi:X\to Y$ be a map satisfying $\psi\circ S=T\circ\psi$. Then $\overline{\psi(C_{\psi})}$ is a minimal subset of $Y$.

Proof.

Denote by $Z:=\overline{\{(x,\psi(x)):x\in X\}}$ the closure of the graph of $\psi$ and note that a fibre $Z_{x}=\{(x,y):y\in Z\}$ is a singleton, if and only if $x\in C_{\psi}$. Let $Z_{0}:=\overline{\{(x,\psi(x)):x\in C_{\psi}\}}$. We claim that $Z_{0}\subseteq A$ whenever $A$ is a non-empty closed $S\times T$-invariant subset of $Z$. Indeed, $\pi_{X}(A)$ is a non-empty closed $S$-invariant subset of $X$, so $\pi_{X}(A)=X$ by minimality of $(X,S)$. In particular, $C_{\psi}\subseteq\pi_{X}(A)$. As all $A_{x}\subseteq Z_{x}$ with $x\in C_{\psi}$ are singletons, $\{(x,\psi(x)):x\in C_{\psi}\}\subseteq A$. Hence also $Z_{0}\subseteq A$.

This shows that $Z_{0}$ is a minimal subset of $X\times Y$ (and, by the way, that it is the only minimal subset of $Z$). It follows that $\pi_{Y}(Z_{0})$ is a minimal subset of $Y$, and so it remains to show that $\psi(C_{\psi})\subseteq\pi_{Y}(Z_{0})$. But, for $x\in C_{\psi}$, $(x,\psi(x))\in Z_{0}$, and so $\psi(x)\in\pi_{Y}(Z_{0})$. ∎

Denote by $C_{\varphi}$ the set of all points in $H$ at which $\varphi:H\to\{0,1\}^{\mathbbm{Z}}$ is continuous.

Lemma 3.10**.**

1. a)

$C_{\varphi}=\left\{h\in H:\ (h+\Delta({\mathbbm{Z}}))\cap\partial W=\emptyset\right\}$. 2. b)

$C_{\varphi}+\Delta(1)=C_{\varphi}$. 3. c)

$\overline{\varphi(C_{\varphi})}$* is the unique minimal subset $M$.*

Proof.

a) This is proved by direct inspection, see e.g. [13, Lemma 6.1].

b) This is obvious.

c) This follows from Lemma 3.9. ∎

Proof of Theorem B.

We start with a list of implications, which, when suitably combined, prove the assertions a) - e) of Theorem B. Most of these implications can be proved without assuming that ${\mathcal{B}}$ is primitive and that $\overline{\Delta({\mathbbm{Z}})\cap W}=W$. Therefore we indicate explicitly, for which implications we use these extra assumptions.

Proof of the equivalence of B1 – B4: These equivalences follow from Proposition 3.2.

Proof of B1 $\Rightarrow$ B6: Observe first that $0\in H$ belongs to $C_{\varphi}$ if and only if $\Delta({\mathbbm{Z}})\cap\partial W=\emptyset$, see Lemma 3.10. But $\Delta({\mathbbm{Z}})\cap\partial W=\Delta({\mathbbm{Z}})\cap\left(\overline{\operatorname{int}(W)}\setminus\operatorname{int}(W)\right)$ in view of B1, and this intersection is empty by Lemma 3.5. As $\operatorname{int}(W)\neq\emptyset$ and as $H=\overline{\Delta({\mathbbm{Z}})}$, $\Delta({\mathbbm{Z}})\cap W\neq\emptyset$ and hence $\varphi(0)\neq(\dots,0,0,0,\dots)$.

Proof of B6 $\Rightarrow$ B5: Let ${\mathcal{B}}=\{b_{1},b_{2},\ldots\}$ and assume (B6) that $0\in C_{\varphi}$, i.e. $\Delta({\mathbbm{Z}})\cap\partial W=\emptyset$, and $\eta\neq(\dots,0,0,0,\dots)$. Now, take $n\in{\mathbbm{Z}}$. Either $n\in{\mathcal{M}}_{{\mathcal{B}}}$ - then $\eta(n)=0$, so $b_{s}\mid n$ for some $s\geq 1$ and $\eta(n+jb_{s})=0$ for each $j\in{\mathbbm{Z}}$. Or $n\in{\mathcal{F}}_{{\mathcal{B}}}$, i.e. $\Delta(n)\in W$. As $\Delta({\mathbbm{Z}})\cap\partial W=\emptyset$ by assumption, this implies $\Delta(n)\in\operatorname{int}(W)$, so that $n\in{\mathcal{E}}=\bigcup_{S\subset{\mathcal{B}}}{\mathcal{F}}_{{\mathcal{A}}_{S}}$ by Lemma 3.1. Hence there is a finite subset $S\subset{\mathcal{B}}$ such that $n\in{\mathcal{F}}_{{\mathcal{A}}_{S}}$. As $\operatorname{lcm}({\mathcal{A}}_{S})=\operatorname{lcm}(S)$, this implies

[TABLE]

Hence $\eta(n+j\operatorname{lcm}(S))=1$ for each $j\in{\mathbbm{Z}}$. This proves that $\eta$ is a Toeplitz sequence different from $(\dots,0,0,0,\dots)$.

Proof of B5 $\Rightarrow$ B1: Assume that $\eta$ is a Toeplitz sequence. Then ${\mathcal{B}}$ is taut by Lemma 3.7, hence $\overline{\Delta({\mathbbm{Z}})\cap W}=W$ by Theorem A. Now B1 follows from the chain of the next three implications.

Proof of B5 $\Rightarrow$ B8: Each Toeplitz sequence is almost periodic [8], [12, Theorem 4], i.e. its orbit closure is minimal.

Proof of B8 $\Rightarrow$ B7: If $X_{\eta}=M$, then $\eta\in M$, and $\eta\neq(\dots,0,0,0,\dots)$, because otherwise the minimality of $X_{\eta}$ implies $X_{\eta}=\{(\dots,0,0,0,\dots)\}$, contradicting $\operatorname{card}(X_{\eta})>1$.

Proof of B7 $\Rightarrow$ B1 (assuming that $\overline{\Delta({\mathbbm{Z}})\cap W}=W$): Assume that $(\dots,0,0,0,\dots)\neq\eta\in M=\overline{\varphi(C_{\varphi})}$. Then $M=X_{\eta}\subseteq Y$ by Lemma 3.8, and there is a sequence $h_{1},h_{2},\dots\in C_{\varphi}$ such that $\eta=\lim_{i\to\infty}\varphi(h_{i})$. Consider $n\in{\mathbbm{Z}}$ with $\Delta(n)\in W$, i.e. such that $\eta(n)=1$. In particular $\eta=\varphi(0)\neq(\dots,0,0,0,\dots)$. Corollary 3.1 implies $\lim_{i\to\infty}h_{i}=\lim_{i\to\infty}\theta(\varphi(h_{i}))=\theta(\eta)=0$. Then $1=\eta(n)=\lim_{i\to\infty}\varphi(h_{i})(n)$, i.e. $h_{i}+\Delta(n)\in W$ for all sufficiently large $i$. As $h_{i}\in C_{\varphi}$, we have $h_{i}+\Delta({\mathbbm{Z}})\cap\partial W=\emptyset$ (Lemma 3.10). Hence $h_{i}+\Delta(n)\in\operatorname{int}(W)$ for all sufficiently large $i$, what implies that $\Delta(n)=\lim_{i\to\infty}h_{i}+\Delta(n)\in\overline{\operatorname{int}(W)}$. This proves that $\Delta({\mathbbm{Z}})\cap W\subseteq\overline{\operatorname{int}(W)}$. Hence $W=\overline{\Delta({\mathbbm{Z}})\cap W}\subseteq\overline{\operatorname{int}(W)}$, i.e. $W$ is topologically regular.

Proof of B7 $\Rightarrow$ B8: As $\eta\in M$, also $X_{\eta}\subseteq M$, and hence $X_{\eta}=M$. As $\eta\neq(\dots,0,0,0,\dots)$, $X_{\eta}$ contains no fixed point. Hence $\operatorname{card}(X_{\eta})>1$.

Proof of B1 $\Rightarrow$ B9 (assuming that ${\mathcal{B}}$ is primitive): The window $W$ is aperiodic because of Proposition 5.1, and it is topologically regular by B1. As B1 $\Rightarrow$ B8, $X_{\eta}$ is minimal. Therefore Corollary 1a) of [13], together with Lemmas 4.5 and 4.6 of the same reference, implies B9.

Proof of B9 $\Rightarrow$ B8: This is trivial. ∎

Proposition 3.3**.**

Assume that the window $W$ is topologically regular. Then $X_{\eta}\subseteq Y$.

Proof.

We start proving that $\eta\in Y$. Assume the contrary, that is, there are $b_{0}\in{\mathcal{B}}$ and $r\in\{1,\ldots,b_{0}-1\}$ such that

[TABLE]

Let $a=\gcd(r,b_{0})$ and $r^{\prime}=r/a$, $b_{0}^{\prime}=b_{0}/a$. (34) yields that for any $k\in{\mathbbm{N}}$ there exists $b_{k}\in{\mathcal{B}}$ such that

[TABLE]

Let $J=\{k\in{\mathbbm{N}}:r^{\prime}+kb_{0}^{\prime}\;\text{is prime}\}$. By Dirichlet Theorem the set $J$ is infinite. As ${\mathcal{B}}$ is primitive, $b_{k}$ does not divide $a=\gcd(r,b_{0})$. Hence

[TABLE]

for any $k\in J$. Since $a$ has only finitely many divisors, there exists a divisor $a^{\prime}$ such that

[TABLE]

for infinitely many $k\in J$. Thus we obtain a contradiction with the condition (B4) of Theorem B, which is equivalent to (B1) $W=\overline{\operatorname{int}W}$. Thus $\eta\in Y$.

Assume now that $x\in X_{\eta}$ and let $b\in{\mathcal{B}}$. As $\eta\in Y$, there is $N_{b}\in{\mathbbm{N}}$ such that $\operatorname{card}\left(\operatorname{supp}(\eta|_{[0:N_{b}]})\text{ mod }b\right)=b-1$. As $X_{\eta}$ is minimal by (B8) of Theorem B, there is $n\in{\mathbbm{N}}$ such that $\operatorname{supp}(x|_{[n:n+N_{b}]})=\operatorname{supp}(\eta|_{[0:N_{b}]})$. Hence

[TABLE]

so that $x\in Y$, because $\operatorname{card}\left(\operatorname{supp}(x)\text{ mod }b\right)\leqslant b-1$ for all $x\in X_{\eta}$, see Footnote 4 to Remark 1.5. ∎

3.3 Proof of Theorem C

The equivalence of C1, C2 and C3 follows from Lemma 3.1. If C1 holds, i.e. if $\operatorname{int}(W)=\emptyset$, then $\varphi(C_{\varphi})=\{(\dots,0,0,0,\dots)\}$ is a shift invariant set [13, Proposition 3.3d with Remark 3.2b], so that $M=\overline{\varphi(C_{\varphi})}=\{(\dots,0,0,0,\dots)\}$. This is C6, and Theorem 3.8 in [4] shows that C4, C5, C6 and C7 are all equivalent.

We finish by proving C5 $\Rightarrow$ C3: Consider any finite $S\subset{\mathcal{B}}$. As ${\mathcal{B}}\subseteq{\mathcal{M}}_{{\mathcal{A}}_{S}}$ by definition of the set ${\mathcal{A}}_{S}$, C5 implies that $1\in{\mathcal{A}}_{S}$.

4 The sequence ${\mathcal{B}}$ and Haar measure

4.1 Measure and density

Lemma 4.1**.**

$m_{H}(W)=1-\underline{d}({\mathcal{M}}_{\mathcal{B}})=\bar{d}({\mathcal{F}}_{\mathcal{B}})$.

Proof.

For $S\subset{\mathcal{B}}$ denote by ${\mathcal{U}}_{S}$ the family of all sets $U_{S}(\Delta(n))$ that are contained in $W^{c}$ and by $\cup{\mathcal{U}}_{S}$ the union of these sets. Then

[TABLE]

by Lemma 2.5a, and similarly

[TABLE]

by Lemma 2.5b. ∎

Corollary 4.1**.**

[4, Theorem 4.1]* ${\mathcal{B}}$ is a Besicovich sequence if and only if ${\mathcal{F}}_{\mathcal{B}}$ is generic for the Mirsky measure. (As $n\in{\mathcal{F}}_{\mathcal{B}}$ iff $\Delta(n)\in W$, it would be more precise to say that the sequence $(\Delta(n))_{n}$ is generic for the Mirsky measure.)*

Proof.

If ${\mathcal{B}}$ is Besicovich, then $d({\mathcal{F}}_{\mathcal{B}})=m_{H}(W)$, so that ${\mathcal{F}}_{\mathcal{B}}$ has maximal density. Hence it is generic for the Mirsky measure, see [13, Theorem 5b]. On the other hand, if ${\mathcal{F}}_{\mathcal{B}}$ is generic for (any) measure, then its frequency of ones converges in particular, which means that its asymptotic density exists. ∎

Lemma 4.2**.**

$m_{H}(\operatorname{int}(W))=\sup_{S}d({\mathcal{F}}_{{\mathcal{A}}_{S}})\leqslant\underline{d}({\mathcal{E}})$**

Proof.

For $S\subset{\mathcal{B}}$ denote by ${\mathcal{U}}^{o}_{S}$ the family of all sets $U_{S}(\Delta(n))$ that are contained in $\operatorname{int}(W)$ and by $\cup{\mathcal{U}}^{o}_{S}$ the union of these sets. Recall from Lemma 3.1a that $\#{\mathcal{U}}^{o}_{S}=\#({\mathcal{F}}_{{\mathcal{A}}_{S}}\cap\{1,\dots,\operatorname{lcm}(S)\})$. Then

[TABLE]

∎

Lemma 4.3**.**

$m_{H}(\partial W)=\inf_{S}\bar{d}({\mathcal{M}}_{{\mathcal{A}}_{S}}\setminus{\mathcal{M}}_{\mathcal{B}})\leqslant\inf_{S}d({\mathcal{M}}_{{\mathcal{A}}_{S}\setminus{\mathcal{B}}})$.

Proof.

[TABLE]

∎

4.2 Regular Toeplitz sequences

Let ${\mathcal{B}}=\{b_{1},b_{2},\ldots\}$. For each $k\geq 1$, consider the sequence

[TABLE]

where

[TABLE]

Then:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Moreover, following Lemma 2.5c, there is an increasing sequence $(k_{n})$ such that

[TABLE]

We assume that $W\subset H$ is topologically regular, so by Remark 1.3, $\eta={\mathbbm{1}}_{{\mathcal{F}}_{{\mathcal{B}}}}$ is a Toeplitz sequence. We set $s_{k}:={\rm lcm}(b_{1},\ldots,b_{k})$ and would like now to examine the sequence $(s_{k})$ as a periodic structure of $\eta$. More precisely, we would like to see for how many $n\in[1,s_{k}]$, we have $\eta(n)=\eta(n+js_{k})$ for each $j\in{\mathbbm{Z}}$. We call any such $n$ to be “good”. Now, if $n\in{\mathcal{F}}_{\mathscr{A}_{\{b_{1},\ldots,b_{k}\}}}$, then $n+s_{k}{\mathbbm{Z}}\subset{\mathcal{F}}_{\mathscr{A}_{\{b_{1},\ldots,b_{k}\}}}$, so $n$ is good. Otherwise, $n\in{\mathcal{M}}_{\mathscr{A}_{\{b_{1},\ldots,b_{k}\}}}$. Then either $n\in{\mathcal{M}}_{b_{1},\ldots,b_{k}}$ and then clearly $\eta(n+js_{k})=0$ for each $j\in{\mathbbm{Z}}$, so again $n$ is good, or

[TABLE]

Only for such $n$, we are not sure that $n$ is good. Moreover, note that in view of (4.2), we have

[TABLE]

so the sequence $(d({\mathcal{M}}_{\{c^{(k)}_{k+i}:\>i\geq 1\}}))_{k}$ is decreasing, and so is the sequence $(\overline{d}({\mathcal{M}}_{\{c^{(k)}_{k+i}:\>i\geq 1\}}\setminus{\mathcal{M}}_{{\mathcal{B}}}))$. Therefore, by taking into account (35), the infimum of this sequence is equal to the liminf, in fact to the limit and we have

[TABLE]

Definition 4.1**.**

Let $\eta={\mathbbm{1}}_{{\mathcal{F}}_{{\mathcal{B}}}}$ be a Toeplitz sequence. It is a regular Toeplitz sequence for the periodic structure $(s_{k})$, $s_{k}={\rm lcm}(b_{1},\ldots,b_{k})$, if the $\liminf$ in (36) is zero.

Now, using Lemma 4.3, the identity in (36) shows the following result.

Proposition 4.1**.**

If $W$ is topologically regular, then $\eta={\mathbbm{1}}_{{\mathcal{F}}_{{\mathcal{B}}}}$ is a regular Toeplitz sequence for the periodic structure $(s_{k})$, $s_{k}={\rm lcm}(b_{1},\ldots,b_{k})$, if and only if $m_{H}(\partial W)=0$.

Example 4.1**.**

Assume that $\{b^{\prime}_{k}:\>k\geq 1\}$ is a coprime set of odd numbers and let $b_{k}=2^{k}b^{\prime}_{k}$. Then $c^{(k)}_{k+i}=2^{k}$ for each $i\geq 1$. Hence, we have even $d({\mathcal{M}}_{\{c^{(k)}_{k+i}:\>i\geq 1\}})\to 0$, in particular $\eta$ is a regular Toeplitz sequence for the periodic structure $(s_{k})$ with $s_{k}=2^{k}b_{1}^{\prime}\cdots b_{k}^{\prime}$. This example comes from [4].

We will now show that we can obtain Toeplitz sequences also in case $m_{H}(\partial W)>0$.

Example 4.2**.**

We will construct ${\mathcal{B}}=\{b_{1},b_{2},\ldots\}$ such that this set is thin (hence taut) and that $\lim_{k\to\infty}\inf(\{c^{(k)}_{k+i}:\>i\geq 1\}\setminus\{b_{1},\ldots,b_{k}\})=\infty$, which, by Proposition 3.2, implies that $W$ is topologically regular (and hence $\eta$ is Toeplitz by Remark 1.3). Let $\delta_{k}>0$ and $\sum_{k\geq 1}\delta_{k}<1/16$.

We start with $b_{1}=2^{3}$ and set $c^{(1)}_{1+i}=2$ for each $i\geq 1$. Suppose that a sequence

[TABLE]

has been defined. We require that this sequence satisfies:

[TABLE]

[TABLE]

[TABLE]

We will now show how to define $c^{(k+1)}_{k+2}$, $c^{(k+1)}_{k+3}$, $\ldots$ and then $b_{k+1}$. Recall an elementary lemma.

Lemma 4.4**.**

Let $F_{1},F_{2}$ be finite sets of natural numbers such that ${\rm gcd}(f_{1},f_{2})=1$ for each $f_{i}\in F_{i}$, $i=1,2$. Then $d({\mathcal{M}}_{F_{1}\cdot F_{2}})=d({\mathcal{M}}_{F_{1}}\cap{\mathcal{M}}_{F_{2}})=d({\mathcal{M}}_{F_{1}})d({\mathcal{M}}_{F_{2}})$.

Choose $P\subset\mathscr{P}\setminus{\rm spec}\,\{b_{1},\ldots,b_{k}\}$, so that (by Lemma 4.4)

[TABLE]

In view of (4.2),

[TABLE]

Let $P=\{q_{1},\ldots,q_{t}\}$. Then set

[TABLE]

for each $j=0,1,\ldots$ If $2\notin\{i\geq 2:\>c^{(k)}_{k+i}=c^{(k)}_{k+1}\}$ then repeat the same construction with the set $\{i\geq 2:\>c^{(k)}_{k+i}=c^{(k)}_{k+2}\}$. Since (by (4.2)) the set $\{c^{(k)}_{k+i}:\>i\geq 1\}$ is finite, our construction of the sequence $(c^{(k+1)}_{k+1+i})_{i}$ is done in finitely many steps. Finally, we set $b_{k+1}:=c^{(k)}_{k+1}\prod_{q\in P}q$ (or, if needed, $b_{k+1}:=c^{(k)}_{k+1}\prod_{q\in P}q^{\alpha_{k+1}}$ for any $\alpha_{k+1}\in{\mathbbm{N}}$). Note that

[TABLE]

since $P\cap\{b_{1},\ldots,b_{k}\}=\emptyset$. More than that, by the construction, we also have

[TABLE]

Moreover, it is not hard to see that the new sequence

[TABLE]

satisfies (4.2)-(4.2). Furthermore, ${\mathcal{B}}=\{b_{1},b_{2},\ldots\}$ satisfies the other requirements mentioned at the beginning of the construction so that $\eta$ is a Toeplitz sequence and $W$ is topologically regular. Note that in our construction $d({\mathcal{M}}_{\{c^{(1)}_{1+i}:\>i\geq 1\}})=1/2$. Moreover, by (4.2)

[TABLE]

for each $k\geq 1$. Finally notice that $d({\mathcal{M}}_{{\mathcal{B}}})\leq\sum_{k\geq 1}1/b_{k}$, which (by construction) can be made smaller than 1/8. It follows that $\lim_{k\to\infty}\overline{d}({\mathcal{M}}_{\mathscr{A}_{\{b_{1},\ldots,b_{k}\}}}\setminus{\mathcal{M}}_{{\mathcal{B}}})>0$, whence $m_{H}(\partial W)>0$.

5 The maximal equicontinuous factor of $X_{\eta}$

5.1 The period groups of $W$ and of $\operatorname{int}(W)$

Given a subset $A\subseteq H$, denote by

[TABLE]

the period group of $A$. The set $A\subseteq H$ is topologically aperiodic, if $H_{A}=\{0\}$. The following simple observations are proved in [14, Lemma 6.1]:

• •

$H_{A}\subseteq H_{\bar{A}}\cap H_{\operatorname{int}(A)}$.

• •

If $A$ is closed, then $H_{\operatorname{int}(A)}=H_{\overline{\operatorname{int}(A)}}$ is closed.

Proposition 5.1**.**

Assume that ${\mathcal{B}}$ is primitive. Then the window $W$ is topologically aperiodic.

Proof.

Suppose that $h=(h_{b})_{b\in{\mathcal{B}}}\neq 0$ and

[TABLE]

Since $h\neq 0$, there is $b\in{\mathcal{B}}$ such that $b$ does not divide $h_{b}$. Let $n:=\gcd(b,h_{b})$. Then $n\in{\mathcal{F}}_{{\mathcal{B}}}$, as otherwise there exists $b^{\prime}\in{\mathcal{B}}$ such that $b^{\prime}\mid n$; but then $b^{\prime}\mid b$ a contradiction (${\mathcal{B}}$ is assumed to be primitive). Hence $\Delta(n)\in W$. There are $x,y\in{\mathbbm{Z}}$ such that $n=xh_{b}+yb$, whence $b\mid n-xh_{b}$. It follows that $\Delta(n)-xh\notin W$, a contradiction with (37). ∎

If $W$ is topologically regular, then clearly $\operatorname{int}(W)$ is topologically aperiodic, as well. Otherwise $H_{\operatorname{int}(W)}$ may be non-trivial, as we will see in the course of this section.

Recall from (1) that for any set $A\subseteq{\mathbbm{N}}$,

[TABLE]

If $A^{prim}$ is finite, then ${\mathcal{M}}_{A}$ is a union of finitely many arithmetic progressions. Let $c_{A}$ denote the period of ${\mathcal{M}}_{A}$, that is, the least natural number such that $c_{A}+{\mathcal{M}}_{A}={\mathcal{M}}_{A}$.

Lemma 5.1**.**

Assume that $A,B\subset{\mathbbm{N}}$ are finite

1. a)

If $c+{\mathcal{M}}_{A}={\mathcal{M}}_{A}$ for some $c\in{\mathbbm{N}}$, then $\operatorname{lcm}(A^{prim})\mid c$. 2. b)

$c_{A}=\operatorname{lcm}(A^{prim})$** 3. c)

if $A\subset{\mathcal{M}}_{B}$ then $c_{B}\mid c_{A}$

Proof.

1. a)

For any $a\in A^{prim}$ we have $a+c{\mathbbm{Z}}\subseteq{\mathcal{M}}_{A^{prim}}$, from which it follows that there exists $a^{\prime}\in A^{prim}$ such that $a^{\prime}\mid\gcd(a,c)$. But, as $A^{prim}$ is primitive, that means that $a^{\prime}=a$ and $a\mid c$. We conclude that $\operatorname{lcm}(A^{prim})\mid c$. 2. b)

Clearly ${\mathcal{M}}_{A}+\operatorname{lcm}(A^{prim})={\mathcal{M}}_{A}$, thus $c_{A}\mid\operatorname{lcm}(A^{prim})$ and the assertion follows from a). 3. c)

If $A\subset{\mathcal{M}}_{B}$ then $A^{prim}\subset{\mathcal{M}}_{B^{prim}}$, hence $\operatorname{lcm}(B^{prim})\mid\operatorname{lcm}(A^{prim})$, and we finish by b).

∎

Lemma 5.2**.**

Assume that $S\subseteq S^{\prime}$ are finite subsets of ${\mathcal{B}}$, then ${\mathcal{A}}_{S}=\{\gcd(a,\operatorname{lcm}(S)):a\in{\mathcal{A}}_{S^{\prime}}\}$.

Proof.

Since $\operatorname{lcm}(S)\mid\operatorname{lcm}(S^{\prime})$, $\gcd(b,\operatorname{lcm}(S))=\gcd(\gcd(b,\operatorname{lcm}(S^{\prime})),\operatorname{lcm}(S))$ for any $b\in{\mathcal{B}}$ and the assertion follows. ∎

Let $S_{1}\subset S_{2}\subset\ldots\subset S_{k}\subset\ldots$ be a filtration of ${\mathcal{B}}$ with finite sets and denote

[TABLE]

By Lemma 5.1 c) we have $c_{l}\mid c_{l+1}$ for any $l$. It follows that, for any $k$, the sequence $(\gcd(s_{k},c_{l}))_{l\geq 1}$ stabilizes on a divisor $d_{k}$ of $s_{k}$. Clearly, since $c_{k}\mid s_{k}$,

[TABLE]

Observe that

[TABLE]

Indeed, there is $l_{0}\in{\mathbbm{N}}$ such that $d_{k+1}=\gcd(s_{k+1},c_{l})$ for all $l>l_{0}$. Since $s_{k}\mid s_{k+1}$, we get

[TABLE]

It follows that $\gcd(s_{k},c_{l})=\gcd(s_{k},d_{k+1})$ for $l>l_{0}$, and (39) follows.

By applying (39) we prove by induction that

[TABLE]

for $j\geq 0$.

Lemma 5.3**.**

Let $(n_{k})_{k\in{\mathbbm{N}}}$ be a sequence of integers. The following are equivalent:

[TABLE]

and

[TABLE]

Proof.

As $c_{k}\mid d_{k}$, (42) implies (41). Conversely, assume that (41) holds. We show inductively that for all $j\geqslant 0$

[TABLE]

and this implies (42) immediately.

For $j=0$, (43) follows from (41), because $c_{k}\mid s_{k}$. So suppose that (43) holds for some $j\geqslant 0$. Then

[TABLE]

Hence $n_{k}=0\text{ mod }\gcd(s_{k},s_{k+1},c_{k+1+j})=\gcd(s_{k},c_{k+j+1})$ for all $k\in{\mathbbm{N}}$, i.e. (43) for $j+1$. ∎

Recall that $H_{\operatorname{int}(W)}=\{h\in H:\operatorname{int}(W)+h=\operatorname{int}(W)\}$ denotes the period group of $\operatorname{int}(W)$.

Proposition 5.2**.**

1. a)

$h\in H_{\operatorname{int}(W)}$* if and only if $h=\lim_{k}\Delta(n_{k})$ for some sequence $(n_{k})_{k}$ satisfying*

[TABLE]

Moreover, sequences $(n_{k})_{k}$ can be defined inductively: For $n_{1}$ there are $s_{1}/d_{1}$ choices and, given $n_{1},\dots,n_{k}$, there are precisely $s_{k+1}/\operatorname{lcm}(s_{k},d_{k+1})$ many choices for $n_{k+1}$. 2. b)

$H_{\operatorname{int}(W)}=\{0\}$* if and only if $s_{k}=d_{k}$ for all $k\in{\mathbbm{N}}$.*

Remark 5.1**.**

Observe that, in view of (39),

[TABLE]

so that

[TABLE]

[TABLE]

Proof of Proposition 5.2.

a) For each $S_{k}$ denote by $W_{k}:=\bigcup_{n\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}}U_{S_{k}}(\Delta(n))$. Then $\operatorname{int}(W)$ is the increasing union of the sets $W_{k}$, see Lemma 3.1, and $U_{S_{k}}(\Delta(n))\subseteq W_{k}$ if and only if $U_{S_{k}}(\Delta(n))\subseteq\operatorname{int}(W)$. Let $h=\lim_{k}\Delta(n_{k})$, where $n_{k}$ stands for $n_{S_{k}}$, which was defined in Lemma 3.1b. Then

[TABLE]

and $h\in H_{\operatorname{int}(W)}$, if and only if

[TABLE]

Indeed, let $k\in{\mathbbm{N}}$, $m\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$, and let $g=(g_{b})_{b\in{\mathcal{B}}}$ be any element from $U_{S_{k}}(\Delta(m))\subseteq\operatorname{int}(W)$. Then $g_{b}=m\text{ mod }b$ for all $b\in S_{k}$. Assume now that $h\in H_{\operatorname{int}(W)}$. Then $g+h\in\operatorname{int}(W)$ and $(g+h)_{b}=m+n_{k}\text{ mod }b$ for all $b\in S_{k}$, so that $g+h\in U_{S_{k}}(\Delta(m+n_{k}))$. Hence $U_{S_{k}}(\Delta(m))+h\subseteq U_{S_{k}}(\Delta(m+n_{k}))=U_{S_{k}}(\Delta(m))+\Delta(n_{k})$. In particular, $U_{S_{k}}(\Delta(m))$ and $U_{S_{k}}(\Delta(m+n_{k}))$ have identical Haar measure, and so do $U_{S_{k}}(\Delta(m))+h$ and $U_{S_{k}}(\Delta(m+n_{k}))$. As both are open sets and one is contained in the other, they must coincide. Hence $U_{S_{k}}(\Delta(m+n_{k}))=U_{S_{k}}(\Delta(m))+h\subseteq\operatorname{int}(W)+h=\operatorname{int}(W)$, so that $m+n_{k}\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$. This proves that ${\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}+n_{k}\subseteq{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$. As ${\mathcal{A}}_{S_{k}}$ is a finite set, this implies ${\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}+n_{k}={\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$.

Conversely, assume that (46) holds, and let $U_{S_{k}}(\Delta(m))\subseteq\operatorname{int}(W)$. Recall that this implies $U_{S_{k}}(\Delta(m))\subseteq W_{k}$, i.e. $m\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$. Hence, by assumption, also $m+n_{k}\in{\mathcal{F}}_{{\mathcal{A}}_{S_{k}}}$, so that $U_{S_{k}}(\Delta(m+n_{k}))\subseteq W_{k}\subseteq\operatorname{int}(W)$. Let $g\in U_{S_{k}}(\Delta(m))$. Then $g_{b}=m\text{ mod }b$ for all $b\in S_{k}$, so that $(g+h)_{b}=m+n_{k}\text{ mod }b$ for all $b\in S_{k}$, i.e. $g+h\in U_{S_{k}}(\Delta(m+n_{k}))$. Hence $U_{S_{k}}(\Delta(m))+h\subseteq U_{S_{k}}(\Delta(m+n_{k}))\subseteq\operatorname{int}(W)$. As this argument applies to all $k$ and all $U_{S_{k}}(\Delta(m))\subseteq\operatorname{int}(W)$, it proves that $\operatorname{int}(W)+h\subseteq\operatorname{int}(W)$. The same Haar measure argument as before, applied to the open set $\operatorname{int}(W)$, shows that $\operatorname{int}(W)+h=\operatorname{int}(W)$, i.e. $h\in H_{\operatorname{int}(W)}$.

Condition (46) is equivalent to

[TABLE]

Invoking Lemma 5.3, we conclude

[TABLE]

This proves the claimed equivalence.

Now we describe all sequences $(n_{k})_{k\in{\mathbbm{N}}}$ which satisfy(44) and $n_{k}\in\{0,\dots,s_{k}-1\}$ for all $k$. Denote $q_{k}:=s_{k}/d_{k}$.

1. $n_{1}$:

Let $n_{1}=m_{1}d_{1}$ for any $m_{1}\in\{0,\dots,q_{1}-1\}$. 2. $n_{2}$:

$n_{2}$ must be chosen such that $n_{2}=0\text{ mod }d_{2}$ and $n_{2}=n_{1}\text{ mod }s_{1}$. As $\gcd(s_{1},d_{2})=d_{1}\mid n_{1}$ in view of (39), the CRT guarantees the existence of at least one solution $n_{2}$, and if $n_{2}$ is one particular solution, then the set of all solutions is precisely $n_{2}+\operatorname{lcm}(s_{1},d_{2})\cdot{\mathbbm{Z}}$. As $n_{2}$ is to be chosen in $\{0,\dots,s_{2}-1\}$, there are exactly $s_{2}/\operatorname{lcm}(s_{1},d_{2})$ possible choices for $n_{2}$. 3. $\vdots$ 4. $n_{k+1}$:

$n_{k+1}$ must be chosen such that $n_{k+1}=0\text{ mod }d_{k+1}$ and $n_{k+1}=n_{k}\text{ mod }s_{k}$. As $\gcd(s_{k},d_{k+1})=d_{k}\mid n_{k}$ in view of (39), the CRT guarantees the existence of at least one solution $n_{k+1}$, and if $n_{k+1}$ is one particular solution, then the set of all solutions is precisely $n_{k+1}+\operatorname{lcm}(s_{k},d_{k+1})\cdot{\mathbbm{Z}}$. As $n_{k+1}$ is to be chosen in $\{0,\dots,s_{k+1}-1\}$, there are exactly $s_{k+1}/\operatorname{lcm}(s_{k},d_{k+1})$ possible choices for $n_{k+1}$.

b) $H_{\operatorname{int}(W)}=\{0\}$ $\Leftrightarrow$ there is unique choice of the numbers $n_{k}$ described in a) $\Leftrightarrow$ $s_{1}/d_{1}=1$ and $s_{k+1}/\operatorname{lcm}(s_{k},d_{k+1})=1$ for any $k$ $\Leftrightarrow$ $d_{k}=s_{k}$ for any $k$, the last equivalence by Remark 5.1. ∎

5.2 Proof of Theorem D

Remark 5.2**.**

If $(S_{k})_{k}$ is a filtration of ${\mathcal{B}}$ by finite sets and if $h=(h_{b})_{b\in{\mathcal{B}}}\in H$, then we write $\lim_{k}\Delta(n_{S_{k}})=h$, whenever $n_{S_{k}}\in{\mathbbm{Z}}$ are numbers such that for every $k\in{\mathbbm{N}}$:

[TABLE]

Let us denote $s_{k}=\operatorname{lcm}(S_{k})$. There is an inverse system of groups

[TABLE]

The homomorphisms are the canonical projections. Observe that $s_{k}|n_{S_{k+1}}-n_{S_{k}}$ for any $k$ and the sequence $(n_{S_{k}}+s_{k}{\mathbbm{Z}})_{k}$ is an element of the inverse limit $\lim\limits_{\leftarrow}{\mathbbm{Z}}/{s_{k}}{\mathbbm{Z}}$. In this way we obtain an isomorphism of topological groups

[TABLE]

given by $(n_{S_{k}}+s_{k}{\mathbbm{Z}})_{k}\mapsto\lim_{k}\Delta(n_{S_{k}})$. Compare Remark 2.32 [4]. In particular, the inverse limit does not depend on the filtration $(S_{k})_{k}$.161616The last statement follows from a general property of inverse limits: the inverse limits of cofinal inverse systems are isomorphic, [9, Chapter II, Section 12].

Proof of Proposition 1.3.

Let $\beta_{k}:{\mathbbm{Z}}/s_{k}{\mathbbm{Z}}\rightarrow{\mathbbm{Z}}/d_{k}{\mathbbm{Z}}$ be the map given by $n+s_{k}{\mathbbm{Z}}\mapsto n+d_{k}{\mathbbm{Z}}$, let $M_{k}$ be the kernel of $\beta_{k}$ and let $\alpha_{k}:M_{k}\rightarrow{\mathbbm{Z}}/s_{k}{\mathbbm{Z}}$ be the canonical embedding. There is a commutative diagram of abelian groups

[TABLE]

where $f_{k}(n+s_{k}{\mathbbm{Z}})=n+s_{k-1}{\mathbbm{Z}}$, $f^{\prime}_{k}$ is the restriction of $f_{k}$ to $M_{k}$ and $f^{\prime\prime}_{k}(n+d_{k}{\mathbbm{Z}})=n+d_{k-1}{\mathbbm{Z}}$.

The columns of the diagram are exact sequences of groups, in other words, the diagram can be interpreted as an exact sequence of inverse systems of abelian groups.

Since inverse limit is a left exact functor, see [9, Chapter II, Theorem 12.3], we obtain an exact sequence

[TABLE]

The condition (39) yields that the homomorphism $\gamma$ in (50) is surjective, thus we have an exact sequence

[TABLE]

Indeed, let $(n_{k}+d_{k}{\mathbbm{Z}})_{k}\in\lim\limits_{\leftarrow}{\mathbbm{Z}}/{d_{k}}{\mathbbm{Z}}$. By induction we construct the numbers $m_{1},m_{2},\ldots$ such that $d_{k}|m_{k}-n_{k}$ and $s_{k}|m_{k+1}-m_{k}$, for any $k$. Then $\beta((m_{k}+s_{k}{\mathbbm{Z}})_{k})=(n_{k}+d_{k}{\mathbbm{Z}})_{k}$. We set $m_{1}=n_{1}$. Assume that $m_{1},\ldots m_{k}$ have been defined. Since $d_{k}\mid n_{k+1}-n_{k}$, $d_{k}\mid m_{k}-n_{k}$ and $\gcd(d_{k+1},s_{k})=d_{k}$, there exists integers $x,y$ such that $xd_{k+1}+ys_{k}=m_{k}-n_{k+1}$. We set $m_{k+1}=m_{k}-ys_{k}$.

There are group isomorphisms $g_{k}:{\mathbbm{Z}}/{\frac{s_{k}}{d_{k}}}{\mathbbm{Z}}\rightarrow M_{k}$ given by $g_{k}(n+\frac{s_{k}}{d_{k}}{\mathbbm{Z}})=d_{k}n+s_{k}{\mathbbm{Z}}$ and making the following diagram commutative

[TABLE]

(the arrows in the upper row represent the canonical projections). It follows that there is an isomorphism

[TABLE]

By Proposition 5.2 a) it follows that $\lim\limits_{\leftarrow}M_{k}$ is isomorphic to $H_{\operatorname{int}(W)}$. There is an isomorphism given by $\sigma\alpha$, where $\sigma$ is the isomorphism defined in Remark 5.2.

Now a), b) and c) follow from (51), (52) and Remark 5.2. In order to prove d) it is enough to note that $s_{k}=d_{k}$ if and only if $s_{k}\mid c_{k+j}$ for some $j\geq 0$. ∎

Proof of Theorem D.

This is an immediate corollary to Proposition 1.3. ∎

5.3 Examples

Remark 5.3**.**

Given a prime number $p$ and $m\in{\mathbbm{Z}}$ we denote by $v_{p}(m)$ be the $p$-valuation of $m$, that is, if $m\neq 0$ then $v_{p}(m)$ is the maximal integer such that $p^{v_{p}(m)}\mid m$ and $v_{p}(0)=+\infty$. Assume that $t=(t_{k})$ is a sequence of natural numbers such that $t_{k}\mid t_{k+1}$ for any $k$. Set $v_{p}(t)=\sup_{k}v_{p}(t_{k})$. The sequence $t$ yields an inverse system of abelian groups

[TABLE]

where the arrows represent the canonical projections $n+t_{k+1}{\mathbbm{Z}}\mapsto n+t_{k}{\mathbbm{Z}}$. The inverse limit $\lim\limits_{\leftarrow}{\mathbbm{Z}}/t_{k}{\mathbbm{Z}}$ of this system is isomorphic to the group

[TABLE]

where $G_{p}={\mathbbm{Z}}/p^{v_{p}(t)}{\mathbbm{Z}}$ if $v_{p}(t)<+\infty$ and $G_{p}=\widehat{{\mathbbm{Z}}}_{p}$ (the group of $p$-adic numbers) otherwise, i.e. when $\lim_{k}v_{p}(t_{k})=+\infty$.

Recall from (6) that

[TABLE]

Our first exaxmple has a finite, non-trivial maximal equicontinuous factor and a finite set ${\mathcal{A}}_{\infty}$.

Example 5.1**.**

${\mathcal{B}}=\{36\}\cup\{2p_{1},2p_{2},\ldots\}\cup\{3q_{1},3q_{2},\ldots\}$, where $p_{1},q_{1},p_{2},q_{2},\ldots$ are pairwise different primes. Let $S_{k}=\{36,2p_{1},\dots,2p_{k},3q_{1},\dots,3q_{k}\}$. Then

[TABLE]

so that

[TABLE]

In particular, the maximal equicontinuous factor of $X_{\eta}$ is the translation by $1$ on ${\mathbbm{Z}}/6{\mathbbm{Z}}$. Moreover, ${\mathcal{A}}_{\infty}=\{2,3\}$, so that $\emptyset\neq\overline{\operatorname{int}(W)}\neq W$ by Theorems B and C.

Our next example has an infinite maximal equicontinuous factor different from $H$ and an infinite set ${\mathcal{A}}_{\infty}$.

Example 5.2**.**

Let $p_{1},q_{1},p_{2},q_{2},\ldots$ be pairwise different primes. Let

[TABLE]

where

[TABLE]

That is,

[TABLE]

for $k\geq 2$.

Let $S_{k}={\mathcal{B}}_{1}\cup\ldots\cup{\mathcal{B}}_{k}$. Then $s_{k}=\operatorname{lcm}(S_{k})=p_{1}p_{2}^{2}\ldots p_{k}^{2}q_{1}q_{2}^{2}\ldots q_{k}^{2}$ and

[TABLE]

so that

[TABLE]

Hence

[TABLE]

so that

[TABLE]

Hence $H_{\operatorname{int}(W)}\cong\prod_{i=2}^{+\infty}{\mathbbm{Z}}/p_{i}q_{i}{\mathbbm{Z}}$ and $H/H_{\operatorname{int}(W)}\cong\prod_{i=1}^{+\infty}{\mathbbm{Z}}/p_{i}q_{i}{\mathbbm{Z}}$ are infinite compact groups. Moreover,

[TABLE]

is infinite and does not contain the number $1$, thus $\emptyset\neq\overline{\operatorname{int}(W)}\neq W$ by Theorems B and C.

We end with a non-trivial example where the maximal equicontinuous factor equals $H$ and ${\mathcal{A}}_{\infty}$ is an infinite set.

Example 5.3**.**

Let $q,p_{1},p_{2},\ldots$ be pairwise different odd primes. Let

[TABLE]

where

[TABLE]

That is,

[TABLE]

for $k\geq 1$. Let $S_{k}={\mathcal{B}}_{1}\cup\ldots\cup{\mathcal{B}}_{k}$. Then $s_{k}=\operatorname{lcm}(S_{k})=qp_{1}\ldots p_{k}$ and

[TABLE]

hence $c_{{\mathcal{A}}_{S_{k}}}=qp_{1}\ldots p_{k}=\operatorname{lcm}(S_{k})$, so that $s_{k}=c_{k}=d_{k}$ for all $k$. In particular $\operatorname{int}(W)$ is aperiodic by Proposition 5.2. Moreover,

[TABLE]

is infinite and does not contain the number $1$, thus $\emptyset\neq\overline{\operatorname{int}(W)}\neq W$ by Theorems B and C.