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\TopicFR{Théorie des nombres}
\TopicEN{Number theory}

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\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}

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