%~Mouliné par MaN_auto v.0.29.1 2023-11-07 19:07:35 \documentclass[CRMATH,Unicode,XML,published]{cedram} \TopicFR{Théorie des nombres} \TopicEN{Number theory} \makeatletter \def\@setafterauthor{% \vglue3mm% % \hspace*{0pt}% \begingroup\hsize=12cm\advance\hsize\abstractmarginL\raggedright \noindent %\hspace*{\abstractmarginL}\begin{minipage}[t]{10cm} \leftskip\abstractmarginL \normalfont\Small \@afterauthor\par \endgroup \vskip2pt plus 3pt minus 1pt } \makeatother \newtheorem{theol}{Theorem} \renewcommand{\thetheol}{\Alph{theol}} \graphicspath{{./figures/}} \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \title{On subsets of asymptotic bases} \author{\firstname{Ji-Zhen} \lastname{Xu}} \address{School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, People's Republic of China} \address{Nanjing Vocational College of Information Technology,Nanjing 210023, People's Republic of China} \email[J.-Z. Xu]{965165607@qq.com} \author{\firstname{Yong-Gao} \lastname{Chen}\IsCorresp} \address[1]{School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, People's Republic of China} \email[Y.-G. Chen]{ygchen@njnu.edu.cn} \thanks{This work is supported by the National Natural Science Foundation of China, Grant No. 12171243} \CDRGrant[National Natural Science Foundation of China]{12171243} \subjclass{11B13, 11B05, 11P99} \begin{abstract} Let $h\ge 2$ be an integer. In this paper, we prove that if $A$ is an asymptotic basis of order $h$ and $B$ is a nonempty subset of $A$, then either there exists a finite subset $F$ of $A$ such that $F\cup B$ is an asymptotic basis of order $h$, or for any $\varepsilon >0$, there exists a finite subset $F_\varepsilon $ of $A$ such that $d_L(h(F_\varepsilon\cup B))\ge hd_L(B)-\varepsilon $, where $d_L(X)$ denotes the lower asymptotic density of $X$ and $hX$ denotes the set of all $x_1+\cdots +x_h$ with $x_i\in X$ $(1\le i\le h)$. This generalizes a result of Nathanson and S\'ark\"ozy. \end{abstract} \dateposted{2024-02-02} \begin{document} \maketitle \section{Introduction} Let $\mathbb{N}_0$ denote the set of all nonnegative integers. Let $h\ge 2$ be an integer. For $A\subseteq \mathbb{N}_0$, let \[ hA=\{ a_1+\cdots +a_h : a_1,\dots, a_h\in A \}. \] We define \[ d_L (A)=\liminf_{x\to +\infty } \frac{A(x)}{x}, \] where $A(x)$ is the number of positive integers in $A$ which do not exceed $x$. Usually, $d_L(A)$ is called the \emph{lower asymptotic density} of $A$. If \[ \lim_{x\to +\infty } \frac{A(x)}{x} \] exists, then the limit value is called the \emph{asymptotic density} of $A$ and denote it by $d(A)$. A set $A$ is called an \emph{asymptotic basis of order $h$} if $hA$ contains all sufficiently large integers. An asymptotic basis $A$ of order $h$ is called \emph{minimal} if no proper subset of $A$ is an asymptotic basis of order $h$. The notation of minimal asymptotic bases was introduced by St\"ohr~\cite{Stohr1955} in 1955. In 1956, H\"artter~\cite{Hartter1956} proved that for each integer $h\ge 2$, there exist minimal asymptotic bases of order $h$. In 1988, Erd\H os and Nathanson~\cite{ErdosNathanson1988} constructed a minimal asymptotic basis $A$ with $d (A)=1/h$. For related research, one may refer to Chen and Chen~\cite{ChenChen2011}, Chen and Tang~\cite{ChenTang2018}, Ja\'nczak and Schoen~\cite{JanczakSchoen}, Nathanson~\cite{Nathanson1974, Nathanson1988}, Sun~\cite{Sun2021} and Tang and Lin~\cite{TangLin2022}. Nathanson and S\'ark\"ozy~\cite{NathansonSarkozy1989} proved the following results: \begin{theol}\label{thmA} If $A$ is an asymptotic basis of order $h$ and $B$ is a subset of $A$ with $d_L (B)>1/h$, then there exists a finite subset $F$ of $A$ such that $F\cup B$ is an asymptotic basis of order $h$. \end{theol} \begin{theol}\label{thmB} If $A$ is a minimal asymptotic basis of order $h$, then $d_L (A)\le 1/h$. \end{theol} In this paper, the following results are proved. \begin{theo}\label{thm1} Let $h\ge 2$ be an integer. If $A$ is an asymptotic basis of order $h$ and $B$ is a nonempty subset of $A$, then either there exists a finite subset $F$ of $A$ such that $F\cup B$ is an asymptotic basis of order $h$, or for any $\varepsilon >0$, there exists a finite subset $F_\varepsilon $ of $A$ such that $d_L(h(F_\varepsilon\cup B))\ge hd_L(B)-\varepsilon $. \end{theo} \begin{rema*} Theorem~A is a corollary of Theorem~\ref{thm1}. Let $A$ be an asymptotic basis of order $h$ and $B_1$ a subset of $A$ with $d_L (B_1)>1/h$. We take $\varepsilon = (hd_L(B_1) -1)/2$. Then $hd_L(B_1)-\varepsilon = (hd_L(B_1) +1)/2 >1\ge d_L(h(E\cup B_1))$ for any finite subset $E$ of $A$. By Theorem~\ref{thm1}, there exists a finite subset $F$ of $A$ such that $F\cup B_1$ is an asymptotic basis of order $h$. \end{rema*} \begin{coro}\label{cor1} Let $h\ge 2$ be an integer and let $B$ be a nonempty set of nonnegative integers. Then either there exists a finite set $F$ of nonnegative integers such that $F\cup B$ is an asymptotic basis of order $h$, or for any $\varepsilon >0$, there exists a finite set $F_\varepsilon $ of nonnegative integers such that $d_L(h(F_\varepsilon\cup B))\ge hd_L(B)-\varepsilon $. \end{coro} \begin{theo}\label{thm2} Let $h\ge 2$ be an integer. If $A$ is a minimal asymptotic basis of order $h$ and $B$ is a nonempty subset of $A$, then for any $\varepsilon >0$, there exists a finite subset $F_\varepsilon $ of $A$ such that $d_L(h(F_\varepsilon\cup B))\ge hd_L(B)-\varepsilon $. \end{theo} \begin{theo}\label{thm3} Let $h\ge 2$ be an integer. If $A$ is a set of nonnegative integers with $d_L(A)>0$, then there exists a subset $B$ of $A$ with $d_L(B)>0$ such that $F\cup B$ is not an asymptotic basis of order $h$ for any finite set $F$. \end{theo} \section{Proofs} We will use a well known result of Kneser. If two sets $X$ and $Y$ of nonnegative integers are coincide from some point on, then we write $X\sim Y$. For any set $X$ of nonnegative integers and any positive integer $g$, let $X^{(g)}$ be the set of all nonnegative integers $n$ with $n\equiv x\pmod g$ for some $x\in X $. In 1953, Kneser~\cite{Kneser1953} proved the following profound result. \begin{lemm}[Kneser~\cite{Kneser1953}] \label{lem1} Let $h\ge 2$ be an integer and $X$ a nonempty set of nonnegative integers. Then either $d_L(hX)\ge hd_L(X)$ or there exists a positive integer $g$ such that $hX\sim hX^{(g)}$ and \[ d_L(hX)\ge hd_L(X)-\frac{h-1}{g}. \] \end{lemm} \begin{proof}[Proof of Theorem~\ref{thm1}] If there exists a finite subset $F$ of $A$ such that $F\cup B$ is an asymptotic basis of order $h$, then we are done. Now we assume that for any finite subset $F$ of $A$, $F\cup B$ is not an asymptotic basis of order $h$. Let $\varepsilon >0$. For any positive integer $g$, let $A_g=\{ a_{g,1}, \dots, a_{g, s_g} \}$ be a subset of $A$ such that for every $a\in A$, there exists $1\le i\le s_g$ with $a\equiv a_{g,i}\pmod{g}$. It is clear that $A_g^{(g)}=A^{(g)}$. Let \[ F_\varepsilon =\bigcup_{1\le g< (h-1)/\varepsilon } A_g. \] Then $F_\varepsilon$ is finite. It is enough to prove that \[ d_L (h(F_\varepsilon \cup B))\ge hd_L (B)-\varepsilon. \] By Lemma~\ref{lem1}, either \[ d_L (h(F_\varepsilon \cup B))\ge hd_L (F_\varepsilon \cup B), \] or there exists a positive integer $g_1$ such that $h(F_\varepsilon \cup B)\sim h((F_\varepsilon \cup B)^{(g_1)})$ and \[ d_L(h(F_\varepsilon \cup B))\ge hd_L (F_\varepsilon \cup B)-\frac{h-1}{g_1}. \] Since $F_\varepsilon$ is finite, it follows that $d_L (F_\varepsilon \cup B)=d_L(B)$. Hence, either \[ d_L (h(F_\varepsilon \cup B))\ge hd_L (B), \] or there exists a positive integer $g_1$ such that $h(F_\varepsilon \cup B)\sim h((F_\varepsilon \cup B)^{(g_1)})$ and \[ d_L(h(F_\varepsilon \cup B))\ge hd_L (B)-\frac{h-1}{g_1}. \] If \[ d_L (h(F_\varepsilon \cup B))\ge hd_L (B), \] then we are done. Now we assume that there exists a positive integer $g_1$ such that $h(F_\varepsilon \cup B)\sim h((F_\varepsilon \cup B)^{(g_1)})$ and \[ d_L(h(F_\varepsilon \cup B))\ge hd_L (B)-\frac{h-1}{g_1}. \] If $(h-1)/g_1\le \varepsilon$, then we are done. Now we assume that $(h-1)/g_1> \varepsilon$. We will derive a contradiction. By $(h-1)/g_1> \varepsilon$, we have $g_1<(h-1)/\varepsilon$. Thus, \[ A\subseteq A^{(g_1)}=A_{g_1}^{(g_1)}\subseteq F_\varepsilon^{(g_1)} \subseteq (F_\varepsilon \cup B)^{(g_1)}. \] Hence \begin{equation}\label{eq1} hA\subseteq h((F_\varepsilon \cup B)^{(g_1)})\subseteq \mathbb{N}_0. \end{equation} Since $A$ is an asymptotic basis of order $h$, we have $hA\sim \mathbb{N}_0$. It follows from~\eqref{eq1} that $h((F_\varepsilon \cup B)^{(g_1)})\sim \mathbb{N}_0$. Noting that $h(F_\varepsilon \cup B)\sim h((F_\varepsilon \cup B)^{(g_1)})$, we have $h(F_\varepsilon \cup B)\sim \mathbb{N}_0$. This means that $F_\varepsilon \cup B$ is an asymptotic basis of order $h$, a contradiction. This completes the proof of Theorem~\ref{thm1}. \end{proof} \begin{proof}[Proof of Corollary~\ref{cor1}] Since $\mathbb{N}_0$ is an asymptotic basis of order $h$ and $B\subseteq \mathbb{N}_0$, Corollary~\ref{cor1} follows from Theorem~\ref{thm1} immediately. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm2}] Since $A$ is a minimal asymptotic basis of order $h$, it follows that $hA\sim \mathbb{N}_0$. So $d_L(hA)=1$. By Theorem~B, $d_L(B)\le d_L(A)\le 1/h$. Thus, $hd_L(B)\le 1$. Let $\varepsilon >0$. If $A\setminus B$ is finite, then for $F_\varepsilon =A\setminus B$, \[ d_L(h(F_\varepsilon\cup B))=d_L(hA)=1\ge hd_L(B)-\varepsilon. \] Now we assume that $A\setminus B$ is infinite. Thus, for any finite subset $F$ of $A$, we have $F\cup B\not= A$. Since $A$ is a minimal asymptotic basis of order $h$, it follows that for any finite subset $F$ of $A$, $F\cup B$ is not an asymptotic basis of order $h$. Now Theorem~\ref{thm2} follows from Theorem~\ref{thm1} immediately. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm3}] Let \[ B=\bigcup_{n=0}^\infty \left(\left((h+1)^{n^2+1}, (h+1)^{(n+1)^2} \right) \cap A \right). \] For a sufficiently large $x$, let $k$ be the integer with \[ (h+1)^{k^2}\le x< (h+1)^{(k+1)^2}. \] Let $t$ be the integer with \[ (h+1)^{t-1}<\frac {h(h+1)+d_L(A)}{(h+1)d_L(A)}\le (h+1)^t. \] It is clear that $t\ge 1$. If $x\le (h+1)^{k^2+t}$, then \begin{align*} \frac{B(x)}x&\ge \frac{B((h+1)^{k^2})}{x}\ge \frac 1{(h+1)^t} \frac{B((h+1)^{k^2})}{(h+1)^{k^2}}\\ &> \frac{d_L(A)}{h(h+1)+d_L(A)} \frac{B((h+1)^{k^2})}{(h+1)^{k^2}}. \end{align*} Since \begin{align*} B((h+1)^{k^2})&\geq A((h+1)^{k^2})-\sum_{n=0}^{k-1} \left((h+1)^{n^2+1}-(h+1)^{n^2}+1\right) \\ &\ge A((h+1)^{k^2})-k(h+1)^{(k-1)^2+1}, \end{align*} it follows that \[ \frac{B((h+1)^{k^2})}{(h+1)^{k^2}}\ge \frac{A((h+1)^{k^2})}{(h+1)^{k^2}} +o(1)\ge d_L(A)+o(1). \] Hence \[ \frac{B(x)}x\ge \frac{d_L(A)}{h(h+1)+d_L(A)} (d_L(A)+o(1)). \] If $x> (h+1)^{k^2+t}$, then \begin{align*} B(x)&\geq A(x)-\sum_{n=0}^k \left((h+1)^{n^2+1}-(h+1)^{n^2}+1\right) \\ &= A(x)-h\sum_{n=0}^k (h+1)^{n^2}-k-1\\ &\ge A(x)-h (h+1)^{k^2}-hk (h+1)^{(k-1)^2}-k-1\\ &> A(x)-\frac h{(h+1)^t}x-\frac{hk}{(h+1)^{2k-1+t}}x-k-1. \end{align*} It follows that \begin{align*} \frac{B(x)}x &\ge \frac{A(x)}x -\frac h{(h+1)^t} -o(1)\\ &\ge d_L(A)-\frac {h(h+1) d_L(A)}{h(h+1)+d_L(A)}-o(1)\\ &=\frac {d_L(A)^2}{h(h+1)+d_L(A)}-o(1). \end{align*} Combining the above arguments, we have \[ d_L(B)\ge \frac {d_L(A)^2}{h(h+1)+d_L(A)} >0. \] Let $F$ be a finite set of nonnegative integers. Then there exists a positive integer $m$ such that \[ F\subseteq [0, (h+1)^{m^2+1}]. \] For any integer $n>m$, by the definition of $B$, \[ [(h+1)^{n^2}, (h+1)^{n^2+1}]\cap (F\cup B)=[(h+1)^{n^2}, (h+1)^{n^2+1}]\cap B=\emptyset. \] Since \[ (h+1)^{n^2+1}>h(h+1)^{n^2}, \] it follows that $(h+1)^{n^2+1}\notin h(F\cup B)$. Therefore, $F\cup B$ is not an asymptotic basis of order $h$ for any finite set $F$. This completes the proof of Theorem~\ref{thm3}. \end{proof} \bibliographystyle{crplain} \bibliography{CRMATH_Xu_20230084} \end{document}