\def\Sigo{\Sigma(\Omega)} \newcommand{\frr}[1]{} \def\Z{{\mathbb Z}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}} \def\N{{\mathbb N}} \def\Zt{\Z[\tau]} \def\Zb{\Z[\beta]} \def\Qt{\Q[\tau]} \def\Qb{\Q[\beta]} \def\tr{\mbox{tr}} \def\pf{\begin{proof}} \def\pfk{\end{proof}} \documentclass{amsart} \usepackage{epsfig} \textwidth 6.5in \textheight 8.5in \topmargin -0.5cm \leftmargin -0.5cm \oddsidemargin 0cm \evensidemargin 0cm \parindent 0.25in \begin{document} \newsymbol\Vdash 130D \newsymbol\supsetneqq 2325 \newsymbol\subsetneqq 2324 \newtheorem{lem}{Lemma}[section] \newtheorem{thm}[lem]{Theorem} \newtheorem{prop}[lem]{Proposition} \newtheorem{coro}[lem]{Corollary} \newtheorem{que}{Question} \theoremstyle{definition} \newtheorem{de}[lem]{\bf Definition} \theoremstyle{remark} \newtheorem{pozn}[lem]{Remark} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{$s$-convexity, model sets and their relation} \author{Zuzana Mas\'akov\'a} \address{Department of Mathematics, Faculty of Nuclear Science and Physical Engineering\\ Czech Technical University, Trojanova 13, Prague 2, 120 00, Czech Republic\\ masakova@km1.fjfi.cvut.cz} \author{Ji\v r\'\i\ Patera} \address{Centre de recherches math\'ematiques, Universit\'e de Montr\'eal, \\ C.P. 6128 Succursale Centre-ville, Montr\'eal, Qu\'ebec, H3C 3J7, Canada\\ patera@crm.umontreal.ca} \author{Edita Pelantov\'a} \address{Department of Mathematics, Faculty of Nuclear Science and Physical Engineering\\ Czech Technical University, Trojanova 13, Prague 2, 120 00, Czech Republic\\ pelantova@km1.fjfi.cvut.cz} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{abstract} The relation of $s$-convexity and sets modeling physical quasicrystals is explained for quasicrystals related to quadratic unitary Pisot numbers. We show that 1-dimensional model sets may be characterized by $s$-convexity for finite set of parameters $s$. It is shown that the three Pisot numbers $\frac12(1+\sqrt5)$, $1+\sqrt2$, and $2+\sqrt3$ related to experimentally observed non-crystallographic symmetries are exceptional with respect to $s$-convexity. \end{abstract} \maketitle %======================================================================= %\bigskip \section{Introduction} \label{sec:prelim} In this paper we study properties of a class of binary operations defined on a real vector space. % by $x\vdash_s y:=sx+(1-s)y$, where $s$ is a real parameter. As we show in this article such operations can be used for characterizations of point sets modeling physical quasicrystals. % The mathematical problems studied in this paper are connected with % a relatively new field of physics, namely the theory of quasicrystals. % We study binary operations which could be considered as a replacement % of ordinary addition for aperiodic point sets, and therefore used for % modeling of growth of physical quasicrystals. Surprisingly, Such operation was first introduced in a purely mathematical article by I.~Calvert in 1978. However, from his paper~\cite{calvert} and absence of references in it, one cannot deduce the motivation for the study of this operations. The problem was then elaborated by R.~G.~E.~Pinch in 1985 in~\cite{pinch}. The connection of this operation with quasicrystals was first recognized by Berman and Moody in~\cite{bm}, where they work with a special case of such operation. % with $s=-\frac12(1+\sqrt5)$. % Berman and Moody call % this operation `a quasiaddition' and show how it can be used to generate % a 1-dimensional quasicrystal from an initial set of seeds. For arbitrary real parameter $s$, Calvert defined a binary operation \begin{equation}\label{eq:oper} x\vdash_s y:=sx+(1-s)y\,, \qquad x,y\in\R^n\,. \end{equation} Pinch in~\cite{pinch} calls a set closed under such operation an {\bf $s$-convex} set. Pinch shows that an $s$-convex set $\Lambda\subset\R$ is either dense in an interval of $\R$, or it is uniformly discrete. A set $\Lambda\subset\R$ is called {\bf uniformly discrete}, if there exists an $\varepsilon >0$, such that $|x-y|>\varepsilon$ for any $x,y\in\Lambda$, $x\neq y$. Pinch studies the question for which parameters $s$ there exists a uniformly discrete $s$-convex set containing at least two points. We shall denote by ${\it Cl}_sA$ the closure of the set $A$ under the operation~\eqref{eq:oper}. A most suitable candidate for a uniformly discrete $s$-convex set ($s$ fixed) is ${\it Cl}_s\{0,1\}$. If $s$ belongs to $(0,1)$, then ${\it Cl}_s\{0,1\}$ is a closure under a ordinary convex combination and therefore is dense in $[0,1]$. Pinch shows that for $s\notin(0,1)$, the closure ${\it Cl}_s\{0,1\}$ may be uniformly discrete only if $s$ is an algebraic integer. On the other hand he proves that if $s$ is a totally real algebraic integer, such that all its algebraic conjugates are in $(0,1)$, then ${\it Cl}_s\{0,1\}$ is uniformly discrete. The connection to sets modeling quasicrystals was observed by Berman and Moody on the example of model sets based on the golden ratio $\tau=\frac12(1+\sqrt5)$, defined by \begin{equation}\label{eq:tau} \Sigma_\tau(\Omega) := \{a+b\tau \mid a,b\in\Z\,,\, a+b\tau' \in \Omega\}\,, \end{equation} where $\tau'=\frac12(1-\sqrt5)$ is the algebraic conjugate of $\tau$ and $\Omega$ is a bounded interval in $\R$. For such $\Omega$, the set $\Sigma_\tau(\Omega)$ is not only uniformly discrete, but also {\bf relatively dense}, which means that the distances of adjacent points in $\Sigma_\tau(\Omega)$ are bounded by a fixed constant. A set which is both uniformly discrete and relative dense is called a {\bf Delone} set. It is easy to show that $\Sigma_\tau(\Omega)$ is closed under the operation~\eqref{eq:oper} for $s=-\tau$. The operation $\vdash_{-\tau}$ has an important property. It has been shown in~\cite{bm} that the quasicrystal $\Sigma_\tau[0,1]$ may be generated by this operation starting from initial seeds $\{0,1\}$, more precisely ${\it Cl}_{-\tau}\{0,1\}= \Sigma_\tau[0,1]$. This means that the operation is `strong enough' so that any uniformly discrete set closed under it has to be a model set of the form~\eqref{eq:tau}. It can be shown that all model sets~\eqref{eq:tau} are $s$-convex for infinitely many $s\in\Zt$. % any $s\in\Sigma_\tau[0,1]$. (Note that $-\tau$ belongs to it as well.) A question arises whether there exists another $s$ `strong enough' to ensure ${\it Cl}_{s}\{0,1\}=\Sigma_\tau[0,1]$. In this article we provide an answer to a more general question. The golden ratio $\tau$ is a quadratic unitary Pisot number related to 5-fold symmetry quasicrystals. Other experimentally observed non-crystallographic symmetries (8-fold and 12-fold) are related also to quadratic unitary Pisot numbers $1+\sqrt2$ and $2+\sqrt3$. We consider here two infinite families of quadratic unitary Pisot numbers $\beta>1$, solutions of the equations \begin{eqnarray*} &&x^2=mx+1\,, \qquad m\in\Z\,, m\geq1\,, \hbox{ or}\\ &&x^2=mx-1\,, \qquad m\in\Z\,, m\geq3\,. \end{eqnarray*} Let us denote by $\beta'$ the algebraic conjugate of $\beta$, i.e.~the second root of the equation. The number $\beta$ is called unitary Pisot, since $|\beta'|<1$, and $\beta\beta'=\mp1$. A model set is then defined as a subset of the ring $\Z[\beta]:=\Z+\Z\beta$, equipped with the Galois automorphism $x=a+b\beta$ $\ \mapsto\ $ $x'=a+b\beta'$. We let \begin{equation}\label{eq:deqc} \Sigma_\beta(\Omega) := \{x\in\Zb \mid x'\in\Omega\}\,, \end{equation} where $\Omega$ is a bounded interval in $\R$. The set $\Sigma_\beta(\Omega)$ is called a model set in $\Zb$ with the acceptance window $\Omega$. Remark: Let us mention that the solution of equation $x^2=3x-1$ gives the same ring $\Zb$ as the equation $x^2=x+1$, and thus we shall consider the equation $x^2=mx-1$ with $m>3$ only. \begin{que}\label{q:1} Let $\beta$ be quadratic unitary Pisot number. Which parameter $s$ ensures that any uniformly discrete $s$-convex set is a model set in $\Zb$? \end{que} We show in this paper that $s=-\frac12(1+\sqrt5)$ is exceptional for model sets in $\Zt$. The number $\tau^2=1-s$ has the same property, since generally $s$-convexity is equivallent to $(1-s)$-convexity. For no other parameters $s$ the operation~\eqref{eq:oper} is `strong enough'. The only other quadratic unitary Pisot numbers, for which a suitable parameter exists are $\beta=1+\sqrt2$ and $2+\sqrt3$. It means that in the majority of cases the invariance with respect to one operation~\eqref{eq:oper} does not imply being a model set. One may further ask the following question. \begin{que}\label{q:2} Let $\beta$ be quadratic unitary Pisot number. Is it possible to find a finite set ${\mathcal N}$ of parameters, such that $s$-convexity of a uniformly discrete set $\Lambda\subset\Zb$ for all $s\in {\mathcal N}$ implies $\Lambda$ being a model set in $\Zb$? \end{que} The answer is YES, and we provide a constructive proof for each $\beta$. %The necessary condition in order that this could be true is that %$s\in\Sigma_\beta[0,1]$ and ${\it Cl}_{s}\{0,1\}=\Sigma_\beta[0,1]$ The solutions to Questions~\ref{q:1} and~\ref{q:2} are formulated as Theorems~\ref{thm:1} and~\ref{thm:2} in Section~\ref{sec:results}. Their proofs require different approaches and different mathematical tools, therefore they are found in separate Sections~\ref{sec:tools} and~\ref{sec:druha}. %======================================================================= \section{Auxiliary lemmas I}\label{sec:tools} Recall that we are working in a ring $\Zb\subset\Qb$, and that the mentioned conjugation $\ '\ $ is a Galois automorphism on the field $\Qb$. One has $(x+y)'=x'+y'$ and $(xy)'=x'y'$ for any two elements $x,y\in\Zb$. With this in mind, we may state that for $s\in\Zb$ and $\Lambda\subset\Zb$, \begin{equation}\label{eq:eqv} \Lambda \hbox{ is } s\hbox{-convex} \qquad\hbox{ if and only if }\qquad \Lambda'=\{x'\mid x\in\Lambda\} \hbox{ is } s'\hbox{-convex.} \end{equation} From the definition of model sets it follows that $\bigl(\Sigma_\beta (\Omega)\bigr)'=\Zb\cap\Omega$. Hence from~\eqref{eq:eqv} we obtain $$ \Sigma_\beta(\Omega) \hbox{ is } s\hbox{-convex} \qquad\hbox{ if and only if } \qquad \Zb\cap\Omega \hbox{ is } s'\hbox{-convex.} $$ Since $\Omega$ is an interval, $s'x+(1-s')y$ has to be a convex combination in the ordinary sense, i.e.~$s'\in[0,1]$, which is equivallent to $s\in\Sigma_\beta[0,1]$. In this formalism, Question~\ref{q:1} may be rewritten as follows: Does there exist $s\in\Sigma_\beta[0,1]$, such that for any uniformly discrete $s$-convex set $\Lambda\subset\Zb$ it holds that \begin{equation}\label{eq:ladnenko} \Lambda'=\Zb\cap\Omega \end{equation} for some $\Omega$? If~\eqref{eq:ladnenko} should be valid for every uniformly discrete $\Lambda$, it must be true also for $\Lambda={\it Cl}_{s}\{0,1\}$. Since $\Lambda'= {\it Cl}_{s'}\{0,1\}$ contains $0'=0$ and $1'=1$ and since $s'x+(1-s')y$ is a convex combination, the only suitable candidate for $\Omega$ in this case is the interval $\Omega=[0,1]$. It is the reason why in the sequel we focus our attention to investigation of equatlity $$ [0,1]\cap\Zb={\it Cl}_{s'}\{0,1\}\,, \qquad \hbox{ for } s'\in(0,1)\,. $$ \begin{lem}\label{xorxj} Let $\beta$ be quadratic unitary Pisot number. Assume that ${\it Cl}_{s'}\{0,1\}=\Zb\cap[0,1]$. Then for any $y\in\Zb\cap[0,1]$, the scaling factor $s'$ divides either $y$ or $y-1$ in the ring $\Zb$. \end{lem} \pf Pinch in~\cite{pinch} has proved that any $y\in{\it Cl}_{s'}\{0,1\}$ may be written in the form \begin{equation}\label{eq:as} y=\sum_{i=0}^n b_i(s')^i(1-s')^{n-i}\,, \qquad b_i\in\Z\,,\, 0\leq b_i \leq {n\choose i}\,, \end{equation} for some non negative integer $n$. Therefore $b_0\in\{0,1\}$. If $b_0$ is equal to 0, then $y$ is divisible by $s'$, otherwise $s'$ divides $y-b_0=y-1$. \pfk Before the following corollary let us mention several number theoretical facts. On the field $\Qb$ on may define a `norm' $N(x):=xx'\in\Q$ and a `trace' ${\rm tr}(x):=x+x'\in\Q$. Since $|N(\beta)|=1$ and ${\rm tr}(\beta)=m$ are both integers, it holds for every $x\in\Zb$ that $N(x)\in\Z$ and ${\rm tr}(x)\in\Z$. A divisor of unity in $\Zb$ is an element $u$ such that $\frac1u\in\Zb$; $u$ is a divisor of unity iff $|N(u)|=1$. \begin{coro}\label{divdva} If ${\it Cl}_{s'}\{0,1\}=\Zb\cap[0,1]$, then both $s'$ and $1-s'$ divide 2 in the ring $\Zb$. \end{coro} \pf Note that the roles of $s'$ and $1-s'$ are symmetric. Thus it suffices to show that $s'$ divides $2$. If $s'$ is a divisor of unity then the assertion is true. Assume the opposite, i.e. $s'$ is not a divisor of unity and $s'$ does not divide $2$. Choose $y=1/\beta^2$. Since $y$ is a divisor of unity, the scaling factor $s'$ does not divide neither $y$, nor $2y$. According to Lemma~\ref{xorxj} this implies $s'|(y-1)$ and $s'|(2y-1)$. Therefore $s'$ divides $(2y-1)-(y-1)=y$, thus a contradiction. \pfk % If $s'$ is a divisor of $2$ in the ring $\Zb$, the same is true for $s$. Therefore % the norm $N(s')=ss'\in\Z$ can take only values $\pm1$, $\pm2$, or $\pm4$. % It can be shown (see the Appendix) that all numbers in $\Zb$ with the norm % equal to $\pm1$, i.e.~divisors of unity, are of the form $\pm\beta^k$, for some % integer $k$. \begin{prop}\label{vylouc} Let $\beta$ be a quadratic unitary Pisot number. Assume that ${\it Cl}_{s'}\{0,1\}=\Zb\cap[0,1]$ is satisfied for some $s'$. Then either of the possibilities below is true: \begin{itemize} \item $\beta=\tau$ (root of $x^2=x+1$) and $s'=-\tau'$ or $s'=1+\tau'$. \item $\beta=1+\sqrt2$ (root of $x^2=2x+1$) and $s'=-\beta'$ or $s'=1+\beta'$. \item $\beta=2+\sqrt3$ (root of $x^2=4x-1$) and $s'=\beta'$ or $s'=1-\beta'$. \end{itemize} \end{prop} \pf From the relation~\eqref{eq:as} it is clear that all elements of ${\it Cl}_{s'}\{0,1\}$ are polynomials in $s'$. Since $s'\in\Zb$, it satisfies a quadratic equation with integer coefficients, namely $x^2=(s+s')x-ss'$, i.e. $x^2={\rm tr(s)}x-N(s)$. Using this quadratic equation any polynomial $y$ from~\eqref{eq:as} may be reduced to the form $y=a+bs'$. Therefore clearly ${\it Cl}_{s'}\{0,1\} \subset\Z[s']$. The condition ${\it Cl}_{s'}\{0,1\}=\Zb\cap[0,1]$ implies that we need $\Zb=\Z[s']$. This can be satisfied only if $s'=\pm\beta+k$ for some integer $k$. The restriction $01$ as an infinite sequence $(x_i)_{k\geq i>-\infty}$ given by the `greedy' algorithm in the following way: $$ x_k:=\left[\frac{x}{\beta^k}\right]\,, $$ where $k$ satisfies $\beta^k \leq x < \beta^{k+1}$. Denote $r_k=x/\beta^k - x_k$. Numbers $x_{i-1}$ and $r_{i-1}$ are computed from $x_i$ and $r_i$ by prescription: $$ x_{i-1}:= [\beta r_i]\,, \qquad r_{i-1} := \beta r_i - x_{i-1}\,. $$ Clearly $x_i\in\{0,1,\dots,[\beta]\}$ for each $i$. In~\cite{renyi} it is proven that $$ x=\sum_{i=-\infty}^k x_i\beta^i\,, $$ i.e.~the sum converges for each positive real $x$ and for each $\beta>1$. Parry in~\cite{parry} answered the question for which sequences $(x_i)_{k\geq i>-\infty}$ there exists a positive real $x$, such that $(x_i)_{k\geq i>-\infty}$ is its $\beta$-expansion. Let the R\'enyi $\beta$-representation of 1 be $$ 1=\frac{a_1}\beta + \frac{a_2}{\beta^2} + \frac{a_3}{\beta^3} + \cdots\,, \qquad a_i\in\{0,1,\dots,[\beta]\}\,, $$ and let $$ x=\sum_{i=-\infty}^k x_i\beta^i\,, \qquad x_i\in\{0,1,\dots,[\beta]\}\,. $$ Then the sequence $(x_i)_{k\geq i>-\infty}$ is a $\beta$-expansion of $x$ if and only if for any integer $j\leq k$, the sequence $x_jx_{j-1}x_{j-2}\dots$ is lexicographically strictly smaller than sequence $a_1a_2a_3\dots$. Let us apply the rule on $\beta$-expansions for quadratic unitary Pisot numbers $\beta$. At first let us discuss the case $\beta^2=m\beta+1$, i.e.~$[\beta]=m$. The R\'enyi representation of 1 is $1=m/\beta + 1/\beta^2$, so that $a_1=m$, $a_2=1$, $a_3=0$, $\dots$. It means that $x=\sum x_i\beta^i$ is a $\beta$-expansion iff any $x_i=m$ occuring in the sequence $(x_i)_{k\geq i>-\infty}$ is followed by $x_{i-1}=0$. Let now $\beta^2=m\beta-1$, so that $[\beta]=m-1$. The R\'enyi representation of 1 in this case is $$ 1=(m-1)/\beta + (m-2)\sum_{k=2}^{\infty} \frac1{\beta^k}\,. $$ Therefore $(x_i)_{k\geq i>-\infty}$ is a $\beta$-expansion of some $x>0$ iff $x_ix_{i-1}x_{i-2}\dots$ is strictly lexicographically smaller than $(m-1)(m-2)(m-2)\dots$ for any $i\leq k$. Let $x\in\R$. If $x$ is negative we put $x_i =-|x|_i$, negatives of $\beta$-expansion coefficients for $|x|$. If the $\beta$-expansion of an $x\in\R$ ends in infinitely many zeros, it is said to be finite, and the zeros at the end are omitted. Denote the set $$ {\rm Fin}(\beta) := \{ \varepsilon x \mid x\in\R^+_0\,,\, \varepsilon=\pm1\,,\, x \text{ has a finite $\beta$-expansion}\} \,. $$ In the sequel, it will be useful to use the relation between the set ${\rm Fin}(\beta)$ of all numbers with finite $\beta$-expansion and the ring $\Zb$. In~\cite{froug} it is proved that for $\beta$, which is a root of the equation $x^2=mx+1$, the set ${\rm Fin}(\beta)$ is a ring. It follows immediately that \begin{equation}\label{eq:burdia2} \Zb = {\rm Fin}(\beta)\,. \end{equation} For $\beta$ being a solution of $x^2=mx-1$, the situation is different. In this case, ${\rm Fin}(\beta)$ is not closed under addition, (for example $\beta-1\not\in {\rm Fin}(\beta)$), while $\Zb$ is. Burd\'\i k et al. in~\cite{burda} prove that for such $\beta$, the set ${\rm Fin}(\beta)$ can be characterized as the set of those elements $x$ of $\Zb$, for which $N(x)=xx'$ is non negative. Therefore \begin{equation}\label{eq:burdia} \Zb \supsetneqq {\rm Fin}(\beta)=\{x\in\Zb \mid xx'\geq0\}\,. \end{equation} %------------- The crucial proposition for the answer to Question~\ref{q:2} is the following one. \begin{prop} \label{jajaj} Let $\beta$ be a quadratic unitary Pisot number, satisfying the equation $x^2=mx\pm1$. Let $\Lambda$ be a uniformly discrete subset of $\Zb$, containing $0,1$. Put $$ {\mathcal M}=\left\{ \frac{i}\beta \biggm| i=1,2,\dots,\left[\frac{m\pm1}2\right] \right\}\,. $$ If $\Lambda'$ is $s$-convex for every $s\in {\mathcal M}$, then there exists a bounded interval $\Omega$, such that $\Lambda'=\Omega\cap\Zb$. \end{prop} The proof of the above proposition is based on the following lemma. It is useful to introduce the notion of a closure under a set of operations $\vdash_s$. We denote by ${\it Cl}_{\mathcal M}A$ the smallest set containing $A$, which is $s$-convex for all $s$ in a finite set ${\mathcal M}$. \begin{lem} \label{eins} Let $\beta$ be a quadratic unitary Pisot number and ${\mathcal M}$ defined in Proposition~\ref{jajaj}. Then $$ {\it Cl}_{\mathcal M}\{0,1\}=[0,1]\cap\Zb $$ \end{lem} \pf Consider the operations $\vdash_s$ with $s=i/\beta$. For simplicity, denote $$ x\Vdash_i y := \frac{i}\beta x + \left(1-\frac{i}\beta\right)y\,,\qquad i=1,2,\dots $$ The statement of the lemma is equivalent to the fact that any point $\Zb$ in the interval $[0,1]$ can be generated from $0$, and $1$ using operations $\Vdash_i $, $i=1,2,\dots,[(m\pm1)/2]$. We first show that $[0,1]\cap\Zb$ can be generated using all operations $\Vdash_1,\dots,\Vdash_r$, with $r=[\beta]$. In the second step we find expression for $\Vdash_i$, $i=[(m\pm1)/2]+1,\dots,[\beta]$, as a composition of operations $\Vdash_i$, $i=1,\dots,[(m\pm1)/2]$. \begin{trivlist} \item{(a)} Consider $\beta$ to be the solution of the equation $x^2=mx+1$, $m\geq1$. According to~\eqref{eq:burdia2}, any $x\in\Zb\cap(0,1)$ has a finite $\beta$-expansion $$ x=\sum_{i=1}^k\frac{x_i}{\beta^i}\,, \qquad k\in\Z\,. $$ By induction on $k$ we show that $x$ belongs to ${\it Cl}_{\mathcal M}\{0,1\}$. Clearly, for the first step of induction we have \begin{equation} \label{eq:1step} \frac{j}\beta=1\Vdash_j 0\,, \end{equation} for any $j=1,\dots,m$. Suppose that the $\beta$-expansion of a point $x\in\Zb\cap(0,1)$ is of the form $$ x=\frac{j}\beta + \sum_{i=2}^k\frac{x_i}{\beta^i}\,, \qquad j=0,\dots,m-1\,. $$ Then $x$ can be rewritten as the combination $$ x=\frac{j}\beta + \frac{z}\beta\,, \qquad z= \sum_{i=1}^{k-1}\frac{x_{i+1}}{\beta^i}\,. $$ By induction hypothesis, $z$ could be generated from $0$ and $1$ using the given operations; i.e. $z\in{\it Cl}_{\mathcal M}\{0,1\}$. Since \begin{equation}\label{eq:afin} (x\Vdash_i y) + c = (x+c)\Vdash_i (y+c)\,,\quad\hbox{and }\quad \frac{x}\beta \Vdash_i \frac{y}\beta = \frac1\beta (x\Vdash_i y)\,, \qquad\hbox{for any } i=1,\dots,m\,, \end{equation} one has the following relation, $$ {\it Cl}_{\mathcal M}\left\{\frac{j}\beta , \frac{j+1}\beta\right\} = \frac{j}\beta + \frac1\beta {\it Cl}_{\mathcal M}\{0,1\}\,. $$ Therefore $$ x \,=\, \frac{j}\beta + \frac{z}\beta \;\in\; \frac{j}\beta + \frac1\beta {\it Cl}_{\mathcal M}\{0,1\} \,=\, {\it Cl}_{\mathcal M}\left\{\frac{j}\beta , \frac{j+1}\beta\right\} \subset {\it Cl}_{\mathcal M}\{0,1\}\,. $$ The later inclusion is valid due to the first step of induction (see~\eqref{eq:1step}). Assume that the coefficient $x_1$ of $x$ in its $\beta$-expansion is equal to $m$. Necessarily, from the properties of $\beta$-expansions, $x_2$ is equal to 0. In this case we use $$ x \,=\, \frac{m}\beta + \frac{z}{\beta^2} \;\in\; \frac{m}\beta + \frac1{\beta^2} {\it Cl}_{\mathcal M}\{0,1\} \,=\, {\it Cl}_{\mathcal M} \left\{ \frac{m}\beta , 1 \right\} \subset {\it Cl}_{\mathcal M}\{0,1\}\,. $$ Thus we have shown that all points in $[0,1]\cap\Zb$ can be generated starting from 0, and 1, using the operations $\Vdash_i$, $i=1,\dots,m$. The following relation can be used to reduce the number of necessary operations from $m$ to $[(m+1)/2]$. The relation is valid for any $k$, $k=1,\dots,m-1$, independently of $x,y$. $$ (x\Vdash_{k+1} y) \Vdash_1 (x\Vdash_k y) = \bigl(1-\frac{m-k}{\beta} \bigr) x + \frac{m-k}{\beta} y = y \Vdash_{m-k} x\,. $$ \item{(b)} Let now $\beta$ be the root of $x^2=mx-1$, $m\geq4$. In this case, if we want to use the $\beta$-expansions of numbers, we have to encounter a complication: There exist points in the ring $\Zb$ which do not have a finite $\beta$-expansion. Recall that an $x\in\Zb$ has a finite $\beta$-expansion if and only if $N(x)=xx'\geq0$, (see~\eqref{eq:burdia}). For us, only $x\in(0,1)$ are of interest. The following statement is valid. Let $x\in\Zb\cap(0,1)$. Then either $x$ or $(1-x)$ has a finite $\beta$-expansion. Indeed, since $N(x)=xx'$ is an integer and $x<1$, necessarily $|x'|>1$. Therefore $x'>0$ if and only if $1-x'<0$. Now we can proceed, similarly as for (a) by induction on the length of the $\beta$-expansion in order to show that any element of ${\rm Fin}(\beta)\cap[0,1]$ can be generated by corresponding operations $\Vdash_i$, $i=1,\dots,m-1$. For elements $x\in(0,1)\cap\bigl(\Zb\setminus{\rm Fin}(\beta)\bigr)$, we have $1-x\in(0,1)\cap{\rm Fin}(\beta)$. Therefore $1-x$ can be generated from 0 and 1 using the given operations. In the corresponding combination we replace all 0 by 1 and vice versa, which gives the desired combination for element $x$. The coefficients in a $\beta$-expansion take one of the values $0,\dots,m-1$. For an $x\in{\rm Fin}(\beta)\cap(0,1)$, whose $x_1=j$, $j=0,\dots,m-2$, the procedure is the same as in the case (a). Consider an $$ x=\frac{m-1}\beta + \frac{x_2}{\beta^2} + \sum_{i=3}^k \frac{x_i}{\beta^i} =1-\frac1\beta + \frac{x_2+1}{\beta^2} + \sum_{i=3}^k \frac{x_i}{\beta^i} =\frac1\beta z + \left(1-\frac1\beta\right)1=z\Vdash_1 1 \,. $$ Since the segment $x_1x_{2}\dots x_{k}$ of a $\beta$-expansion is strictly lexicographically smaller than the sequence $(m-1)(m-2)(m-2)\dots$, the string $x_2x_3\dots x_k$ is strictly smaller than $(m-2)(m-2)\dots$. Therefore $(x_2+1)x_3\dots x_k$ is the $\beta$-expansion of $z$. By induction hypothesis $z\in{\it Cl}_{\mathcal M}\{0,1\}$, therefore $x=z\Vdash_1 1\in{\it Cl}_{\mathcal M}\{0,1\}$. By that we have shown that any $x\in[0,1]\cap\Zb$ can be generated by given operations $\Vdash_i$, $i=1,\dots,m-1$. The number of operations can be reduced from $m-1$ to $[(m-1)/2]$ by the following relation, which is valid for any $k$, $k=1,\dots,m-2$, for all $x,y$. $$ (x\Vdash_k y) \Vdash_1 (x\Vdash_{k+1} y) = \left(1-\frac{m-k-1}{\beta} \right) x + \frac{m-k-1}{\beta} y = y \Vdash_{m-k-1} x\,. $$ \end{trivlist} \vskip-0.2in \pfk Due to~\eqref{eq:afin}, the Lemma~\ref{eins} can be generalized in the following way. If $c\in\Zb$, we have \begin{equation}\label{eq:skoky} {\it Cl}_{\mathcal M}\{c,c+1\} = [c,c+1]\cap\Zb\,. \end{equation} \pf[Proof of Proposition~\ref{jajaj}] In order to prove the statement of the proposition, we have to find $\Omega$, such that $\Lambda'=\Omega\cap\Zb$. We show that this property is satisfied by the convex hull of $\Lambda'$, so we put $\Omega=\langle \Lambda' \rangle$. We have to justify that any element of $\Zb\cap\langle\Lambda'\rangle$ can be generated using the given operations starting from points of $\Lambda'$. Consider $x\in\Zb\cap\langle\Lambda'\rangle$. Since $0,1\in\Lambda$, then also $0,1\in\Lambda'$, and hence $\Lambda'\supset([0,1]\cap\Zb)$, as a consequence of Lemma~\ref{eins}. Therefore, if $x\in[0,1]$, then $x$ belongs to $\Lambda'$. Without loss of generality let $x>1$. Observe that the points of $\Lambda'$ cover densely the interval $\langle\Lambda'\rangle$. Therefore one can find a finite sequence of intervals $[y_1-1,y_1],\dots,[y_k-1,y_k]$, such that the elements $y_i$ belong to $\Lambda'$, and it holds that $0