\begin{document}

\title{Generalized heredity in ${\mathcal B}$-free systems

}
\author{Gerhard Keller
\thanks{I am indebted to Aurelia Dymek for several very helpful remarks and corrections to a previous version of this manuscript.}
\
\thanks{
The final version of these notes is based upon work supported by the Swedish Research Council under grant no. 2016-06596 while the author was in residence at Institut Mittag-Leffler in Djursholm, Sweden during the workshop
``Thermodynamic Formalism -- Applications to Geometry, Number Theory, and Stochastics'' (2019).}
}
\affil{Department of Mathematics, University of Erlangen-N\"urnberg, \\
 Cauerstr. 11, 91058 Erlangen, Germany
 \par Email: keller@math.fau.de}
\date{Version of \today}

\maketitle

{\begin{abstract}
Let ${\mathcal B}\subseteq{\mathbbm N}$ be a primitive set. We complement results on heredity of the ${\mathcal B}$-free subshift $X_\eta$ from \cite{BKKL2015} in two directions: In the proximal case we prove that a subshift $X_\varphi$, which might be slightly larger than the subshift $X_\eta$, is always hereditary. (There is no need to assume that the set ${\mathcal B}$ has light tails as in \cite{BKKL2015}, but if ${\mathcal B}$ is taut, in particular if it has light tails, then $X_\varphi=X_\eta$ \cite{KKL2016,Keller2019}.) We also generalize the concept of heredity to the non-proximal (and hence non-hereditary) case by proving that $X_\varphi$ is always ``hereditary away from its unique minimal subsystem'' (which is alway Toeplitz). Finally we characterize regularity of this Toeplitz subsystem equivalently by the condition $m_H\left(\partial(\overline{\operatorname{int}(W)})\right)=0$, where $W$ (``the window'') is a subset of a compact abelian group $H$ canonically associated with the set ${\mathcal B}$, and $m_H$ denotes Haar measure on $H$. Throughout, results from \cite{KKL2016} are heavily used.
\end{abstract}}
\blfootnote{\emph{MSC 2010 classification:} 37A35, 37A45, 37B05.}
\blfootnote{\emph{Keywords:} ${\mathcal B}$-free dynamics, sets of multiples, heredity.}

\section{Introduction and results}\label{sec:introduction}
For any given non-empty set ${\mathcal B}\subseteq{\mathbbm N}=\{1,2,\dots\}$ one can define its \emph{set of multiples}
\begin{equation*}
{\mathcal M}_{\mathcal B}:=\bigcup_{b\in{\mathcal B}}b{\mathbbm Z}
\end{equation*}
and the set of \emph{${\mathcal B}$-free numbers}
\begin{equation*}
{\mathcal F}_{\mathcal B}:={\mathbbm Z}\setminus{\mathcal M}_{\mathcal B}\ .
\end{equation*}
The investigation of structural properties of ${\mathcal M}_{\mathcal B}$ or, equivalently, of ${\mathcal F}_{\mathcal B}$ has a long history (see the monograph \cite{hall-book} and the recent paper \cite{BKKL2015} 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
$(\{0,1\}^{\mathbbm Z},S)$, where $S$ 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 \cite{Sa} which originated
 this point of view by studying the set of square-free numbers, and also
\cite{Peckner2012}, \cite{Ab-Le-Ru}, \cite{BKKL2015}, \cite{KKL2016} for later contributions.

\subsection{Motivation and main results} 
In this note we always assume that ${\mathcal B}$ is \emph{primitive}, i.e. that there are no $b,b'\in{\mathcal B}$ with $b\mid b'$. We say that $X_\eta$ is \emph{hereditary}, if $y\in\{0,1\}^{\mathbbm Z}$ belongs to $X_\eta$ whenever there is $x\in X_\eta$ with $y\leqslant x$.
For certain hereditary ${\mathcal B}$-free systems the structure of invariant measures, in particular the uniqueness of the measure of maximal entropy, was studied in \cite{KLW2015} and \cite{BKKL2015}, see also \cite{KPM2019}. Here we apply some of the tools developed in \cite{KKL2016} to generalize the concept of heredity.

Obviously, if $X_\eta$ is hereditary, then
$(\dots,0,0,\dots)\in X_\eta$,
in which case this fixed point of the shift is the unique minimal subset of $(X_\eta,S)$ \cite[Thm.~B]{BKKL2015}.
Theorem~3.7 of \cite{BKKL2015} and Theorem~C of \cite{KKL2016} provide lists of properties which are all equivalent to 
``$(\dots,0,0,\dots)\in X_\eta$'' and which also motivate to call ${\mathcal B}$  \emph{proximal} in this case.
It is noteworthy that one of these equivalent properties is ``${\mathcal B}$ contains an infinite pairwise co-prime subset''.

Recall that the set ${\mathcal B}\subseteq{\mathbbm N}$ is \emph{taut} if \footnote{Indeed, tautness of ${\mathcal B}$ is defined in \cite{hall-book} by $\boldsymbol{\delta}({\mathcal M}_{{\mathcal B}\setminus\{b\}})<\boldsymbol{\delta}({\mathcal M}_{\mathcal B})$ 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 and to coincide with $\underline{d}({\mathcal M}_{\mathcal B})$ by the Theorem of Davenport and Erd\"os \cite{DE1936,DE1951}.}
\begin{equation}
\forall b\in{\mathcal B}:\ \underline{d}\left({\mathcal M}_{{\mathcal B}\setminus\{b\}}\right)<\underline{d}\left({\mathcal M}_{\mathcal B}\right),
\end{equation}
and that it has \emph{light tails}, if
\begin{equation}
\lim_{K\to\infty}\overline{d}\left({\mathcal M}_{\{b\in{\mathcal B}:b>K\}}\right)=0\ ,
\end{equation}
where $\underline{d}({\mathcal A}):=\liminf_{N\to\infty}N^{-1}\operatorname{card}({\mathcal A}\cap[1,N])$ and
$\overline{d}({\mathcal A}):=\limsup_{N\to\infty}N^{-1}\operatorname{card}({\mathcal A}\cap[1,N])$. 
If ${\mathcal B}$ has light tails, then 
${\mathcal B}$ is taut, but the converse does not hold \cite[Sec.~4.3]{BKKL2015}.

It is proved in \cite[Theorem~D]{BKKL2015} that, for sets ${\mathcal B}$ with light tails,
proximality is not only necessary for heredity, but also sufficient, and this result is extended to taut sets in \cite[Thm.~3]{Keller2019}.
Our main result generalizes the concept of heredity to the non-proximal case, and it highlights the role of the tautness assumption in this result.
In order to formulate it, we 
recall some notions from \cite{KKL2016}.
\begin{itemize}
\item $\Delta:{\mathbbm Z}\to\prod_{b\in{\mathcal B}}{\mathbbm Z}/b{\mathbbm Z}$, $\Delta(n)=(n,n,\dots)$, denotes the canonical diagonal embedding.
\item $H:=\overline{\Delta({\mathbbm Z})}$ is a compact abelian group, and we denote by $m_H$ its normalised Haar measure.

\item The \emph{window} associated to ${\mathcal B}$  is defined as
\begin{equation}\label{eq:W}
W:=\{h\in H: h_b\neq0\ (\forall b\in{\mathcal B})\}.
\end{equation}
\item For an arbitrary subset $A\subseteq H$ we define the coding function
$\varphi_A:H\to\{0,1\}^{\mathbbm Z}$ by $\varphi_A(h)(n)=1$ if  and only if $h+\Delta(n)\in A$. Of particular interest are the coding functions $\varphi:=\varphi_W$,
$\varphi_{W'}$ and $\varphi_{\operatorname{int}(W)}$, where $W':=\overline{\operatorname{int}(W)}$.
Observe that $\varphi_{\operatorname{int}(W)}\leqslant\varphi_{W'}\leqslant\varphi$, and that $\varphi_{\operatorname{int}(W)}$ is lower semicontinuous, while $\varphi_{W'}$ and $\varphi$ are upper semicontinuous.
\item
Observe that
$\varphi(h)(n)=1$ if  and only if  $h_b+n\neq0$ mod $b$ for all $b\in{\mathcal B}$.
\item With this notation $\eta=\varphi(\Delta(0))$ and $X_\eta=\overline{\varphi(\Delta({\mathbbm Z}))}$, so that
$X_\eta\subseteq X_\varphi:=\overline{\varphi(H)}$.
\end{itemize}
Finally we need to introduce some further notation:
\begin{equation}
[\varphi_{W'},\varphi] :=\left\{x\in\{0,1\}^{\mathbbm Z}: \exists h\in H\text{ s.t. }\varphi_{W'}(h)\leqslant x\leqslant\varphi(h)\right\},
\end{equation}
and for any closed subshift $X\subseteq\{0,1\}^{\mathbbm Z}$,
\begin{equation}
\tilde X:=\{x'\in\{0,1\}^{\mathbbm Z}: \exists\, x\in X\text{ s.t. }x'\leqslant x\}.
\end{equation}
is the \emph{hereditary closure} of $X$.
Observe that
$X$ is {hereditary}, if and only if $\tilde{X}=X$.

\begin{remark}\label{rem:closed-strip}
$[\varphi_{W'},\varphi]$ is a shift-invariant subset of $\{0,1\}^{\mathbbm Z}$. It is not necessarily closed, because in general $\varphi_{W'}$ is only upper semicontinuous. It is closed when $\varphi_{W'}$ is continuous, i.e. when $W'$ is open and closed.
This applies in particular when ${\mathcal B}$ is proximal, i.e. when $\operatorname{int}(W)=\emptyset$ 
\cite{KR2015,KKL2016} and hence also $W'=\emptyset$, in which case $\varphi_{W'}=0$. In that case $[\varphi_{W'},\varphi]=\tilde{X}_\varphi$, see e.g. \cite[Lemma~5.4]{KR2015}.
\end{remark}

\begin{theorem}
\label{theo:hereditary-ext}
If  ${\mathcal B}$ is primitive, then $\varphi(H)\subseteq[\varphi_{W'},\varphi] \subseteq X_\varphi$, in particular
$X_\varphi=\overline{[\varphi_{W'},\varphi]}$.
\end{theorem}
\noindent This theorem is proved in Subsection~\ref{subsec:main-proofs}.

\begin{corollary}\label{coro:prox}
If ${\mathcal B}$ is primitive and proximal, then $X_\varphi=\tilde{X}_\varphi$.
\end{corollary}
\begin{proof}
If ${\mathcal B}$ is primitive and proximal, then $[\varphi_{W'},\varphi]=\tilde{X}_\varphi$ is closed, see Remark~\ref{rem:closed-strip}. Hence $X_\varphi=\overline{[\varphi_{W'},\varphi]}=\tilde{X}_\varphi$ by Theorem~\ref{theo:hereditary-ext}. 
\end{proof}

\begin{corollary}\label{coro:hereditary-ext}
If  ${\mathcal B}$ is taut (and hence primitive), then $X_\eta=X_\varphi=\overline{[\varphi_{W'},\varphi]}$.
\end{corollary}
\begin{proof}
As $X_\eta=X_\varphi$ whenever ${\mathcal B}$ is taut (see \cite[Thm.~4]{Keller2019},
which is based on \cite[Prop.~2.2]{KKL2016} and \cite[Prop.~5.11]{BKKL2015}), this follows from Theorem~\ref{theo:hereditary-ext}.
\end{proof}

\begin{remark}
If ${\mathcal B}$ is taut and proximal, then $X_\eta=X_\varphi$ 
by Corollary~\ref{coro:hereditary-ext},
and Corollary~\ref{coro:prox} implies $X_\eta=\tilde{X}_\eta$, which was conjectured in \cite{BKKL2015}, where the authors prove that $X_\eta=\tilde{X}_\eta$ when ${\mathcal B}$ is proximal and has light tails \cite[Thm.~D]{BKKL2015}.

Hence, if ${\mathcal B}$ is taut and proximal, then
some of the results from \cite{Kwietniak2013} apply: $(X_\eta,S)=(X_\varphi,S)$ is distributionally chaotic of type 2
\cite[Thm.~23]{Kwietniak2013}, but not of type 1 \cite[Thm.~24]{Kwietniak2013}. Moreover, for each $t\geqslant0$, the set of all ergodic invariant measures on $X_\eta$ with entropy not exceeding $t$ is arcwise connected w.r.t.~the $\bar d$-metric on the set of all invariant measures \cite[Thm.~6]{KKK2018}.
\end{remark}

\subsection{Notation and further results}
In order to formulate some more detailed corollaries to Theorem~\ref{theo:hereditary-ext} and to prove this theorem, we need to
recall some further notions from the theory of sets of multiples \cite{hall-book}
and also from \cite{KKL2016}.
Let ${\mathcal B}$ be a non-empty subset of ${\mathbbm N}$.
\begin{itemize}
\item The measure $\nu_\eta:=m_H\circ\varphi^{-1}$ on $X_\varphi$ is called the \emph{Mirsky measure} of the system. The point $\eta\in X_\eta$ is quasi-generic for $\nu_\eta$, in particular $\operatorname{supp}(\nu_\eta)\subseteq X_\eta$ \cite[Prop.~E and Thm.~F]{BKKL2015}.

\item If ${\mathcal B}$ is taut, then $X_\eta=\operatorname{supp}(\nu_\eta)$ \cite[Thm.~4]{Keller2019}. (The reverse implication is proved in \cite[Cor.~1.8]{KPM2019}.)
\end{itemize}
Combined with Theorem~\ref{theo:hereditary-ext} and Corollary~\ref{coro:hereditary-ext}, this proves the following corollary.

\begin{corollary}\label{coro:all-X-ext}
\begin{compactenum}[a)]
\item
If ${\mathcal B}$ is primitive, then 
\begin{equation}\label{eq:all-inclusions}
\operatorname{supp}(\nu_\eta)
\subseteq 
X_\eta
\subseteq 
X_\varphi
=
\overline[\varphi_{W'},\varphi] .
\end{equation}
\item If ${\mathcal B}$ is taut, then 
the inclusions in \eqref{eq:all-inclusions} are identities.
\end{compactenum}
\end{corollary}

\begin{question}
It is proved in \cite[Cor.~1.8]{KPM2019} that the first inclusion in \eqref{eq:all-inclusions} is strict, when ${\mathcal B}$ is not taut.
To which extent can the second inclusion in \eqref{eq:all-inclusions} be strict when ${\mathcal B}$ is not taut? What can be said about invariant probability measures assigning full mass to any of these set-differences? Are there any such measures?
\end{question}

\begin{remark}
A particular class of non-taut examples to which Corollary~\ref{coro:prox} applies are Behrend sets ${\mathcal B}$, i.e.~sets for which $\underline d({\mathcal M}_{\mathcal B})=1$. For such sets $\overline d({\mathcal F}_{\mathcal B})=m_H(W)=0$, in particular $\operatorname{int}(W)=\emptyset$, i.e.~${\mathcal B}$ is proximal, and hence $X_\varphi$ is hereditary by the corollary. 
Moreover, $\operatorname{supp}(\nu_\eta)=\{(\dots,0,0,0,\dots)\}$, which illustrates how far the first inclusion in \eqref{eq:all-inclusions} can be from equality, and I expect that also the second inclusion is strict.
Here is a particular example:
Denote by ${\mathcal P}$ the set of prime numbers. Then ${\mathcal B}:=\{pq: p,q\in{\mathcal P}\}$ is a Behrend set, and indeed ${\mathcal F}_{\mathcal B}={\mathcal P}\cup(-{\mathcal P})\cup\{-1,1\}$ \cite[p.~173]{Meyer1973}.
\end{remark}

We continue to recall some notation and results from \cite{KKL2016}.
By $S,S'\subset{\mathcal B}$ we always denote \emph{finite} subsets. For $S\subset{\mathcal B}$ let
\begin{equation}\label{eq:A_S-def}
{\mathcal A}_S:=\{\gcd(b,\operatorname{lcm}(S)): b\in{\mathcal B}\},
\end{equation}
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 ${\mathcal E}:=\bigcup_{S\subset{\mathcal B}}{\mathcal F}_{{\mathcal A}_S}\subseteq{\mathcal F}_{\mathcal B}$. Then 
\begin{equation}\label{eq:E-complement}
{\mathbbm Z}\setminus{\mathcal E}
=
\bigcap_{S\subset{\mathcal B}}{\mathcal M}_{{\mathcal A}_S},
\end{equation}
and
\begin{equation}\label{eq:E-intW}
{\mathcal E}=\{n\in{\mathbbm Z}: \Delta(n)\in\operatorname{int}(W)\}
\end{equation}
by \cite[Lemma~3.1c]{KKL2016}. Observe that $\varphi_{{\operatorname{int}(W)}}(\Delta(k))= S^k(1_{\mathcal E})$ for all $k\in{\mathbbm Z}$.
In view of \cite[Lemma 3.5]{KKL2016}, which states $\Delta({\mathbbm Z})\cap\left(W'\setminus\operatorname{int}(W)\right)=\emptyset$, we have $\varphi_{W'}(\Delta(k))=\varphi_{{\operatorname{int}(W)}}(\Delta(k))=S^k(1_{\mathcal E})$ for all $k\in{\mathbbm Z}$,
in particular $\varphi_{W'}$ and $\varphi_{\operatorname{int}(W)}$ are continuous at all points from $\Delta({\mathbbm Z})\subset H$.

\begin{corollary}\label{coro:cE}
$1_{\mathcal E}=\varphi_{W'}(\Delta(0))\in X_\varphi$.
\end{corollary}

\begin{proof}
$1_{\mathcal E}=\varphi_{W'}(\Delta(0))\in[\varphi_{W'},\varphi] $ by definition, and $[\varphi_{W'},\varphi]\subseteq X_\varphi$ by Theorem~\ref{theo:hereditary-ext}.
\end{proof}

Denote the orbit closure of $1_{\mathcal E}\in\{0,1\}^{\mathbbm Z}$ by $X_{\mathcal E}$.
Then $X_{\mathcal E}=\overline{\varphi_{{\operatorname{int}(W)}}(\Delta({\mathbbm Z}))}=\overline{\varphi_{W'}(\Delta({\mathbbm Z}))}$ is 
minimal in view of \cite[Lemma~3.9]{KKL2016}. As $1_{\mathcal E}\in X_\varphi$, this proves the following corollary.

\begin{corollary}
$X_{\mathcal E}$ is the unique minimal subset of $X_\varphi$ (and hence also of $X_\eta$).
\end{corollary}

\begin{remark}
As $\varphi_{\operatorname{int}(W)}\leqslant\varphi_{W'}$ and as $\varphi_{\operatorname{int}(W)}$ is lower semicontinuous,
we have
\begin{equation*}
[\varphi_{W'},\varphi]
\subseteq
X_\varphi
=\overline{[\varphi_{W'},\varphi]}
\subseteq
\overline{[\varphi_{\operatorname{int}(W)},\varphi]}
=
[\varphi_{\operatorname{int}(W)},\varphi].
\end{equation*}
Assume now that $m_H(\partial W')=0$. (We will see in Proposition~\ref{prop:toeplitz} that this happens if and only if $1_{\mathcal E}$ is a regular Toeplitz sequence.)
Then $\varphi_{\operatorname{int}(W)}=\varphi_{W'}$ $m_H$-a.s., so that
$[\varphi_{W'},\varphi]\cap(\pi^{\scriptscriptstyle H})^{-1}\{h\}=X_\varphi\cap(\pi^{\scriptscriptstyle H})^{-1}\{h\}$ for $m_H$-a.a. $h\in H$.
Hence all invariant measures on $X_\varphi$ are supported by $[\varphi_{W'},\varphi]$, if $m_H(\partial W')=0$.
\end{remark}

\begin{conjecture}
Suppose that $m_H(\partial W')=0$. 
\begin{compactenum}[a)]
\item
For any invariant probability measure $\nu$ on $(X_\varphi,S)$ there exists an invariant probability measure $\rho$ on $(H\times \{0,1\}^{\mathbbm Z},R_{\Delta(1)}\timesS)$, whose projection onto the first coordinate equals $m_H$ and such that, for any measurable $A\subseteq X_\varphi$,
\begin{equation}\label{eq:convolution-1}
\nu(A)=\int_{H \times\{0,1\}^{\mathbbm Z}} 1_A(\varphi_{W'}(h)+ x\cdot(\varphi(h)-\varphi_{W'}(h)))\,d\rho(h,x).
\end{equation}
\item
$(X_\varphi,S)$ has a unique measure

of maximal entropy. It can be represented as in \eqref{eq:convolution-1} with $\rho=m_H\otimes B_{(1/2,1/2)}$, where $B_{(1/2,1/2)}$ is the Bernoulli measure.

\end{compactenum}
For the hereditary case this is proved in \cite[Theorems~ I and J]{BKKL2015}.
\end{conjecture}

\subsection{The Toeplitz subsystem}
We finish this introduction with a result that sheds some more light on the
unique minimal set $X_{\mathcal E}=\overline{\varphi_{W'}(\Delta({\mathbbm Z}))}$. Observe that $\overline{\operatorname{int}(W')}=W'$. If $W'$ were an ``arithmetic window'', just as the window $W$ defined in \eqref{eq:W}, then Theorem B of \cite{KKL2016} would provide much additional information on $X_{\mathcal E}$, e.g. that $\varphi_{W'}(\Delta(0))$ is a Toeplitz sequence. However, $W'$ is not ``arithmetic'' in the strict sense. But one can modify the set ${\mathcal B}$ into a set ${\mathcal B}^*$ in such a way that $1_{{\mathcal F}_{{\mathcal B}^*}}=\varphi_{W'}(\Delta(0))$.
To that end we recall the definition of the set $\cA_\infty$ from \cite{KKL2016},
\begin{equation}\label{eq:Ainf-def}
\cA_\infty:=\{n\in{\mathbbm N}: \forall_{S\subset{\mathcal B}}\ \exists_{S': S\subseteq S'\subset{\mathcal B}}: n\in{\mathcal A}_{S'}\setminus S'\},
\end{equation}
and one of its important properties \cite[Lemma~3.3]{KKL2016},
\begin{equation}\label{eq:E-identities}
{\mathcal E}={\mathcal F}_{{\mathcal B}\cup{\cA_\infty}}={\mathcal F}_{\mathcal B}\cap{\mathcal F}_{\cA_\infty}.
\end{equation}
Let
\begin{equation}\label{eq:B_0-def}
{\mathcal B}_0:={\mathcal B}\setminus{\mathcal M}_{\cA_\infty},\quad
{\mathcal B}^*:={\mathcal B}_0\cup\prim\cA_\infty,
\end{equation}
where $\prim\cA_\infty$ denotes the set of primitive elements of $\cA_\infty$, i.e. the set of those $a\in\cA_\infty$ which are not multiples of any other element of $\cA_\infty$.
In Subsection~\ref{subsec:lemmas-proof} we prove the following lemmas:

\begin{lemma}\label{lemma:Ainf}
$n\in \cA_\infty$ if and only if there is an infinite pairwise co-prime set ${\mathcal C}\subseteq{\mathbbm N}\setminus\{1\}$ such that
$n\,{\mathcal C}\subseteq{\mathcal B}$.
\end{lemma}

Let $\Delta^*:{\mathbbm Z}\to\prod_{b^*\in{\mathcal B}^*}{\mathbbm Z}/b^*{\mathbbm Z}$ denote again the canonical diagonal embedding and let $H^*:=\overline{\Delta^*({\mathbbm Z})}$.
Define a group homomorphism $\Gamma:\Delta({\mathbbm Z})\to \Delta^*({\mathbbm Z})$ by $\Gamma(\Delta(n))=\Delta^*(n)$. 
\begin{lemma}\label{lemma:Gamma-continuous}
$\Gamma$ is uniformly continuous on $\Delta({\mathbbm Z})$ and extends uniquely to a
continuous surjective group homomorphism $\Gamma:H\to H^*$.
\end{lemma}

Let $W^*\subseteq H^*$ denote the corresponding window and $\varphi^*:H^*\to\{0,1\}^{\mathbbm Z}$ the corresponding coding map. 

\begin{lemma}\label{lemma:B^*}(See also \cite[Thm.~B]{KKL2016}.)
\begin{compactenum}[a)]
\item The set ${\mathcal B}^*$ is primitive.
\item ${\mathcal M}_{{\mathcal B}^*}={\mathcal M}_{{\mathcal B}\cup\cA_\infty}=\bigcap_{S\subset{\mathcal B}}{\mathcal M}_{{\mathcal A}_S}$. Equivalently, ${\mathcal F}_{{\mathcal B}^*}={\mathcal F}_{{\mathcal B}\cup\cA_\infty}={\mathcal E}$.
\item There are no $n\in{\mathbbm N}$ and no infinite pairwise
co-prime set ${\mathcal C}\subseteq{\mathbbm N}\setminus\{1\}$ such that $n\,{\mathcal C}\subseteq{\mathcal B}^*$.
\item The window $W^*\subseteq H^*$ constructed from ${\mathcal B}^*$ is topologically regular, i.e. $W^*=\overline{\operatorname{int}(W^*)}$.
\item $1_{{\mathcal F}_{{\mathcal B}^*}}=\varphi_{W'}(\Delta(0))=1_{\mathcal E}$ is a Toeplitz sequence.
\end{compactenum}
\end{lemma}

\begin{lemma}\label{lemma:Gamma}
\begin{compactenum}[a)]
\item $\Gamma(W')=W^*$.
\item 
$\Gamma(H\setminus W')=H^*\setminus W^*$
\item If $(\varphi_{W'}(h))_i=0$ for some $h\in H$ and $i\in{\mathbbm Z}$, then 
\begin{equation*}
\begin{split}
&\exists b\in{\mathcal B}\ \exists b^*\in{\mathcal B}^*: b\in b^*{\mathbbm Z},\;(h_b+i)\in b^*{\mathbbm Z},
\text{ and [$b^*\in\prim\cA_\infty$ or $b^*= b$].}
\end{split}
\end{equation*}
\end{compactenum}
\end{lemma}

In view of \cite[Proposition~1.2]{KKL2016}, the Toeplitz sequence $1_{{\mathcal F}_{{\mathcal B}^*}}$ is regular if and only if $m_{H^*}(\partial W^*)=0$, where $m_{H^*}$ denotes the Haar measure on $H^*$. The following proposition characterizes the regularity of  
$1_{{\mathcal F}_{{\mathcal B}^*}}$ in terms of $m_H(\partial W')$.
It is proved in Subsection~\ref{subsec:toeplitz-proofs}.

\begin{proposition}\label{prop:toeplitz}
The Toeplitz sequence $1_{{\mathcal F}_{{\mathcal B}^*}}=\varphi_{W'}(\Delta(0))=1_{\mathcal E}$ is regular if and only if $m_H(\partial W')=0$.
\end{proposition}

\section{Proofs}
\subsection{Proofs of Lemmas~\ref{lemma:Ainf} -~\ref{lemma:Gamma}}\label{subsec:lemmas-proof}
All sets and quantities derived from ${\mathcal B}^*$ instead of ${\mathcal B}$ also carry a $^*$ superscript.

\begin{proof}[Proof of Lemma~\ref{lemma:Ainf}]
Suppose first that $n\in\cA_\infty$. A repeated application of the definition~\eqref{eq:Ainf-def} of this set allows to produce a strictly increasing sequence $S_1\subset S_2\subset\dots$ of finite subsets of ${\mathcal B}$ and a strictly increasing sequence $b_1<b_2<\dots$ of elements of ${\mathcal B}$ such that
\begin{equation*}
n=\gcd\left(\operatorname{lcm}(S_k),b_k\right) \text{ and $b_k\in S_{k+1}$ for all $k\in{\mathbbm N}$.}
\end{equation*}
Let ${\mathcal C}:=\{b_1/n,b_2/n,\dots\}$. Then $n\,{\mathcal C}=\{b_1,b_2,\dots\}\subseteq{\mathcal B}$, and ${\mathcal C}$ is an infinite pairwise co-prime subset of ${\mathbbm N}$, because, for any $j<k$,
\begin{equation*}
\gcd(b_j/n,b_k/n)=n^{-1}\gcd(b_j,b_k)\mid n^{-1}\gcd(\operatorname{lcm}(S_k),b_k)=1\ .
\end{equation*}
Finally, $b_j/n\neq 1$ for all $j$, because otherwise $b_{j+1}=b_j/n\cdot b_{j+1}=b_{j+1}/n\cdot b_j$ is a multiple of $b_j$, which is impossible, because ${\mathcal B}$ is primitive.

Conversely, suppose that $n\,{\mathcal C}\subseteq{\mathcal B}$, where ${\mathcal C}$
is an infinite pairwise co-prime subset of ${\mathbbm N}\setminus\{1\}$. 
Then $n\not\in{\mathcal B}$, because ${\mathcal B}$ is primitive.
Let $S\subset{\mathcal B}$ and choose any $c\in{\mathcal C}$.
Let $S':=S\cup\{nc\}$. As $S'\subset{\mathcal B}$
is finite,
one can choose some $c'\in{\mathcal C}$ which is co-prime to $\operatorname{lcm}(S')$.
Then $nc'\in{\mathcal B}$, and as $n\mid\operatorname{lcm}(S')$, we conclude that
$n=\gcd\left(\operatorname{lcm}(S'),nc'\right)\in{\mathcal A}_{S'}$. As $n\not\in{\mathcal B}$, this shows that $n\in{\mathcal A}_{S'}\setminus S'$. As $S\subset{\mathcal B}$ could be chosen arbitrarily and as $S'\supseteq S$, this proves that $n\in\cA_\infty$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:Gamma-continuous}]
As $\Gamma$ is a group homomorphism, it suffices to prove continuity at $\Delta(0)\in\Delta({\mathbbm Z})$ in order to guarantee uniform continuity:
 To see this, fix $b^*\in {\mathcal B}^*$. If $b^*\in{\mathcal B}_0$, then $(\Gamma(\Delta(n)))_{b^*}=(\Delta^*(n))_{b^*}=(\Delta(n))_{b^*}$ for all $n\in{\mathbbm Z}$.
If $b^*\in \prim\cA_\infty$,
then there is $S\subset{\mathcal B}$ such that $b^*\mid\operatorname{lcm}(S)$.
Let $n\in{\mathbbm Z}$. In view of the CRT, the residue $n\text{ mod }\operatorname{lcm}(S)$ is uniquely determined by the $S$-tuple $(\Delta(n)_b)_{b\in S}$. Hence, as $b^*\mid\operatorname{lcm}(S)$, also 
$(\Gamma(\Delta(n)))_{b^*}=(\Delta^*(n))_{b^*}=n\text{ mod } b^*$ is uniquely determined by this $S$-tuple.

Uniform continuity allows to extend $\Gamma$ uniquely to a continuous group homomorphism on $H=\overline{\Delta({\mathbbm Z})}$. In particular, $\Gamma(H)\supseteq \overline{\Gamma(\Delta({\mathbbm Z}))}=\overline{\Delta^*({\mathbbm Z})}=H*$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:B^*}]
a)\;Let $b,b'\in{\mathcal B}^*$, $b'=rb$ for some $r\in{\mathbbm N}$. We show that $r=1$:
\begin{compactitem}[-]
\item If $b',b\in\prim\cA_\infty$, then $r=1$, because $\prim\cA_\infty$ is primitive.
\item If $b'\in\prim\cA_\infty$ but $b\in{\mathcal B}_0\subseteq{\mathcal B}$, then there is some $\tilde b\in{\mathcal B}$ such that $rb=b'\mid\tilde b$.
As ${\mathcal B}$ is primitive, this implies $r=1$.
\item If $b'\not\in\prim\cA_\infty$, then $b'\in{\mathcal B}_0={\mathcal B}\setminus{\mathcal M}_{\cA_\infty}$ so that $b\not\in\prim\cA_\infty$. But then $b\in{\mathcal B}_0$, and again $r=1$ because ${\mathcal B}$ is primitive.
\end{compactitem}
b)\;Recall that ${\mathcal B}^*=({\mathcal B}\setminus{\mathcal M}_{\cA_\infty})\cup\prim\cA_\infty$.
As ${\mathcal M}_{\prim\cA_\infty}={\mathcal M}_{\cA_\infty}={\mathcal M}_{{\mathcal M}_{\cA_\infty}}$, we conclude that
\begin{equation*}
\begin{split}
{\mathcal M}_{{\mathcal B}^*}
&=
{\mathcal M}_{{\mathcal B}\setminus{\mathcal M}_{\cA_\infty}}\cup{\mathcal M}_{\prim\cA_\infty}
=
{\mathcal M}_{{\mathcal B}\setminus{\mathcal M}_{\cA_\infty}}\cup{\mathcal M}_{{\mathcal M}_{\cA_\infty}}
=
{\mathcal M}_{{\mathcal B}\cup{\mathcal M}_{\cA_\infty}}\\
&=
{\mathcal M}_{{\mathcal B}}\cup{\mathcal M}_{{\mathcal M}_{\cA_\infty}}
=
{\mathcal M}_{{\mathcal B}}\cup{\mathcal M}_{{\cA_\infty}}
=
{\mathcal M}_{{{\mathcal B}\cup\cA_\infty}}.
\end{split}
\end{equation*}
Taking complements this is equivalent to ${\mathcal F}_{{\mathcal B}^*}={\mathcal F}_{{\mathcal B}\cup\cA_\infty}={\mathcal E}$, where we used \eqref{eq:E-identities} for the last identity. Finally note that ${\mathbbm Z}\setminus{\mathcal E}={\mathbbm Z}\setminus\bigcup_{S\subset{\mathcal B}}{\mathcal F}_{{\mathcal A}_S}=\bigcap_{S\subset{\mathcal B}}{\mathcal M}_{{\mathcal A}_S}$.\\
c)\;Suppose for a contradiction that there
are $n\in{\mathbbm N}$ and an infinite pairwise
co-prime set ${\mathcal C}\subseteq{\mathbbm N}\setminus\{1\}$ such that $n\,{\mathcal C}\subseteq{\mathcal B}^*$. If $n\in\cA_\infty$, then 
$n\,{\mathcal C}\subseteq{\mathcal B}^*\cap{\mathcal M}_{\cA_\infty}=\prim\cA_\infty$, so that ${\mathcal C}=\{1\}$, a contradiction. Hence $n\not\in\cA_\infty$. 
Then Lemma~\ref{lemma:Ainf} implies that 
$(n\,{\mathcal C})\cap{\mathcal B}_0\subseteq (n\,{\mathcal C})\cap{\mathcal B}$ is finite, so that 
$(n\,{\mathcal C})\cap\prim\cA_\infty$ is infinite. This leads to the contradiction $n\in\cA_\infty$, as can be seen as follows: Let $\{nc_1<nc_2<\dots\}:=(n\,{\mathcal C})\cap\prim\cA_\infty$. For each $nc_k$ there is an infinite pairwise co-prime set ${\mathcal C}_k\subseteq{\mathbbm N}\setminus\{1\}$ such that $nc_k{\mathcal C}_k\subseteq{\mathcal B}$ (Lemma~\ref{lemma:Ainf}).
Inductively we choose elements $c_j'\in{\mathcal C}_{k_j}$ for $j=1,2,\dots$ and suitable $k_1<k_2<\dots$ in the following way: $k_1=1$ and
$c_1'$ can be any element of ${\mathcal C}_1$. Suppose that $k_1<\dots<k_j$ and $c_1',\dots,c_j'$ are chosen. 
Then choose $k_{j+1}>k_j$ and $c_{j+1}'\in {\mathcal C}_{k_{j+1}}$ in such a way that $c_{k_{j+1}}$ and
$c_{j+1}'$ are co-prime to the finite set
$\{c_{k_1}c_{1}',\dots,c_{k_j}c_{j}'\}$. In this way we obtain an infinite pairwise co-prime
set ${\mathcal C}':=\{c_{k_1}c_{1}',c_{k_2}c_{2}',\dots\}\subseteq{\mathbbm N}\setminus\{1\}$. It satisfies $n\,{\mathcal C}'\subseteq{\mathcal B}$, because 
$n\cdot c_{k_j}c_j'\in nc_{k_j}{\mathcal C}_{k_j}\subseteq{\mathcal B}$. Hence $n\in\cA_\infty$ by Lemma~\ref{lemma:Ainf}.\\
d) \;As ${\mathcal B}^*$ is primitive (see assertion a), this assertion as well as the Toeplitz property in e) are equivalent to assertion c), see \cite[Thm.~B]{KKL2016}.\\
e) \; The two identities follow from assertion b) and Corollary~\ref{coro:cE}.

\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:Gamma}]
a) 
Let $k\in{\mathbbm Z}$. Then 
\begin{equation*}
\begin{split}
\Delta^*(k)\in\operatorname{int}(W^*)
&\Leftrightarrow
k\in{\mathcal E}^*\hspace*{1.2cm}\text{by \cite[Lemma~3.1]{KKL2016},}\\
&\Leftrightarrow
k\in{\mathcal F}_{{\mathcal B}^*}\hspace*{1cm}\text{by Lemma~\ref{lemma:B^*}d and \cite[Prop.~3.2]{KKL2016},}\\
&\Leftrightarrow
k\in{\mathcal E}\hspace*{1.35cm}\text{by Lemma~\ref{lemma:B^*}b.}
\end{split}
\end{equation*}
Observing the topological regularity of $W^*$ (see Lemma~\ref{lemma:B^*}d), we conclude
\begin{equation*}
W^*
=
\overline{\operatorname{int}(W^*)\cap\Delta^*({\mathbbm Z})}
=
\overline{\Delta^*({\mathcal E})}
=
\overline{\Gamma(\Delta({\mathcal E}))}
=
\Gamma(\overline{\Delta({\mathcal E})}),
\end{equation*}
where we used the continuity of $\Gamma$ and the compactness of $\overline{\Delta({\mathcal E})}$ for the last identity. Hence
\begin{equation*}
W^*
=
\Gamma(\overline{\Delta({\mathcal E})})
=
\Gamma(\overline{\operatorname{int}(W)\cap\Delta({\mathbbm Z})})
=
\Gamma(\overline{\operatorname{int}(W)})
=
\Gamma(W').
\end{equation*}
b)
Let $h\in H\setminus W'$ and suppose for a contradiction that $\Gamma(h)\in W^*$. We will use the following facts:
\begin{compactenum}[(1)]
\item  Because of part a) of this lemma, there is $h'\in W'$ such that $\Gamma(h')=\Gamma(h)$. As $W'$ is topologically regular by construction and as $\Delta({\mathbbm Z})$ is dense in $H$, there are $n'_1<n'_2<\dots$ such that $h'=\lim_{j\to\infty}\Delta(n'_j)$ and hence $\Gamma(h')=\lim_{j\to\infty}\Gamma(\Delta(n'_j))=\lim_{j\to\infty}\Delta^*(n'_j)$, where all $\Delta(n'_j)$ belong to $\operatorname{int}(W')=\operatorname{int}(W)$. Then \cite[Lem.~3.1c]{KKL2016} implies $n'_j\in{\mathcal E}$, so that all $n'_j\in{\mathcal F}_{{\mathcal B}^*}$ by Lemma~\ref{lemma:B^*}b.
\item Fix some sequence $n_1<n_2<\dots$ such that $h=\lim_{j\to\infty}\Delta(n_j)$.
As $W'$ is closed, there exists some neighbourhood of $h$ which is disjoint from $W'$. Hence there are a finite subset $S$ of ${\mathcal B}$ and some integer $j_1$ such that for all $j\geqslant j_1$
\begin{compactitem}[-]
\item $(\Delta(n_j))_b=h_b$ for all $b\in S$, and
\item if for some $k\in{\mathbbm Z}$ holds $(\Delta(n_j))_b=(\Delta(k))_b$ for all $b\in S$, then $\Delta(k)\not\in W'$ and hence $k\in{\mathcal M}_{{\mathcal B}^*}$.
\end{compactitem}
(For the latter conclusion recall that
$k\in{\mathcal M}_{{\mathcal B}^*}={\mathbbm Z}\setminus{\mathcal E}$ if and only if $\Delta(k)\not\in\operatorname{int}(W)=\operatorname{int}(W')$ by \cite[Lem.~3.1c]{KKL2016}.)
\item As $\lim_{j\to\infty}\Delta^*(n_j)=\lim_{j\to\infty}\Gamma(\Delta(n_j))=\Gamma(h)=\Gamma(h')=\lim_{j\to\infty}\Delta^*(n'_j)$, there is $j_2\geqslant j_1$ such that $(\Delta^*(n_j))_{b^*}=(\Delta^*(n'_j))_{b^*}$ for all $b$ from the finite set ${\mathcal B}^*_S:=\{b^*\in{\mathcal B}^*:b^*\mid\operatorname{lcm}(S)\}$ and all $j\geqslant j_2$.
\end{compactenum}
We first exploit fact (2): Let $j\geqslant j_1$. For each $k=n_j+t\cdot\operatorname{lcm}(S)$
there are $r_{j,t}\in{\mathbbm Z}$ and $b_{j,t}^*\in{\mathcal B}^*$ such that $k=r_{j,t}\cdot b^*_{j,t}$. Let $s_j:=\gcd(n_j,\operatorname{lcm}(S))$. By Dirichlet's theorem on primes in arithmetic progressions, given $j\geqslant j_1$, there are infinitely many primes $p_1<p_2<\dots$ and integers $t_i$ such that $s_j\,p_i=n_j+t_i\,\operatorname{lcm}(S)=r_{j,t_i}\,b^*_{j,t_i}$. For each $p_i$ at least one of the following two possibilities occurs:
\begin{compactitem}[-]
\item $p_i\mid r_{j,t_i}$. Then $b^*_{j,t_i}$ divides $s_j$ and hence $\operatorname{lcm}(S)$.
\item $p_i\mid b^*_{j,t_i}$. Then $r_{j,t_i}$ divides $s_j$ so that $q_i:=\frac{s_j}{r_{j,t_i}}\mid\operatorname{lcm}(S)$ and $q_i=\frac{b^*_{j,t_i}}{p_i}$.
\end{compactitem}
If, for any given $j\geqslant j_1$, the  second possibility occurs for infinitely many $p_i$, then there is at least one $q$ such that $q=q_i$ for infinitely many $p_i$. Hence $q\,p_i=b^*_{j,t_i}\in{\mathcal B}^*$ for all these $p_i$, which implies that $W^*$ is not topologically regular \cite[Thm.~B]{KKL2016} and thus contradicts
our Lemma~\ref{lemma:B^*}d. Otherwise, for all $j\geqslant j_1$, the first possibility occurs  for all but at most finitely many $p_i$. In particular, for each $j\geqslant j_1$, there exists $i_j$ such that $n_j=r_{j,t_{i_j}}\,b^*_{j,t_{i_j}}-t_{i_j}\,\operatorname{lcm}(S)$, and $b^*_{j,t_{i_j}}\mid\operatorname{lcm}(S)$. Abbreviate $b^*_j:=b^*_{j,t_{i_j}}$. It follows that $n_j\in b^*_j{\mathbbm Z}$ and $b_j^*\in{\mathcal B}^*_S$, i.e.~$(\Delta^*(n_j))_{b_j^*}=0$ and $b_j^*\in{\mathcal B}^*_S$. Now fact (3) implies that also $(\Delta^*(n'_j))_{b_j^*}=0$, i.e.~$n'_j\in b_j^*{\mathbbm Z}\subseteq{\mathcal M}_{{\mathcal B}^*}$, which contradicts fact (1).
\\
c) Let $h\in H$. Then $h\in H\setminus W'$ if and only if $\Gamma(h)\in H^*\setminus W^*$ because of assertions a) and b) of this lemma.
Hence, for any $i\in{\mathbbm Z}$,

\begin{equation*}
\begin{split}
(\varphi_{W'}(h))_i=0
\Leftrightarrow&\;
h+\Delta(i)\in H\setminus W'\\
\Leftrightarrow&\;
\Gamma(h+\Delta(i))\in  H^*\setminus W^*\\
\Leftrightarrow&\;
\exists b^*\in{\mathcal B}^*: (\Gamma(h+\Delta(i)))_{b^*}=0\\
\Leftrightarrow&\;
\exists b^*\in{\mathcal B}^*: (\Gamma(h))_{b^*}+i\in b^*{\mathbbm Z}.
\end{split}
\end{equation*}
If $b^*\in\prim\cA_\infty$, there is some $b\in{\mathcal B}$ such that $b^*\mid b$. If $b^*\in{\mathcal B}_0$, choose $b=b^*$. In any case we have $b\in b^*{\mathbbm Z}$.
Pick a sequence $(m_j)_j$ in ${\mathbbm Z}$ such that $h=\lim_{j\to\infty}\Delta(m_j)$.
Then, for all sufficiently large $j\in{\mathbbm N}$,
\begin{equation*}
(\Gamma(h))_{b^*}
=
(\Gamma(\Delta(m_j)))_{b^*}
=
(\Delta^*(m_j))_{b^*}
=
\left(m_j\text{ mod }{b^*}\right)
=
\left(m_{j}\text{ mod }{b}\right)\text{ mod } b^*
=
h_b\text{ mod } b^*,
\end{equation*}
so that
\begin{equation*}
(\varphi_{W'}(h))_i=0
\Leftrightarrow\;
(h_b+i)\in b^*{\mathbbm Z}.
\end{equation*}
\end{proof}

\subsection{Proofs of Theorem~\ref{theo:hereditary-ext} and Corollary~\ref{coro:all-X-ext}}
\label{subsec:main-proofs}
The proof of Theorem~\ref{theo:hereditary-ext} is inspired by the proof of Theorem~D in \cite{BKKL2015} and combines ideas from that proof with observations from \cite{KKL2016}. The two main differences compared to the proofs in \cite{BKKL2015} are:
\begin{itemize}[-]
\item The proximality assumption enters the proof from \cite{BKKL2015} immediately after eqn.~(58) of that paper: for each finite $S\subset{\mathcal B}$ there exists $b\in{\mathcal B}$ such that $\gcd(\operatorname{lcm}(S),b)=1$, i.e. $1\in{\mathcal A}_S$ and hence ${\mathbbm Z}={\mathcal M}_{{\mathcal A}_S}$. Without that assumption we can only use our general knowledge of the set ${\mathcal A}_S$,
namely that ${\mathbbm Z}\setminus{\mathcal E}\subseteq{\mathcal M}_{{\mathcal A}_S}$.
\item The light tails assumption is used in the proof of \cite[Lemma~5.21]{BKKL2015}, from which Proposition~5.11 of that reference is deduced. 
This proposition 
basically asserts that for certain cylinder sets in $H$ there exist $k'\in{\mathbbm Z}$ such that $S^{k'}\eta=\varphi(\Delta(k'))$ belongs to that cylinder set. This is proved by showing that the set of integers $k'$ with this property has positive density.

Here we prove the existence by a topological argument using compactness of $H$ and the Heine-Borel theorem. The price to pay is that the existence can no longer 
be guaranteed in the set $\Delta({\mathbbm Z})$ but only in its closure $H$.
\end{itemize}
The following notation will be used repeatedly:
\begin{itemize}
\item By $S,S'\subset{\mathcal B}$ we always mean \emph{finite} subsets.
\item The topology on $H$ is generated by the (open and closed) cylinder sets
\begin{equation*}
U_S(h):=\{h'\in H: \forall b\in S: h_b=h'_b\},\text{ defined for finite }S\subset{\mathcal B}\text{ and }h\in H\ .
\end{equation*}
\item For $n\geqslant1$, let ${\mathcal B}^{(n)} := \{b\in{\mathcal B} : p\leqslant n\text{ for any }p\in\operatorname{Spec}(b)\}$,
where $\operatorname{Spec}(b)$ stands for the set of all prime divisors of $b$. 
\item For $n\in{\mathbbm N}$ and $b\in{\mathcal B}$ let $H_b^N:=\bigcap_{i=-N}^N\{h\in H: h_b+i\neq 0\text{ mod } b\}$. 
All sets $H_b^N$ are open and closed by definition. If  $b\not\in{\mathcal B}^{(2N+1)}$, then $b$ has at least one prime divisor greater than $2N+1$ so that 
${\mathbbm Z}\not \subseteq\bigcup_{i=-N}^N(b{\mathbbm Z}+i)$, in particular
$H_b^N\neq\emptyset$.
For ${\mathcal C}\subseteq{\mathcal B}$ denote $H_{\mathcal C}^N:=\bigcap_{b\in{\mathcal C}}H_b^N$.
\end{itemize}
We start with a modification of some lemmas from \cite{BKKL2015}.

\begin{lemma}\label{lemma:new-1}
Let $N\in{\mathbbm N}$, $n\geqslant 2N+1$, ${\mathcal B}^{(n)}\subseteq {\mathcal A}\subset{\mathcal B}$, ${\mathcal A}$ finite, $p>n$ a prime number not dividing $\operatorname{lcm}({\mathcal A})$,  and $S\subset{\mathcal B}\setminus{\mathcal A}$ be finite. Denote $S_0:=\{b\in S: p\not\mid b\}$. If 
$U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N\neq\emptyset$ for some $k\in{\mathbbm Z}$, then also 
$U_{\mathcal A}(\Delta(k))\cap H_S^N\neq\emptyset$.
\end{lemma}

\begin{proof}
If $S_0=S$, there is nothing to prove. So suppose that $S\setminus S_0\neq \emptyset$.
By definition,
\begin{equation*}
U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N
\subseteq
(U_{\mathcal A}(\Delta(k))\cap H_S^N)\cup\bigcup_{b\in S\setminus S_0}\bigcup_{i=-N}^N\{h\in U_{\mathcal A}(\Delta(k)): h_b+i\in p{\mathbbm Z}\}.
\end{equation*}
Fix any $b'\in S\setminus S_0$. As all $b\in S\setminus S_0$ are multiples of $p$, we have 
$h_b-h_{b'}\in p{\mathbbm Z}$ for such $b$, so that
$h_b+i\in p{\mathbbm Z}$ if and only if $h_{b'}+i\in p{\mathbbm Z}$ for all $h\in H$ and $i=-N,\dots,N$. Hence
\begin{equation*}
U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N
\subseteq
(U_{\mathcal A}(\Delta(k))\cap H_S^N)\cup\bigcup_{i=-N}^N\{h\in U_{\mathcal A}(\Delta(k)): h_{b'}+i\in p{\mathbbm Z}\}.
\end{equation*}

Suppose for a contradiction that $U_{\mathcal A}(\Delta(k))\cap H_S^N=\emptyset$. Then
\begin{equation}\label{eq:assumption-1}
U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N
\subseteq
\bigcup_{i=-N}^N\{h\in U_{\mathcal A}(\Delta(k)): h_{b'}+i\in p{\mathbbm Z}\}.
\end{equation}
As $p>n\geqslant 2N+1$, there is $j\in{\mathbbm Z}$ such that $j+i\not\in p{\mathbbm Z}$ for all $i=-N,\dots,N$.
Let $h\in U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N$. 
As $\gcd(p,b)=1$ for all $b\in{\mathcal A}\cup S_0$, the CRT guarantees the existence of an integer $m$ such that 
\begin{compactitem}[-]
\item $m=h_b=k\text{ mod } b$ for all $b\in {\mathcal A}$,
\item $m=h_b\text{ mod } b$ for all $b\in S_0$, and
\item $m=j\text{ mod } p$.
\end{compactitem}
Hence $\Delta(m)\in U_{\mathcal A}(\Delta(k))\cap H_{S_0}^N$, but since $p|b'$, we have
$\Delta(m)|_{b'}+i\text{ mod } p=j+i\text{ mod } p\neq0$ for all $i=-N,\dots,N$ in contradiction to \eqref{eq:assumption-1}.

\end{proof}

\begin{lemma}\label{lemma:new-2}
Let $N\in{\mathbbm N}$, $n\geqslant 2N+1$, and ${\mathcal B}^{(n)}\subseteq {\mathcal A}\subset{\mathcal B}$, ${\mathcal A}$ finite.
Denote 
\begin{equation}\label{eq:cC-def}
{\mathcal C}={\mathcal C}({\mathcal A},n):=\{b\in{\mathcal B}\setminus{\mathcal A}: \text{$b$ has at least one prime divisor $p>n$  co-prime to $\operatorname{lcm}({\mathcal A})$}\}.
\end{equation}
Then, for each $h\in H$, there is $h'\in U_{\mathcal A}(h)$ such that
$h'_b+i\neq 0\text{ mod } b$ for all $b\in{\mathcal C}$ and $i\in\{-N,\dots,N\}$.
\end{lemma}

\begin{proof}
We must show that $U_{\mathcal A}(h)\cap H_{{\mathcal C}}^N\neq\emptyset$. As all $H_b^N$ are compact, 
it suffices to show that $U_{\mathcal A}(h)\cap H_S^N\neq\emptyset$ 
for each finite set $S\subset{\mathcal C}$. 
We proceed by induction on the cardinality of the set $S$.
\begin{itemize}[-]
\item  If $S=\emptyset$, there is nothing to prove, because $U_{\mathcal A}(h)\neq\emptyset$.
\item  For the inductive step, let 
$\emptyset\neq S\subseteq{\mathcal C}$, $S$ finite, and note that there exists at least one prime number $p>n$ which is co-prime to $\operatorname{lcm}({\mathcal A})$ and divides at least one element $b\in S$. To this $p$ and its associated set $S_0\subseteq S\setminus\{b\}$ we apply Lemma~\ref{lemma:new-1}.
\end{itemize}
\end{proof}

\begin{lemma}(See also \cite[Prop.~5.10]{BKKL2015}.)\label{lemma:ext-1}\\
Let $h\in H$, $N\in{\mathbbm N}$ and $I_0\subseteq\{-N\dots,N\}$ be such that 
$(\varphi_{W'}(h))_{i}=0<1=(\varphi(h))_i$ for all $i\in I_0$. Define $\omega\in\{0,1\}^{\{-N,\dots,N\}}$ by
\begin{equation*}
\omega_i
:=
\begin{cases}
(\varphi(h))_i&\text{ if }i\not\in I_0\\
0&\text{ if }i\in I_0.
\end{cases}
\end{equation*}
Then, for each $S\subset{\mathcal B}$, there is $h'\in U_S(h)$ such that $(\varphi(h'))_i=\omega_i$ for all $i=-N,\dots,N$.
\end{lemma}

\begin{proof}
\textbf{Step 1:}\quad This step applies to any $h\in H$ and $n\geqslant2N+1$.
For any set $\tilde{\mathcal B}$ with ${\mathcal B}^{(n)}\subseteq\tilde{\mathcal B}\subseteq{\mathcal B}$ denote
\begin{equation*}
\begin{split}
I_N^{\tilde{\mathcal B}}(h)
:=&
\{i\in\{-N,\dots,N\}: \exists b\in{\tilde{\mathcal B}}\text{ s.t. }h_b+i\in b{\mathbbm Z}\}.
\end{split}
\end{equation*}
Observe that for ${\tilde{\mathcal B}}={\mathcal B}$, 
\begin{equation*}
\begin{split}
I_N^{\mathcal B}(h)
=
\{i\in\{-N,\dots,N\}:h+\Delta(i)\not\in W\}
=
\{i\in\{-N,\dots,N\}: (\varphi(h))_i=0\}.
\end{split}
\end{equation*}
In particular, $I_0\cap I_N^{{\mathcal B}}(h)=\emptyset$.

For each $i\in I_N^{\tilde{\mathcal B}}(h)$ we fix some $b_i\in{\tilde{\mathcal B}}$ with  $(h_{b_i}+i)\in b_i{\mathbbm Z}$.
Let
\begin{equation*}
{\tilde{\mathcal B}}_1:={\mathcal B}^{(n)}\cup\{b_i: i\in I_N^{\tilde{\mathcal B}}(h)\}\subseteq{\tilde{\mathcal B}}.
\end{equation*}
By \cite[Lemma~5.14]{BKKL2015}, the set ${\tilde{\mathcal B}}_1$ is finite. 

Let $\tilde\beta_1:=\operatorname{lcm}({\tilde{\mathcal B}}_1)$. Then, for any $i\in I_N^{\tilde{\mathcal B}}(h)$ and $\ell\in{\mathbbm Z}$,
$(h_{b_i}+i+\ell\tilde\beta_1)\in b_i{\mathbbm Z}$, i.e.
$I_N^{\tilde{\mathcal B}}(h)\subseteq I_N^{{\tilde{\mathcal B}}_1}(h+\Delta(\ell\tilde\beta_1))$.
Assume for a contradiction that there exists $j\in I_N^{{\tilde{\mathcal B}}_1}(h+\Delta(\ell\tilde\beta_1))\setminus I_N^{\tilde{\mathcal B}}(h)$. Then there is $b\in{\tilde{\mathcal B}}_1$ such that $(h_{b}+\ell\tilde\beta_1+j)\in b{\mathbbm Z}$, but $(h_{b}+j)\not\in b{\mathbbm Z}$, a contradiction to $b\mid\tilde\beta_1$. Therefore,
\begin{equation}\label{eq:I_N-identity}
I_N^{\tilde{\mathcal B}}(h)= I_N^{{\tilde{\mathcal B}}_1}(h+\Delta(\ell\tilde\beta_1))\quad\text{for all }\ell\in{\mathbbm Z}.
\end{equation}
\textbf{Step 2:}\quad
From now on we choose $n\geqslant2N+1$ large enough that $S\subseteq{\mathcal B}^{(n)}$, and we consider the particular element $h\in H$ from the assumption of the lemma.

For each $i\in I_0$ we have $(\varphi_{W'}(h))_i=0$. 
{
Hence
one can choose  $\tilde b_i\in{\mathcal B}$ and $\tilde b_i^*\in{\mathcal B}^*$ 
as in Lemma~\ref{lemma:Gamma}c
such that $\tilde b_i\in\tilde b_i^*{\mathbbm Z}$ and
$h_{\tilde b_i}+i\in\tilde b_i^*{\mathbbm Z}$. But $h_{\tilde b_i}+i\not\in\tilde b_i{\mathbbm Z}$, because $(\varphi(h))_i=1$.
In particular, $\tilde b_i^*\neq \tilde b_i$ so that $\tilde b_i^*\in\prim\cA_\infty$ by Lemma~\ref{lemma:Gamma}c. 
Hence, in view of Lemma~\ref{lemma:Ainf}, there is an infinite pairwise co-prime set ${\mathcal D}_i\subseteq{\mathbbm N}\setminus\{1\}$ such that $\tilde b_i^*{\mathcal D}_i\subseteq{\mathcal B}$. 
}

We are going to apply Step 1 to $\tilde{\mathcal B}={\mathcal B}$. This yields the set ${\mathcal B}_1={\mathcal B}^{(n)}\cup\{b_i:i\in I_N^{\mathcal B}(h)\}$ and the number $\beta_1:=\operatorname{lcm}({\mathcal B}_1)$.
As $I_0$ is a finite set, we can choose $d_i\in{\mathcal D}_i$ $(i\in I_0)$ pairwise co-prime and co-prime to $\beta_1$ and all $\tilde b_j$, $j\in I_0$, 
{
in particular also co-prime to all $\tilde b_j^*$, $j\in I_0$.
Denote $b_i:=d_i\cdot\tilde b_i^*$ for $i\in I_0$.} 
\footnote{Observe that the choice of the $b_i$, $i \in I_0$, is not in conflict with the $b_i$ chosen in Step 1, because in that step the indices $i$ are in the set $I_N^{{{\mathcal B}}}(h)$, which is disjoint from $I_0$.} Then $b_i\in{\mathcal B}$; and
$b_i>n$, because otherwise $b_i\in{\mathcal B}^{(n)}\subseteq{\mathcal B}_1$, 
so that $d_i\mid b_i\mid\beta_1$ in contradiction to the choice of the $d_i$.
As $\tilde b_i^*\mid\gcd(b_i,\tilde b_i)$, we see that $\tilde b_i^*\mid (h_{\tilde b_i}-h_{b_i})$.
It follows that $h_{b_i}+i\in\tilde b_i^*{\mathbbm Z}$.

We claim that there is $x\in{\mathbbm Z}$ such that
\begin{align*}
x&=-(h_{b_i}+i)\text{ mod } b_i\quad\text{for all }i\in I_0,\\
x&=0\hspace{1.27cm}\text{ mod }\beta_1.
\end{align*}
Indeed, this follows from the CRT, because
{
\begin{compactitem}[-]
\item $\gcd(b_i,b_j)=\gcd(d_i\,\tilde{b}_i^*,d_j\,\tilde{b}_j^*)=\gcd(\tilde{b}_i^*,\tilde{b}_j^*)
\mid (h_{b_j}+j)-(h_{b_i}+i)$
for all $i,j\in I_0$, and
\item $\gcd(b_i,\beta_1)=\gcd(d_i\,\tilde{b}_i^*,\beta_1)=\gcd(\tilde b_i^*,\beta_1)\mid \tilde b_i^*\mid b_i\mid (h_{b_i}+i)$.
\end{compactitem}
}
Hence $x=\ell\beta_1$ for some $\ell\in{\mathbbm Z}$, and 
\begin{equation}\label{eq:unique-i-1}
(h+\Delta(x)+\Delta(i))_{b_i}=0\quad\text{for all $i\in I_0$.}
\end{equation}
As $b_i>n\geqslant 2N+1$,
\begin{equation}\label{eq:unique-i-2}
(h+\Delta(x)+\Delta(k))_{b_i}\neq0\quad\text{for all }i\in I_0\text{ and } k\in\{-N,\dots,N\}\setminus\{i\}.
\end{equation}

Let ${\mathcal B}^\operatorname{diam}ondsuit:={\mathcal B}_1\cup\{b_i:i\in I_0\}$ and $h^\operatorname{diam}ondsuit:=h+\Delta(x)$. We claim that
\begin{equation}\label{eq:I_N-identity-2}
I_N^{\mathcal B}(h)\cup I_0=I_N^{{\mathcal B}_1}(h+\Delta(x))\cup I_0=I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit).
\end{equation}
Indeed, the first identity follows from \eqref{eq:I_N-identity}, and the ``$\subseteq$''-direction of the second one follows from \eqref{eq:unique-i-1}.
So let $k\in I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)$, and fix  $b\in{\mathcal B}^\operatorname{diam}ondsuit$ such that $h^\operatorname{diam}ondsuit_b+k\in b{\mathbbm Z}$. If $b\in{\mathcal B}_1$, then $k\in I_N^{{\mathcal B}_1}(h^\operatorname{diam}ondsuit)$, and we are done. Otherwise there is $i\in I_0$ such that $b=b_i$, i.e. $(h+\Delta(x)+\Delta(k))_{b_i}= h^\operatorname{diam}ondsuit_{b}+k\text{ mod } b\in b{\mathbbm Z}$.
In view of \eqref{eq:unique-i-2} this implies $k=i\in I_0$.

\textbf{Step 3:}\quad
In this  step we  replace the set ${\mathcal B}^\operatorname{diam}ondsuit$ in \eqref{eq:I_N-identity-2} by ${\mathcal B}$. The price to pay will be to replace also $h^\operatorname{diam}ondsuit$ by another element from $U_S(h)$.

As ${\mathcal B}^{(n)}\subseteq{\mathcal B}_1\subseteq{\mathcal B}^\operatorname{diam}ondsuit\subseteq{\mathcal B}$, we can apply Step 1 to $\tilde{\mathcal B}={\mathcal B}^\operatorname{diam}ondsuit$ and $h=h^\operatorname{diam}ondsuit$.
This yields a set ${\mathcal B}^{(n)}\subseteq{\mathcal B}^\operatorname{diam}ondsuit_1\subseteq{\mathcal B}^\operatorname{diam}ondsuit$ such that
\begin{equation}\label{eq:I_N-identity-3}
I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)= I_N^{{\mathcal B}^\operatorname{diam}ondsuit_1}(h^\operatorname{diam}ondsuit).
\end{equation}
Denote $\beta_1^\operatorname{diam}ondsuit:=\operatorname{lcm}({\mathcal B}^\operatorname{diam}ondsuit_1)$, and
define
\begin{equation}\label{eq:B3-def}
{\mathcal B}_2:=\{b\in{\mathcal B}\setminus{\mathcal B}_1^\operatorname{diam}ondsuit: \text{$b$ has no prime divisor $p>n$ co-prime to $\beta_1^\operatorname{diam}ondsuit$}\}.
\end{equation}
As ${\mathcal B}^{(n)}\subseteq{\mathcal B}_1^\operatorname{diam}ondsuit$, the set ${\mathcal B}$ is the disjoint union of ${\mathcal B}_1^\operatorname{diam}ondsuit$, ${\mathcal B}_2$ and ${\mathcal C}({\mathcal B}_1^\operatorname{diam}ondsuit,n)$, where ${\mathcal C}({\mathcal B}_1^\operatorname{diam}ondsuit,n)$ is defined in \eqref{eq:cC-def}. Observe that
a prime $p>n$ is co-prime to $\beta_1^\operatorname{diam}ondsuit=\operatorname{lcm}({\mathcal B}_1^\operatorname{diam}ondsuit)$, if and only if it is co-prime to $\operatorname{lcm}({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2)$.
Hence
\begin{equation}
\begin{split}
{\mathcal C}({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2,n)
&=
\{b\in{\mathcal B}\setminus({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2): \text{$b$ has at least one prime divisor $p>n$ co-prime to $\beta_1^\operatorname{diam}ondsuit$}\}\\
&=
\{b\in{\mathcal B}\setminus{\mathcal B}_1^\operatorname{diam}ondsuit: \text{$b$ has at least one prime divisor $p>n$ co-prime to $\beta_1^\operatorname{diam}ondsuit$}\}\\
&=
{\mathcal C}({\mathcal B}_1^\operatorname{diam}ondsuit,n)
=
{\mathcal B}\setminus({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2).
\end{split}
\end{equation}
As in the proof of Proposition~5.11 in \cite{BKKL2015} one
shows that ${\mathcal B}_2$, and hence also ${\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2$, is finite
\footnote{If $p$ is a prime divisor of any
$b'\in{\mathcal B}_2$ then either $p\leqslant n$, or $p > n$ and $p\mid\beta_1^\operatorname{diam}ondsuit$. Hence 
$|\operatorname{Spec}({\mathcal B}_2)| < \infty$, and since ${\mathcal B}_2$ is primitive (it is a subset of the primitive set ${\mathcal B}$), it must be finite \cite[Lemma~ 5.14]{BKKL2015}.}
 and that
\begin{equation}\label{eq:like68}
\begin{split}
I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2}(h^\operatorname{diam}ondsuit)
&=
I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit).
\end{split}
\end{equation}
Indeed, let $i\in I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2}(h^\operatorname{diam}ondsuit)$. 
Then there is $b\in{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2$ such that $h^\operatorname{diam}ondsuit_{b}+i\in b{\mathbbm Z}$. 
If $b\in{\mathcal B}_1^\operatorname{diam}ondsuit$, then $i\in I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)$. Otherwise,
$b\in{\mathcal B}_2\subseteq{\mathcal B}\setminus{\mathcal B}_1^\operatorname{diam}ondsuit\subseteq{\mathcal B}\setminus{\mathcal B}^{(n)}$ has a prime divisor $p>n$. 
By the definition of ${\mathcal B}_2$ in \eqref{eq:B3-def}, $p\mid \beta_1^\operatorname{diam}ondsuit=\operatorname{lcm}({\mathcal B}_1^\operatorname{diam}ondsuit)$. As ${\mathcal B}_1^\operatorname{diam}ondsuit={\mathcal B}^{(n)}\cup\{b_j: j\in I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)\}$, it follows that
$p\mid b_j$ for some $j\in I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)$.
Then $p\mid(h^\operatorname{diam}ondsuit_{b_j}+j)$, and since $p\mid\gcd(b,b_j)\mid(h^\operatorname{diam}ondsuit_{b}-h^\operatorname{diam}ondsuit_{b_j})$, it follows that $p\mid(h^\operatorname{diam}ondsuit_{b}+j)$. But we also have $p\mid b\mid (h^\operatorname{diam}ondsuit_{b}+i)$,  so that $p\mid(i-j)$, which contradicts $p>n\geqslant2N+1$. 

Thus having established \eqref{eq:like68},
we apply Lemma~\ref{lemma:new-2} to the set 
${\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2$

So there is $h'\in U_{{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2}(h^\operatorname{diam}ondsuit)$ with $h'_b+ i\neq 0\text{ mod } b$ for all $b\in{\mathcal C}({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2,n)={\mathcal B}\setminus({\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2)$ and $i\in\{-N,\dots,N\}$.
Combined with \eqref{eq:I_N-identity-2}, \eqref{eq:I_N-identity-3} and \eqref{eq:like68} this shows
\begin{equation}\label{eq:I_N-final}
I_N^{\mathcal B}(h)\cup I_0
=
I_N^{{\mathcal B}^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)
=
I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit}(h^\operatorname{diam}ondsuit)
=
I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2}(h^\operatorname{diam}ondsuit)
=
I_N^{{\mathcal B}_1^\operatorname{diam}ondsuit\cup{\mathcal B}_2}(h')
=
I_N^{{\mathcal B}}(h').
\end{equation}
Observe that $h'\in U_S(h^\operatorname{diam}ondsuit)=U_S(h)$, because $S\subseteq{\mathcal B}^{(n)}\subseteq{\mathcal B}_1\cap{\mathcal B}_1^\operatorname{diam}ondsuit$, so that, for all $b\in S$, $h'_b=h^\operatorname{diam}ondsuit_b=h_b+\ell\beta_1\text{ mod } b=h_b$.

To conclude the proof, notice  that, in view of \eqref{eq:I_N-final}, for any $i\in\{-N,\dots,N\}$
\begin{equation*}
(\varphi(h'))_i=0
\;\Leftrightarrow\;
i\in I_N^{\mathcal B}(h')
\;\Leftrightarrow\;
i\in I_N^{\mathcal B}(h)\cup I_0
\;\Leftrightarrow\;
(\varphi(h))_i=0\text{ or }i\in I_0
\;\Leftrightarrow\;
\omega_i=0.
\end{equation*}
\end{proof}

\begin{proposition}\label{prop:hereditary-ext}
If $\varphi_{W'}(h)\leqslant x\leqslant\varphi(h)$ for some $x\in\{0,1\}^{\mathbbm Z}$ and $h\in H$, then there are $h_1,h_2,\dots\in H$ such that $h=\lim_{N\to\infty}h_N$ and
$x=\lim_{N\to\infty}\varphi(h_N)$.
\end{proposition}

\begin{proof}
Fix a filtration $S_1\subset S_2\subset\dots$ of ${\mathcal B}$ by finite sets $S_k$.
For $N\in{\mathbbm N}$, let $I_N:=\{i\in\{-N,\dots,N\}: x_i=0, (\varphi(h))_i=1\}$. 
By Lemma~\ref{lemma:ext-1}, there are $h_N\in U_{S_N}(h)$ $(N\in{\mathbbm N})$ such that $x_i=(\varphi(h_N))_i$ for all $i=-N,\dots,N$. Then $\lim_{N\to\infty}h_N=h$ and 
$\lim_{N\to\infty}\varphi(h_N)=x$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{theo:hereditary-ext}]
$\varphi(H)\subseteq[\varphi_{W'},\varphi] $ is obvious. So
let $x\in[\varphi_{W'},\varphi] $. 
There is $h\in H$ with $\varphi_{W'}(h)\leqslant x\leqslant\varphi(h)$.
Hence $x\in X_\varphi$ because of Proposition~\ref{prop:hereditary-ext}.
\end{proof}

\subsection{Proof of Proposition~\ref{prop:toeplitz}}\label{subsec:toeplitz-proofs}
All sets and quantities derived from ${\mathcal B}^*$ instead of ${\mathcal B}$ also carry a $^*$ superscript.
For $S\subset{\mathcal B}$ define
\begin{equation}
{\mathcal M}_{{\mathcal B}^*|S}:={\mathcal M}_{\{b\in{\mathcal B}^*: b\,\mid\,\operatorname{lcm}(S)\}},
\end{equation}
and observe that ${\mathcal M}_{{\mathcal B}^*}=\bigcup_{S\subset{\mathcal B}}{\mathcal M}_{{\mathcal B}^*|S}$.

\begin{lemma}\label{lemma:B*S}
${\mathcal M}_{{\mathcal B}^*|S}=\bigcap_{k\in{\mathbbm Z}}\left({\mathcal M}_{{\mathcal B}^*}+k\cdot\operatorname{lcm}(S)\right)$ and
${\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S}={\mathcal E}+\operatorname{lcm}(S)\cdot{\mathbbm Z}$.
\end{lemma}

\begin{proof}
Let $n\in{\mathcal M}_{{\mathcal B}^*|S}$, so $n=\ell b$ for some $\ell\in{\mathbbm Z}$ and  $b\in{\mathcal B}^*$, $b\mid\operatorname{lcm}(S)$. Then $b\mid n-k\cdot\operatorname{lcm}(S)$ for all $k\in{\mathbbm Z}$, i.e. 
$n\in\bigcap_{k\in{\mathbbm Z}}\left({\mathcal M}_{{\mathcal B}^*}+k\cdot\operatorname{lcm}(S)\right)$. 

Conversely suppose that $n\in\bigcap_{k\in{\mathbbm Z}}\left({\mathcal M}_{{\mathcal B}^*}+k\cdot\operatorname{lcm}(S)\right)$.
Then $n+\operatorname{lcm}(S)\cdot{\mathbbm Z}\subseteq{\mathcal M}_{{\mathcal B}^*}$, in particular $n\in{\mathcal M}_{{\mathcal B}^*}$.
Let $m:=\gcd(\operatorname{lcm}(S),n)$, $\hat n:=n/m$ and $\hat s:=\operatorname{lcm}(S)/m$. Then $\hat n$ and $\hat s$ are co-prime, and by Dirichlet's theorem, ${\mathcal P}':={\mathcal P}\cap(\hat n+\hat s\cdot{\mathbbm Z})$ is an infinite set of prime numbers. Observe that $m\,{\mathcal P}'\subseteq{\mathcal M}_{{\mathcal B}^*}$, hence for all $p\in{\mathcal P}'$ there are $k_p\in{\mathbbm Z}$ and $b_p\in{\mathcal B}^*$ such that $mp=k_pb_p$.

Suppose for a contradiction that for all $p\in{\mathcal P}'$ there is $\ell_p\in{\mathbbm Z}$ such that $b_p=\ell_pp$. Then $m=k_p\ell_p$, and as ${\mathcal P}'$ is infinite, there are $\ell\in{\mathbbm Z}$ and an infinite subset ${\mathcal P}_0'$ of ${\mathcal P}'$ such that $\ell_p=\ell$ for all $p\in{\mathcal P}_0'$.
Hence $\ell\,{\mathcal P}_0'=\{b_p:p\in{\mathcal P}_0'\}\subseteq{\mathcal B}^*$, which is impossible in view of Lemma~\ref{lemma:B^*}c.
So there is some $p\in{\mathcal P}'$ that
does not divide $b_p$. Hence $m=k_p/p\cdot b_p\in b_p{\mathbbm Z}$, and so $n=\hat{n}m\in b_p{\mathbbm Z}$ and $\operatorname{lcm}(S)=\hat{s}m=(\hat{s}\cdot k_p/p)\cdot b_p$, i.e. $b_p\mid\operatorname{lcm}(S)$.

We proved ${\mathcal M}_{{\mathcal B}^*|S}=\bigcap_{k\in{\mathbbm Z}}\left({\mathcal M}_{{\mathcal B}^*}+k\cdot\operatorname{lcm}(S)\right)$. Taking complements this yields
${\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S}=\bigcup_{k\in{\mathbbm Z}}\left({\mathcal F}_{{\mathcal B}^*}+k\cdot\operatorname{lcm}(S)\right)={\mathcal E}+\operatorname{lcm}(S)\cdot{\mathbbm Z}$, where we used also Lemma~\ref{lemma:B^*}b for the last identity.
\end{proof}

\begin{lemma}\label{lemma:bdW'}
\begin{compactenum}[a)]
\item Let $n\in{\mathbbm Z}$ and $S\subset{\mathcal B}$. Then
$U_S(\Delta(n))\subseteq\operatorname{int}(W')$ if and only if $n\in{\mathbbm Z}\setminus{\mathcal M}_{{\mathcal A}_S}$, and
$U_S(\Delta(n))\cap W'\neq\emptyset$ if and only if $n\in{\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S}$.
\item 
$m_H(\partial W')=\inf_{S\subset{\mathcal B}}d({\mathcal M}_{{\mathcal A}_S}\setminus{\mathcal M}_{{\mathcal B}^*|S})$.
\end{compactenum} 
\end{lemma}

\begin{proof} a)\;Recall that $U_S(\Delta(n))$ is open. As $\operatorname{int}(W')=\operatorname{int}(W)$, the first assertion follows from \cite[Lemma~3.1a]{KKL2016}.
For the second one observe first that $U_S(\Delta(n))\cap W'\neq\emptyset$ if and only if $U_S(\Delta(n))\cap \operatorname{int}(W')\neq\emptyset$. In view of \cite[Lemma~3.1a]{KKL2016} this is equivalent to: there are $S\subset S'\subset{\mathcal B}$ and $\ell\in n+\operatorname{lcm}(S)\cdot{\mathbbm Z}$ such that $\ell\in{\mathcal F}_{{\mathcal A}_{S'}}$, which can be equivalently rewritten as
\begin{equation*}
\begin{split}

&
\bigcup_{S\subseteq S'\subset{\mathcal B}} {\mathcal F}_{{\mathcal A}_{S'}}\cap\left(n+\operatorname{lcm}(S)\cdot{\mathbbm Z}\right)\neq\emptyset\\
\Leftrightarrow\quad&
{\mathcal E}\cap\left(n+\operatorname{lcm}(S)\cdot{\mathbbm Z}\right)\neq\emptyset\\
\Leftrightarrow\quad&
n\in\left({\mathcal E}+\operatorname{lcm}(S)\cdot{\mathbbm Z}\right)
=
{\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S}\quad(\text{Lemma~\ref{lemma:B*S}}).
\end{split}
\end{equation*}
b)\;Let $S_1\subset S_2\subset\dots$ be any filtration of ${\mathcal B}$ by finite sets $S_k$. Then part a) of the lemma implies
\begin{equation*}
m_H(\operatorname{int}(W'))
=
\lim_{k\to\infty}d({\mathbbm Z} \setminus{\mathcal M}_{{\mathcal A}_{S_k}})
=
\sup_{S\subset{\mathcal B}}d({\mathbbm Z}\setminus{\mathcal M}_{{\mathcal A}_{S}})
\end{equation*}
and
\begin{equation*}
m_H(W')
=
\lim_{k\to\infty}d({\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S_k})
=
\inf_{S\subset{\mathcal B}}d({\mathbbm Z}\setminus{\mathcal M}_{{\mathcal B}^*|S}).
\end{equation*}
Observing that ${\mathcal M}_{{\mathcal B}^*|S}\subseteq{\mathcal M}_{{\mathcal B}^*}\subseteq{\mathcal M}_{{\mathcal A}_{S'}}$ (Lemma~\ref{lemma:B^*}) for all $S,S'\subset{\mathcal B}$, we conclude that
\begin{equation*}
m_H(\partial W')
=
m_H(W')-m_H(\operatorname{int}(W'))
=
\lim_{k\to\infty}d({\mathcal M}_{{\mathcal A}_{S_k}}\setminus{\mathcal M}_{{\mathcal B}^*|S_k})
=
\sup_{S\subset{\mathcal B}}d({\mathcal M}_{{\mathcal A}_S}\setminus{\mathcal M}_{{\mathcal B}^*|S}).
\end{equation*}
\end{proof}

For $S\subset{\mathcal B}$ let $S^*:=(S\cap{\mathcal B}_0)\cup\{a\in\prim\cA_\infty: S\cap a\cdot({\mathbbm N}\setminus\{1\})\neq\emptyset\}$, where ${\mathcal B}_0={\mathcal B}\setminus{\mathcal M}_{\cA_\infty}$ as in \eqref{eq:B_0-def}.
Then $S^*\subset{\mathcal B}^*$.

\begin{lemma}\label{lemma:inclusions}
For any $S\subset{\mathcal B}$ there is $S'\subset{\mathcal B}^*$ with $S^*\subseteq S'$ such that
${\mathcal M}_{{\mathcal A}^*_{S'}}\subseteq{\mathcal M}_{{\mathcal A}_S}\subseteq{\mathcal M}_{{\mathcal A}^*_{S^*}}$.
\end{lemma}
\begin{proof}
Let $S\subset{\mathcal B}$.
For $a\in{\mathbbm N}$ define ${\mathcal B}_a:=\{b\in{\mathcal B}^*:\gcd(\operatorname{lcm}(S),b)=a\}$, and let ${\mathcal D}_S:=\{a\in{\mathbbm N}:{\mathcal B}_a\neq\emptyset\}$. Then ${\mathcal D}_S\subseteq\{1,\dots,\operatorname{lcm}(S)\}$ is finite and ${\mathcal B}^*=\bigcup_{a\in{\mathcal D}_S}{\mathcal B}_a$. 
For each $a\in{\mathcal D}_S$ fix some $b_a\in{\mathcal B}_a$ and let 
\begin{equation*}
S':=S^*\cup\{b_a:a\in{\mathcal D}_S\}.
\end{equation*}
Then $S'\subset{\mathcal B}^*$, and we claim that ${\mathcal A}^*_{S'}\subseteq{\mathcal M}_{{\mathcal A}_S}$, so that
${\mathcal M}_{{\mathcal A}^*_{S'}}\subseteq{\mathcal M}_{{\mathcal A}_S}$. Indeed: Let $b\in{\mathcal B}^*$. There is $a\in{\mathbbm N}$ such that $b\in{\mathcal B}_a$. As $b_a\in{\mathcal B}_a$ as well, we have $a\mid\gcd(b_a,b)\mid\gcd(\operatorname{lcm}(S'),b)$,
so that $\gcd(\operatorname{lcm}(S'),b)\in a{\mathbbm Z}$. It remains to show that $a\in{\mathcal A}_S$.
\begin{compactitem}[-]
\item If $b\in{\mathcal B}_a\cap{\mathcal B}_0$, then $a=\gcd(\operatorname{lcm}(S),b)\in{\mathcal A}_S$.
\item If $b\in{\mathcal B}_a\cap\prim\cA_\infty$, then
there is an infinite pairwise co-prime set ${\mathcal C}\subseteq{\mathbbm N}\setminus\{1\}$ such that $b\,{\mathcal C}\subseteq{\mathcal B}$ (Lemma~\ref{lemma:Ainf}).
Choose any $c\in{\mathcal C}$ co-prime to $\operatorname{lcm}(S)$. Then $bc\in{\mathcal B}$,
so that $a=\gcd(\operatorname{lcm}(S),b)=\gcd(\operatorname{lcm}(S),bc)\in{\mathcal A}_S$.
\end{compactitem}

For the the proof of the inclusion ${\mathcal M}_{{\mathcal A}_S}\subseteq{\mathcal M}_{{\mathcal A}^*_{S^*}}$, let $a=\gcd(\operatorname{lcm}(S),b)\in{\mathcal A}_S$ with $b\in{\mathcal B}$. Then $\operatorname{lcm}(S^*)\mid\operatorname{lcm}(S)$ by definition of $S^*$, and, by definition of ${\mathcal B}^*$, there is $b^*\in{\mathcal B}^*$ which divides $b$.
Hence $a\in \gcd(\operatorname{lcm}(S^*),b^*)\cdot{\mathbbm Z}\subseteq{\mathcal M}_{{\mathcal A}^*_{S^*}}$. This proves ${\mathcal A}_S\subseteq{\mathcal M}_{{\mathcal A}^*_{S^*}}$ and hence ${\mathcal M}_{{\mathcal A}_S}\subseteq{\mathcal M}_{{\mathcal A}^*_{S^*}}$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:toeplitz}]
We proved in Lemma~\ref{lemma:B^*}e that $1_{{\mathcal F}_{{\mathcal B}^*}}=\varphi_{W'}(\Delta(0))=1_{\mathcal E}$ is a Toeplitz sequence. It is regular if and only if $m_{H^*}(\partial W^*)=0$, see \cite[Prop.~1.2]{KKL2016}, and in view of \cite[Lemma~4.3]{KKL2016} \footnote{There is an obvious misprint in this lemma and its proof: All occurences of $\underline d$ must be replaced by $\bar d$.} this is equivalent to
\begin{equation*}
\inf_{S''\subset{\mathcal B}^*}\ \bar d({\mathcal M}_{{\mathcal A}^*_{S''}}\setminus{\mathcal M}_{{\mathcal B}^*})=0.
\end{equation*}
Here $\bar d$ denotes the upper asymptotic density $\bar d(M):=\limsup_{n\to\infty}n^{-1}\operatorname{card}(M\cap[1,n])$. Similarly, the lower asymptotic density is denoted by $d(M)$, and if the limit exists, which is the case for all sets $M={\mathcal M}_A$ with finite $A$, it is denoted by $d(M)$.

On the other hand, $m_H(\partial W')=0$ if and only if 
\begin{equation*}
\inf_{S\subset{\mathcal B}}\ d({\mathcal M}_{{\mathcal A}_{S}}\setminus{\mathcal M}_{{\mathcal B}^*|S})=0
\end{equation*}
by Lemma~\ref{lemma:bdW'}.
The equivalence of both conditions will follow, once we have proved the (more general) identity
\begin{equation}\label{eq:identity}
\inf_{S''\subset{\mathcal B}^*}\ \bar d({\mathcal M}_{{\mathcal A}^*_{S''}}\setminus{\mathcal M}_{{\mathcal B}^*})
=
\inf_{S\subset{\mathcal B}}\ d({\mathcal M}_{{\mathcal A}_{S}}\setminus{\mathcal M}_{{\mathcal B}^*|S})\ .
\end{equation}

So let $S\subset{\mathcal B}$. Choose $S^*\subseteq S'\subset{\mathcal B}^*$ as in Lemma~\ref{lemma:inclusions}.
Then, for any $S'\subseteq S''\subset{\mathcal B}^*$,
\begin{equation*}
{\mathcal M}_{{\mathcal A}^*_{S''}}\subseteq{\mathcal M}_{{\mathcal A}^*_{S'}}
\subseteq
{\mathcal M}_{{\mathcal A}_S},
\end{equation*}
so that ${\mathcal M}_{{\mathcal A}^*_{S''}}\setminus{\mathcal M}_{{\mathcal B}^*}\subseteq{\mathcal M}_{{\mathcal A}_S}\setminus{\mathcal M}_{{\mathcal B}^*|S}$.
This proves the ``$\leqslant$'' direction in \eqref{eq:identity}.

Conversely, let $S''\subset{\mathcal B}^*$. Choose any $S\subset{\mathcal B}$, for which $S''\subseteq S^*$.
Then, by Lemma~\ref{lemma:inclusions},
\begin{equation*}
{\mathcal M}_{{\mathcal A}_S}\subseteq
{\mathcal M}_{{\mathcal A}^*_{S^*}}
\subseteq
{\mathcal M}_{{\mathcal A}^*_{S''}},
\end{equation*}
so that ${\mathcal M}_{{\mathcal A}_S}\setminus{\mathcal M}_{{\mathcal B}^*|S}\subseteq({\mathcal M}_{{\mathcal A}^*_{S''}}\setminus{\mathcal M}_{{\mathcal B}^*})\cup({\mathcal M}_{{\mathcal B}^*}\setminus{\mathcal M}_{{\mathcal B}^*|S})$. 
Hence
\begin{equation*}
\begin{split}
d({\mathcal M}_{{\mathcal A}_S}\setminus{\mathcal M}_{{\mathcal B}^*|S})
&\leqslant
\bar d({\mathcal M}_{{\mathcal A}^*_{S''}}\setminus{\mathcal M}_{{\mathcal B}^*})+\underline{d}({\mathcal M}_{{\mathcal B}^*}\setminus{\mathcal M}_{{\mathcal B}^*|S}).

\end{split}
\end{equation*}
This proves the
``$\geqslant$'' direction in \eqref{eq:identity}, because $\underline{d}({\mathcal M}_{{\mathcal B}^*})=\sup_{S\subset{\mathcal B}}d({\mathcal M}_{{\mathcal B}^*|S})$ by the theorem of Davenport and Erd\"os, see \cite[eqn.~(0.67)]{hall-book}.
\end{proof}

\end{document}