\section{\bf Introduction and notations}


In the field of additive combinatorics a popular topic is to compare the densities of different sets (of, say, positive integers). The well-known theorem of Kneser  gives a description of the sets $A$ having lower density $\alpha$ such that the density of $A+A:=\{a+b\,:\, a,b\in A\}$ is less than $2\alpha$ (see for instance \cite{RO}). The analogous question with the product set $A^2:=\{ab\,:\, a,b\in A\}$ is apparently more complicated.

For any set $A\subset \mathbb{N}$ of natural numbers, we define the lower asymptotic density $\underline{\d}A$, the upper asymptotic density
$\overline{\d}A$ and the asymptotic density $\d A$ in the natural way:
$$
\underline{\mathbf{d}}A= \liminf_{n\to\infty}\frac{|A\cap[1,n]|}{n},
\quad
\overline{\mathbf{d}}A= \limsup_{n\to\infty}\frac{|A\cap[1,n]|}{n}
$$
known as the lower and upper asymptotic densities.
If the two values coincide, then we denote by $\mathbf{d}A$ the common value and call it the \textit{asymptotic density} of $A$.

Throughout the paper $\mathbb{N}$ denotes the set of positive integers and $\mathbb{N}_0:=\mathbb{N}\cup \{0\}$. We will use the notion
$A(x)=\{n\in A: n\leq x\}$ for $A\subseteq \mathbb{N}$ and $x\in \mathbb{R}$. For functions $f,g:\mathbb{N}\to \mathbb{R}_+$ we write $f=O(g)$ (or $f\ll g$), if there exists some $c>0$ such that $f(n) \leq cg(n)$ for large enough $n$.

In Section 2 we investigate the connection between the (upper-, lower-, and asymptotic) density of a set of integers and the density of its sumset. In Section 3 we give a partial answer to a question of Erd\H os by giving a necessary condition for the existence of the asymptotic density of the set of subset sums of a given set of integers.
Finally, in Section 4 we consider analogous problems for product sets.


\section{\bf Density of sumsets}

 For subsets $A,B$ of an additive monoid $G$,
the sumset $A+B$ is defined to be the set of all
sums $a+b$ with $a\in A$, $b\in B$. For $G=(\mathbb{N}_0,+)$ the following clearly hold:
\begin{align*}
\underline{\d}A&\le\overline{\d}A,\\
\underline{\d}A&\le\underline{\d}(A+A),\\
\overline{\d}A&\le\overline{\d}(A+A).
\end{align*}



\smallskip
We shall assume that our sets $A$ are normalized in the sense that $A$ contains $0$
and $\gcd(A)=1$.

First observe that there exists a set of integers $A$ not having an asymptotic density
such that its sumset $A+A$ has a density:
for instance $A=\{0\}\cup\bigcup_{n\ge0}[2^{2n},2^{2n+1}]$ has lower density $1/3$, upper density $2/3$
and its sumset $A+A$ has density $1$, since it contains every nonnegative integer.
For this kind of sets $A$, we denote
respectively
\begin{align*}
\underline{\d}A&=:\alpha_A,\\
\overline{\d}A&=:\beta_A,\\
\d(A+A)&=:\gamma_A,\\
(\alpha_A,\beta_A,\gamma_A)&=:p_A,
\end{align*}
and we have
$$
\alpha_A \le \beta_A \le \gamma_A.
$$
The first question arising from this is to decide whether or not for any
$p=(\alpha,\beta,\gamma)$ such that $0\leq \alpha\le \beta \le \gamma\leq 1$ there exists
a set $A$ of integers such that $p=p_A$. This question has no positive answer in general, though the following
weaker statement holds.


\begin{prop}\label{p1}
Let $0\le \alpha\le 1$. There exists a normalized set $A\subset\mathbb{N}$ such that
$\d A=\alpha$ and $\d(A+A)=1$.
\end{prop}



\begin{proof}
 Let $0\in B$ be a thin additive basis, that is, a basis containing 0 and satisfying $|B(x)|=o(x)$ as $x\to\infty$. Now, let $A=B\cup\{\lfloor n/\alpha\rfloor,\ n\ge1\}$. Then $A$ is a normalized set satisfying
 $A+A=\mathbb{N}_0$ and $\d A=\alpha$.
 
 (Note that $B=\{0,1,2,\dots,\lfloor 1/\alpha \rfloor\}$ is also an appropriate choice for $B$.) 
\end{proof}


\begin{remark}
We shall mention that  Faisant et al \cite{Fa} proved the following related result:  for any $0\le  \alpha \le 1$ and any positive integer $k$, there exists a sequence $A$ such that
 $\d (jA)= j\alpha /k$, $j=1,\dots,k$.
\end{remark}


After a conjecture stated by Pichorides,  the related question about the characterisation of the two-dimensional
domains $\{(\underline{\d}B,\overline{\d}B)\,:\, B\subset A\}$ has been solved
(see \cite {GG} and \cite{FH}).


\smallskip
Note that if the density   $\gamma_A$ exists, then $\alpha_A$, $\beta_A$ and $\gamma_A$  have to satisfy some strong conditions.
For instance, by Kneser's theorem, we know that if for some set $A$ we have $\gamma_A<2\alpha_A$,
then $A+A$ is, except possibly a finite number of elements, a union of
arithmetic progressions in $\mathbb{N}$ with the same difference. This implies that
$\gamma_A$ must be a rational number. From the same theorem of Kneser, we
also deduce that if $\gamma_A<3\alpha_A/2$, then $A+A$ is an arithmetic progression
from some point onward. It means that $\gamma_A$ is a unit fraction, hence $A$ contains
any sufficiently large integer, if we assume that $A$ is normalized.


Another strong connection between $\alpha_A$ and $\gamma_A$ can be deduced from
Freiman's theorem on the addition of sets (cf. \cite{Fr}). Namely, every normalized set $A$ satisfies
$$
\gamma_A\ge\frac{\alpha_A}{2}+\min\Big(\alpha_A,\frac12\Big).
$$



A related but more surprising statement is the following:

\begin{prop}\label{p3}
There is a set of positive integers for which $\d(A)$ does exist and $\d(A+A)$ does not exist.
\end{prop}

\begin{proof}
Let us take $U=\{0,2,3\}$ and $V=\{0,1,2\}$,
then observe that
$$
U+(U\cup V)=\{0,1,2,3,4,5,6\}\quad
V+(U\cup V)=\{0,1,2,3,4,5\}.
$$
Let $(N_k)_{k\ge 0}$ be a sufficiently quickly increasing sequence of integers with $N_0=0$, $N_1=1$,
and define $A$ by
$$
A=(U\cup V)\cup\bigcup_{k\ge1}\Big((U+7\mathbb{Z})\cap[7N_{2k},7N_{2k+1}]\cup (V+7\mathbb{Z})\cap[7N_{2k+1},7N_{2k+2}]\Big).
$$
Then $A$ has density $3/7$. Moreover, for any $k\ge0$
$$
[7N_{2k},7N_{2k+1}]\subset A+A,
$$
thus $\overline{\d}(A+A)=1$, if we assume $\lim_{k\to\infty}N_{k+1}/N_k=\infty$.

 \noindent
We also have
$$
(A+A)\cap [14N_{2k-1},7N_{2k}]=(\{0,1,2,3,4,5\}+7\mathbb{N})\cap
[14N_{2k-1},7N_{2k}],
$$
hence $\underline{\d}(A+A) = 6/7$ using again the assumption that  $\lim_{k\to\infty}N_{k+1}/N_k=\infty$.
\end{proof}

\medskip


For any set $A$ having a density, let
\begin{align*}
{\d}A&=:\alpha_A,\\
\underline{\d}(A+A)&=:\underline{\gamma}_A,\\
\overline{\d}(A+A)&=:\overline{\gamma}_A,\\
(\alpha_A,\underline{\gamma}_A,\overline{\gamma}_A)&=:q_A,
\end{align*}
then we have
$$
\alpha_A\le \underline{\gamma}_A\le \overline{\gamma}_A.
$$
A question similar to the one   asked for $p_A$ can be stated as follows: given $q=(\alpha, \underline{\gamma}, \overline{\gamma})$
such that $0\le \alpha\le \underline{\gamma}\le \overline{\gamma}\leq 1$, does there exist a set $A$ such that $q=q_A$ ? 

\smallskip
We further mention an interesting question of Ruzsa: does there exist $0<\nu<1$ and a constant $c=c(\nu)>0$ such that for any set $A$ having a  density,
$$
\underline{\d}(A+A)\geq c\cdot (\overline{\d}(A+A))^{1-\nu} (\d A)^{\nu}\,?
$$
Ruzsa proved (unpublished) that in case of an affirmative answer, we  necessarily have $\nu\ge 1/2$.




\section{\bf Density of subset sums}


\medskip

Let $A=\{a_1 < a_2 < \cdots \}$ be a sequence of positive integers. Denote the set of all subset sums of $A$ by
$$
P(A):=\Big\{\sum_{i=1}^k\varepsilon_ia_i \,:\, k\ge0,\ \varepsilon_i\in \{0,1\}\ (1\le i\le k) \Big\}.
$$
Zannier conjectured and Ruzsa proved that the condition $a_n\leq 2a_{n-1}$ implies that the density $\mathbf{d}(P(A))$ exists (see \cite{r1}).
Ruzsa also asked the following questions:
\begin{itemize}
\item[i)] Is it true that for every pair of real numbers $0\leq \alpha\leq \beta\leq 1$, there exists a sequence of integers for which $\underline{\mathbf{d}}(P(A))=\alpha; \ \overline{\mathbf{d}}(P(A))=\beta$\,? This question was answered positively in \cite{r2}. 

\item[ii)] Is it true that the condition $a_n\leq a_1+a_2+\dots +a_{n-1}+c$ also implies that $\mathbf{d}(P(A))$ exists\,?

\end{itemize}

\noindent
We shall prove the following statement.

\begin{prop}\label{p33}
Let $(a_n)_{n=1}^\infty$ be a sequence of positive integers. Assume that for some function $\theta$ satisfying $\theta(k)\ll \frac{k}{(\log k)^2}$ we have $$|a_n-s_{n-1}|=\theta(s_{n-1})\text{ for every }n,$$
where $s_{n-1}:=a_1+a_2+\dots+a_{n-1}$.\\
\noindent
 Then $\mathbf{d}(P(A))$ exists.
\end{prop}

\begin{proof}
We first prove that there exists a real number $\delta$ such that
$$
|P(A)(s_n)|=\big(\delta+o(1)\big)s_n\quad \text{as $n \to \infty$.}
$$
Let $n\ge2$ be large enough. Then
$$
P(A)\cap[1,s_n]= \Big(P(A)\cap[1,s_{n-1}]\Big)\cup
\Big(P(A)\cap(s_{n-1},s_{n}-\theta(s_{n-1}))\Big).
$$
Since $a_n\ge s_{n-1}-\theta(s_{n-1})$, we have $P(A)\cap(s_{n-1},s_{n}]\supseteq a_n+P(A)\cap(\theta(s_{n-1}),s_{n-1}]$, thus
\begin{equation}\label{eq_low}
\Big|P(A)\cap[1,s_n]\Big|\ge 2\Big|P(A)\cap[1,s_{n-1}]\Big|-2\theta(s_{n-1})-1.
\end{equation}
On the other hand,
$$
P(A)\cap[1,s_n]\subseteq \left(P(A)\cap[1,s_{n-1}]\right)\cup \left(a_n+P(A)\cap[1,s_{n-1}]\right)\cup [s_n-\theta(s_n),s_n],
$$
since $a_{n+1}\geq s_n-\theta(s_n)$. Therefore,
\begin{equation}\label{eq_upp}
\Big|P(A)\cap[1,s_n]\Big|\le 2\Big|P(A)\cap[1,s_{n-1}]\Big|+\theta(s_{n})+1.
\end{equation}
Observe that $s_n=a_n+s_{n-1}\le 2s_{n-1}+\theta(s_{n-1})$,
hence letting
$$
\delta_n=\frac{\Big|P(A)\cap[1,s_n]\Big|}{s_n},
$$
we obtain from \eqref{eq_low} and \eqref{eq_upp} that
\begin{equation}\label{delta}
\delta_n-\delta_{n-1}=O\left(\frac{\theta(s_{n})}{s_n}\right).
\end{equation}
Now, we show that $s_n\gg 2^{n}$. Since 
\begin{equation}\label{thetabecs}
s_n=s_{n-1}+a_n\geq 2s_{n-1}-\theta(s_{n-1})=s_{n-1}\left( 2-\frac{\theta(s_{n-1})}{s_{n-1}} \right),\end{equation}
the condition $\theta(k)\ll \frac{k}{(\log k)^2}$ implies that from \eqref{thetabecs} we obtain that $s_n\gg 1.5^n$. Therefore, in fact, for large enough $n$ we have $s_n\geq s_{n-1} \left(2- \frac{c}{n^2} \right)$ with some $c>0$. Now, let $10c<K$ be a fixed integer. For $K<n$ we have 
$$s_n\geq s_K\prod\limits_{i=K+1}^{n} \left(2- \frac{c}{i^2} \right) \geq s_K \left[ 2^{n-k}-2^{n-k-1} \sum\limits_{i=K+1}^{n} \frac{c}{i^2}   \right]\gg 2^n,$$
since $\sum\limits_{i=K+1}^{n} \frac{c}{i^2}<1/10$.
Hence, $s_n\gg 2^{n}$ indeed holds.

Therefore, using the assumption on $\theta$ we obtain that $\frac{\theta(s_{n})}{s_n}\ll \frac{1}{n^2}$. So \eqref{delta} yields that
$$
\delta_n-\delta_{n-1}=O(n^{-2}).
$$
Therefore, the sequence $\delta_n$ has a limit which we denote by $\delta$.
Furthermore, observe that  
\begin{equation}\label{deltakonv}
\delta_n=\delta+O(1/n).
\end{equation}

The next step is to consider an arbitrary sufficiently large positive integer $x$ and decompose it as
$$
x=a_{n_1+1}+a_{n_2+1}+\cdots+a_{n_j+1}+z,
$$
where $n_1>n_2>\cdots>n_j>k$ and $0\le z$ are defined in the following way. (Here $k$ is a fixed, sufficiently large positive integer.) The index $n_1$ is chosen in such a way that $a_{n_1+1}\le x<a_{n_1+2}$. If $x-a_{n_1+1}\geq a_{n_1}$, then $n_2=n_1-1$, otherwise $n_2$ is the largest index for which $x-a_{n_1+1}\geq a_{n_2+1}$. The indices $n_3,n_4,\dots$ are defined similarly. We stop at the point when the next index would be at most $k$ and set $z:=x-a_{n_1+1}-a_{n_2+1}-\cdots-a_{n_j+1}$. As $z\leq \theta(s_{n_1+1})+s_k$, we have 
\begin{equation}\label{zkicsi}
z=o(x).
\end{equation}

Furthermore, let
$$
b_i=a_{n_1+1}+a_{n_2+1}+\cdots+a_{n_i+1},\quad i=0,1,\dots,j.
$$
(The empty sum is $b_0:=0$, as usual.)



Let $X_0:=(0,s_{n_1}-\theta(s_{n_1}))$ and for $1\leq i\leq j-1$ let $X_i:=(b_i+\theta(s_{n_i}),b_i+s_{n_{i+1}}-\theta(s_{n_{i+1}}))$ and consider 
\begin{multline*}
X:=X_0\cup X_1 \cup \dots \cup X_{j-1}=\\
=(0,s_{n_1}-\theta(s_{n_1}))\cup (b_1+\theta(s_{n_1}),b_1+s_{n_2}-\theta(s_{n_2}))\cup
\dots \cup (b_{j-1}+\theta(s_{n_{j-1}}),b_{j-1}+s_{n_j}-\theta(s_{n_j})).
\end{multline*}

Note that in this union each element  appears at most once, since according to the definition of $\theta$ the sets $X_0,X_1,\dots,X_{j-1}$ are pairwise disjoint as
$$b_i+s_{n_{i+1}}-\theta(  s_{n_{i+1}}   )\leq b_{i+1}=b_i+a_{n_{i+1}+1}$$
holds for every $0\leq i\leq j-2$.

The set of those elements of $[1,x]$ that are not covered by $X$ is:
\begin{multline*}
[1,x]\setminus X=[s_{n_1}-\theta(s_{n_1}),b_1+\theta(s_{n_1})]\cup[b_1+s_{n_2}-\theta(s_{n_2}),b_2+\theta(s_{n_2})]\cup\dots \\
\cup [b_{j-2}+s_{n_{j-1}}-\theta(s_{n_{j-1}}),b_{j-1}+\theta(s_{n_{j-1}})]\cup [b_{j-1}+s_{n_j}-\theta(s_{n_j}),x].
\end{multline*}

Therefore,
$$|[1,x]\setminus X|\leq 3\sum\limits_{i=1}^{j} \theta(s_{n_i})+z.$$

Using $\sum\limits_{i=1}^{j} \theta(s_{n_i})\ll \sum\limits_{i=1}^{j} \frac{s_{n_i}}{n_i^2}\ll\frac{x}{k^2}$ and \eqref{zkicsi}, we obtain that $\left|[1,x]\setminus X\right|\leq (\varepsilon_k+o(1))x$, where $\varepsilon_k\to 0$ (as $k\to \infty$). (Note that $\varepsilon_k\ll 1/k^2$.)

That is, the set $X$ covers $[1,x]$ with the exception of a ``small'' portion of size $O(x/k^2)$. Therefore, by letting $k\to \infty$ the density of the uncovered part tends to 0.




Let us consider $P(A)\cap X_i$. If a sum is contained in $P(A)\cap X_i$, then the sum of the elements with indices larger than $n_{i+1}$ is $b_i$. Otherwise, the sum is either at most $b_i+\theta(s_{n_i})$ or at least $b_{i}+s_{n_{i+1}}-\theta(s_{n_{i+1}})$.

Therefore, $P(A)\cap X_i=(b_i+P(\{a_1,a_2,\dots,a_{n_{i+1}}\}))\cap X_i$.


Hence,
$$\delta_{n_{i+1}}s_{n_{i+1}}-2\theta(s_{n_{i+1}})-1\leq |P(A)\cap X_i|\leq \delta_{n_{i+1}}s_{n_{i+1}}.$$


Therefore,  
\begin{multline}\label{PAalso}
|P(A)\cap[x]|\geq \sum\limits_{i=0}^{j-1} \left( \delta_{n_{i+1}}s_{n_{i+1}}-2\theta(s_{n_{i+1}})-1 \right) \geq \\
\geq \delta x -\delta z + \delta \sum\limits_{i=0}^{j-1} (s_{n_{i+1}}-a_{n_{i+1}+1})+\sum\limits_{i=0}^{j-1} (\delta_{n_{i+1}}-\delta)s_{n_{i+1}}-2\sum\limits_{i=0}^{j-1} \left(\theta(s_{n_{i+1}})+1 \right)
\end{multline}
and
\begin{multline}\label{PAfelso}
|P(A)\cap[x]|\leq \sum\limits_{i=0}^{j-1}  \delta_{n_{i+1}}s_{n_{i+1}} \leq \\
\leq \delta x -\delta z + \delta \sum\limits_{i=0}^{j-1} (s_{n_{i+1}}-a_{n_{i+1}+1})+\sum\limits_{i=0}^{j-1} (\delta_{n_{i+1}}-\delta)s_{n_{i+1}}
\end{multline}
Now, observe that
\begin{itemize}
    \item $|z|=o(x)$ by \eqref{zkicsi},
    \item $\sum\limits_{i=0}^{j-1} |s_{n_{i+1}}-a_{n_{i+1}+1}|=o(x)$ by using $|s_{n_{i+1}}-a_{n_{i+1}+1}|=\theta(s_{n_{i+1}})$ and $\sum\limits_{i=0}^{j-1} a_{n_{i+1}+1}\leq x$,
    \item $\sum\limits_{i=0}^{j-1} (\delta_{n_{i+1}}-\delta)s_{n_{i+1}}\ll x/k$ by using \eqref{deltakonv}. Letting $k\to \infty$ this term is also of size $o(x)$.
\end{itemize}
Hence, we obtain from \eqref{PAalso} and \eqref{PAfelso} that $|P(A)\cap[x]|=\delta x+o(x)$.
























\end{proof}




\section{\bf Density of product sets}



For any semigroup $G$
and any subset $A\subseteq G$,  we denote by $A^2$ the product set
$$
A^2=A\cdot A=\{ab\,:\, a,b\in A\}.
$$

In this section we focus on the case $G=(\mathbb{N},\cdot)$, the semigroup (for multiplication) of all positive integers. The restricted case $G=\mathbb{N}\setminus\{1\}$ is even more interesting, since $1\in A$ implies $A\subset A^2$.



The sets of integers satisfying the small doubling
hypothesis $\mathbf{d}(A+A)=\mathbf{d}A$ are well described
through Kneser's theorem.  The similar question for the product set does not plainly lead to a strong description. We can restrict our attention to sets $A$ such that
$\gcd(A)=1$, since by setting $B:=\frac{1}{\gcd{A}}A$ we have $\mathbf{d}A=\frac{1}{\gcd(A)}\mathbf{d}B$ and $\mathbf{d}A^2=\frac{1}{(\gcd(A))^2}\mathbf{d}(B^2)$.

\begin{exe}
i) Let $A_{\mathrm{nsf}}$ be the set of all non-squarefree integers. Letting $A=\{1\}\cup A_{\mathrm{nsf}}$ we have  $A^2=A$ and
$$
\mathbf{d}A=1-\zeta(2)^{-1}.
$$
ii) However, while $\gcd(A_{\mathrm{nsf}})=1$, we have
$$
\mathbf{d}A_{\mathrm{nsf}}^2<\mathbf{d}A_{\mathrm{nsf}}=1-\zeta(2)^{-1}.
$$
iii) Furthermore, the set $A_{\mathrm{sf}}$ of all squarefree integers satisfies
$$
\mathbf{d}A_{\mathrm{sf}}=\zeta(2)^{-1} \text{ and }
\mathbf{d}A_{\mathrm{sf}}^2=\zeta(3)^{-1},
$$
since $A_{\mathrm{sf}}^2$ consists of all cubefree integers.\\[0.5em]
iv) Given a positive integer $k$, the set $A_k=\big\{n\in\mathbb{N}\,:\,
\gcd(n,k)=1\big\}$ satisfies
$$
A_k^2=A_k\quad\text{and}\quad \mathbf{d}A_k=\frac{\phi(k)}k,
$$
where $\phi$ is Euler's totient function.
\end{exe}

We have the following result:

\begin{prop}
For any positive $\alpha<1$ there exists a set $A\subset \mathbb{N}$ such that
$\mathbf{d}A>\alpha$ and  $\mathbf{d}A^2<\alpha$.
\end{prop}

\begin{proof}
Assume first that $\alpha<1/2$.\\
For $k\ge 1$ let $A_k=k\mathbb{N}=\{kn,\ n\ge1\}$, then $A_k^2=k^2\mathbb{N}$. Therefore, $\mathbf{d} A_k=1/k$ and $\mathbf{d} (A_k^2)=1/k^2$.
If $1/(k+1) \le \alpha < 1/k$, then $A_k$ satisfies the requested condition. Since $\bigcup\limits_{k\geq 2}\left[ \frac{1}{k+1},\frac{1}{k} \right)=(0,1/2)$, an appropriate $k$ can be chosen for every $\alpha\in (0,1/2).$ \\[0.5em]
Assume now that $1>\alpha\ge1/2$.\\
Let $p_1<p_2<\cdots$ be the increasing sequence of prime numbers and
$$
B_r:=\bigcup_{i=1}^rp_i\mathbb{N}.
$$
The complement of the set $B_r$ contains exactly those positive integers that are not divisible by any of $p_1,p_2,\dots,p_r$, thus we have
$$\mathbf{d}(B_r)=1-\prod_{i=1}^r\left(1-\frac1{p_i}\right)=:\gamma_r.$$
Similarly, the complement of the set $B_r^2$ contains exactly those positive integers that are not divisible by any of $p_1,\dots,p_r$ or can be obtained by multiplying such a number by one of $p_1,\dots,p_r$. Hence, we obtain that
$$\mathbf{d}(B_r^2)=1-\left(1+\sum\limits_{i=1}^r \frac{1}{p_i}\right)\prod_{i=1}^r\left(1-\frac1{p_i}\right)=:\beta_r.$$
Note that 
\begin{equation}\label{intersect}
\beta_{r+1}=1-\left(1+\sum\limits_{i=1}^{r+1} \frac{1}{p_i}\right)\left(1-\frac{1}{p_{r+1}}\right)\prod_{i=1}^r\left(1-\frac1{p_i}\right)<1-\frac{3}{2}\cdot\frac{2}{3}\cdot\prod_{i=1}^r\left(1-\frac1{p_i}\right)=\gamma_r.
\end{equation}
As $(\beta_1,\gamma_1)=(1/4,1/2)$, moreover $(\beta_r)_{r=1}^\infty$ and $(\gamma_r)_{r=1}^\infty$ are increasing sequences satisfying \eqref{intersect} and $\lim \gamma_r =1$, we obtain that $[1/2,1)$ is covered by $\bigcup\limits_{r=1}^\infty (\beta_r,\gamma_r)$. That is, for every $\alpha\in[1/2,1)$ we have $\alpha\in (\beta_r,\gamma_r)$ for some $r$, and then $A=B_r$ is an appropriate choice.








\end{proof}

We pose two questions about the densities of $A$ and $A^2$.

\begin{quest}
 If $1\in A$ and $\mathbf{d}A=1$, then $\mathbf{d}(A^2)=1$, too.
Given two integers $k,\ell$, the set
 $$
 \{n\in\mathbb{N}\,:\, \gcd(n,k)=1\}\cup k\ell \mathbb{N}
 $$
 is multiplicatively stable.
What are the sets $A$ of positive integers such that $A^2=A$ or less
restrictively
$$
1\in A\text{ and }1>\mathbf{d}A^2=\mathbf{d}A>0\,?
$$
\end{quest}

\begin{quest}
It is clear that  $\mathbf{d}A>0$ implies  $\mathbf{d}A^2>0$, since
$A^2\supset (\min A)A$.

 For any $\alpha\in(0,1)$ we denote
 $$
 f(\alpha):=\inf_{A\subset\mathbb{N}; \ \mathbf{d}A=\alpha}\mathbf{d}A^2.
 $$
 Is it true that $f(\alpha)=0$ for any $\alpha$ or at least for $\alpha<\alpha_0$\,?
\end{quest}


The next result shows that the product set of a set having density 1 and satisfying a technical condition must also have density 1.

\begin{prop}\label{pp1}
Let $1\notin A$ be a set of positive integers with asymptotic density $\mathbf{d}A=1$.
Furthermore, assume that $A$ contains
an infinite subset of mutually coprime integers $a_1<a_2<\cdots$ such tha
$$
\sum_{i\ge1}\frac{1}{a_i}=\infty.
$$
Then the product set $A^2$ also has density $\mathbf{d}(A^2)=1$.
\end{prop}

\begin{proof}
Let $\varepsilon>0$ be arbitrary and choose a large enough $k$ such that
\begin{equation}\label{ep}
\sum_{i=1}^k \frac1a_i >-\log\varepsilon.
\end{equation}
Let $x$ be a large integer.
For any $i=1,\dots,k$, the set $A^2(x)$ contains all the products $a_ia$ with $a\le x/a_i$. We shall use a sieve argument. Let $A'$ be a finite subset of $A$ and  $X=[1,x]\cap\mathbb{N}$ for some $x>\max(A')$. For any $a\in A'$, let
$$
X_a=\Big\{n\le x\,:\, a\nmid n\text{ or }\frac na\not \in A\Big\}.
$$
Observe that
$$
X\setminus X_a=(aA)(x).
$$
Then
$$
(A'A)(x)=\bigcup_{a\in A'}\left(X\setminus X_a\right).
$$
By the inclusion-exclusion principle we obtain
$$
|(A'A)(x)|=\sum_{k=1}^{|A'|}(-1)^{j-1}\sum_{\substack{B\subseteq A'\\|B|=j}}\Big|\bigcap_{b\in B}\left(X\setminus X_b\right)\Big|,
$$
whence
\begin{equation}\label{szita}
\Big|\bigcap_{a\in A'}X_a\Big|= \sum_{j=0}^{|A'|}(-1)^{j}\sum_{\substack{B\subseteq A'\\|B|=j}}\Big|\bigcap_{b\in B}\left(X\setminus X_b\right)\Big|,
\end{equation}
where the empty intersection $\bigcap_{b\in \emptyset}\left(X\setminus X_b\right)$ denotes the full set $X$.

For any finite set of integers $B$ we denote by $\mathrm{lcm}(B)$ the least common multiple of the elements of $B$.
Now, we consider
$$
\bigcap_{b\in B}\left(X\setminus X_b\right)=\Big\{n\le x\,:\, \mathrm{lcm}(B)\mid n\text{ and }  \frac{n}{b}\in A \ (\forall b\in B) \Big\}.
$$
By the assumption $\mathbf{d}A=1$ we immediately get
$$
\Big|\bigcap_{b\in B}\left(X\setminus X_b\right)\Big|=\frac{x}{\mathrm{lcm}(B)}(1+o(1)).
$$
Plugging this into \eqref{szita}:
$$
\Big|\bigcap_{a\in A'(x)}X_a\Big|= x\sum_{k=0}^{|A'|}(-1)^{j}\sum_{\substack{B\subseteq A'\\|B|=j}}\frac1{\mathrm{lcm}(B)}+o(x).
$$

Since the elements of $A'=\{a_1,a_2,\dots,a_k\}$ are mutually coprime,
$$
x- |A'A(x)| = x\sum_{j=0}^k(-1)^j\sum_{1\le a_{i_1}<\cdots<a_{i_j}\le k}
\frac{1}{a_{i_1}a_{i_2}\dots a_{i_j}}+o(x)=x\prod_{i=1}^k\Big(1-\frac1{a_i}\Big)+o(x).
$$
(Note that for $j=0$ the empty product is defined to be 1, as usual.)
Since $1-u\le \exp(-u)$ we get
$$
x- |A'A(x)|  \le x\exp\Big(-\sum_{i=1}^k \frac1{a_i}\Big)+o(x)<\varepsilon x+o(x)
$$
by our assumption \eqref{ep}. Thus finally
$$
|A^2(x)|\ge |A'A(x)|>x(1-\varepsilon-o(1)).
$$
This ends the proof.
\end{proof}


\begin{remark}
Specially, the preceding result applies when $A$ contains a
sequence of prime numbers $p_1<p_2<\cdots$
such that
$\sum_{i\ge1}1/p_i=\infty$.
For this it is enough to assume that
$$
\liminf_{i\to\infty}\frac{i\log i}{p_i}>0.
$$
\end{remark}






However, we do not know how to avoid the assumption on the mutually coprime integers having infinite reciprocal sum. We thus pose the following question:

\begin{quest}
Is it true that $\mathbf{d}A=1$ implies $\mathbf{d}(A^2)=1$?
\end{quest}



\medskip


\subsection*{An example for a set $A$ such that $\mathbf{d}(A)=0$ and $\mathbf{d}(A^2)=1$.}

According to the fact that the multiplicative properties of the elements play an important role, one can build a set whose elements are characterized by their number of prime factors. Let
$$
A=\big\{n\in\mathbb{N}\,:\,\Omega(n)\le 0.75 \log\log n+1\big\},
$$
where $\Omega(n)$ denotes the number of prime factors (with multiplicity) of $n$.
An appropriate generalisation of the Hardy-Ramanujan theorem (cf. \cite{HR} and \cite{GT}) shows that the normal order of $\Omega(n)$ is $\log\log n$ and the Erd\H os-Kac theorem asserts that
$$
\mathbf{d}\left\{n\in\mathbb{N}\,:\, \alpha<\frac{\Omega(n)-\log\log n}{\sqrt{\log\log n}}<\beta\right\}=\frac1{\sqrt{2\pi}}\int_{\alpha}^{\beta}e^{-t^2/2}dt,
$$
which implies $\mathbf{d}A=0$. Now we prove that $\mathbf{d}A^2=1$. The principal feature in the definition of $A$ is that $A^2$  must contain almost all integers $n$ such that $\omega(n)\le 1.2 \log\log n$.

For $n\in\mathbb{N}$ let
$$
P_+(n):=\max\big\{p: p\text{ is a prime divisor of } n\big\}.
$$
Let us consider first the density of the integers $n$ such that
\begin{equation}\label{p+}
P_+(n) > n\exp(-(\log n)^{4/5}).
\end{equation}
Let $x$ be a large number and write
$$
\left|
\Big\{n\le x\,:\, P_+(n)\le n\exp(-(\log n)^{4/5})
\Big\}\right|=\left|
\Big\{n\le x\,:\, P_+(n)\le x\exp(-(\log x)^{4/5})
\Big\}\right|+o(x).
$$
By a theorem of Hildebrand (cf. \cite{Hi}) on the estimation  of $\Psi(x,z)$, the number of $z$-friable integers  up to
$x$, we conclude that the above cardinality is $x+o(x)$.
Hence, we may avoid the integers $n$ satisfying \eqref{p+}. By the same estimation we may also avoid those integers $n$ for which $P_+(n) < \exp((\log n)^{4/5})$.

Let $n$ be an integer such that $\Omega(n)\leq 1.2\log\log n$ and
$$
\exp((\log n)^{4/5})\le P_+(n)\le n\exp(-(\log n)^{4/5}).
$$
Our goal is to find a decomposition $n=n_1n_2$ with $\Omega(n_i)\le 0.75\log\log n_i+1$, $i=1,2$.

Let
$$
n=p_1p_2\dots p_{t-1}P_+(n),
$$
where $t=\Omega(n)$. We also assume that $p_1\le p_2\le \cdots\le p_{t-1}\le P_+(n)$. Let $m=\frac{n}{P_+(n)}$. Then
$$
\exp((\log n)^{4/5})\le m\le n\exp(-(\log n)^{4/5}).
$$
Let
$$
n_1=p_1p_2\dots p_{u-1}P_+(n)\text{ and }
n_2=p_u\dots p_{t-1},
$$
where $u=\lfloor(t-1)/2\rfloor$. Then
$n_2\ge \sqrt{m}$, which yields
$$
\log\log n_2\ge \log\log m -\log 2\ge 0.8\log\log n-\log 2.
$$
On the other hand,
$$
\Omega(n_2)=t-u\le \frac t2+1 \le 0.6 \log\log n+1\le 0.75 \log\log n_2+\frac{3\log 2}{4}.
$$
Now $n_1\ge  P_+(n)\ge \exp((\log n)^{4/5})$, hence
$$
\log\log n_1\ge 0.8\log\log n
$$
and
$$
\Omega(n_1)\le \frac {t-1}{2}\le 0.6 \log\log n\le 0.75 \log\log n_1$$
Therefore, the following statement is obtained:

\begin{prop}\label{pp3}
The set
$$
A=\big\{n\in\mathbb{N}\,:\,\Omega(n)\le 0.75 \log\log n+1\big\}
$$
has  density $0$ and its product set $A^2$ has density $1$.
\end{prop}

By a different approach we may extend the above result as follows.

\begin{thm}
For every $0\leq \alpha\leq \beta\leq 1$ there exists a set $A\subseteq \mathbb{N}$ such that $\mathbf{d}A=0$, $\underline{\d}(A\cdot A)=\alpha$ and $\overline{\d}(A\cdot A)=\beta$.
\end{thm}

\begin{proof}
We start with defining a set $Q$ such that $\d(Q)=0$ and $ \d(Q\cdot Q)=\beta$.
Let us choose a subset $P_0$ of the primes such that $\prod\limits_{p\in P_0}(1-1/p)=\beta$. Such a subset can be chosen, since $\sum 1/p=\infty$.
Now, let $p_k$ denote the $k$-th prime and let
$$
P_1=\{p_i: i\text{ is odd}\}\setminus P_0,
$$
$$
P_2=\{p_i: i\text{ is even}\}\setminus P_0.
$$
Furthermore, let
 $$
 Q_1=\{n: \text{all prime divisors of $n$ belong to $P_1$}\}
 $$
 and
 $$
 Q_2=\{n: \text{all prime divisors of $n$ belong to $P_2$}\}.
 $$
 Let $Q=Q_1\cup Q_2$. Clearly, $Q\cdot Q=Q_1\cdot Q_2$ contains exactly those numbers that do not have any prime factor in $P_0$, so $\d(Q\cdot Q)=\beta$. For $i\in\{1,2\}$ and $x\in\mathbb{R}$ the probability that an integer does not have any prime factor being less than $x$ from $P_i$ is $\prod\limits_{p< x, p\in P_i}(1-1/p)\leq \frac{1}{\beta}\prod\limits_{p< x, p\in P_i\cup P_0}(1-1/p)\leq \frac{1}{\beta}\exp\left\{-\sum\limits_{\substack{j:\ p_j< x, \\ j\equiv i\pmod{2}}}\frac{1}{p_j}\right\}=O\left( \frac{1}{\beta\sqrt{\log x}}\right)$. Therefore, $\d(Q_1)=\d(Q_2)=0$, and consequently $\d(Q)=0$ also holds.
  If $\alpha=\beta$, then $A=Q$ satisfies the conditions. From now on let us assume that $\alpha<\beta$.


Our aim is to define a subset $A\subseteq Q$ in such a way that $\underline{\d}(A\cdot A)=\alpha$ and $ \overline{\d}(A\cdot A)=\beta$. As $A\subseteq Q$ we will have  $\d(A)=0$ and $\overline{\d}(A\cdot A)\leq \beta$. The set $A$ is defined recursively. We  will define an increasing sequence of integers $(n_j)_{j=1}^\infty$ and sets $A_j$ ($j\in\mathbb{N}$)  satisfying the following conditions (and further conditions to be specified later):
\begin{itemize}
\item[(i)] $A_j\subseteq A_{j-1}$,
\item[(ii)] $A_j\cap [1,n_{j-1}]=A_{j-1}\cap [1,n_{j-1}]$,
\item [(iii)]$A_j\cap [n_j+1,\infty]=Q\cap [n_j+1,\infty]$.
\end{itemize}
That is, $A_j$ is obtained from $A_{j-1}$ by dropping out some elements of $A_{j-1}$ in the range $[n_{j-1}+1,n_j]$.
Finally, we  set $A=\bigcap\limits_{j=1}^\infty A_j$.

Let $n_1=1$ and  $A_1=Q$.
We  define the sets $A_j$ in such a way that the following condition holds for every $j$ with some $n_0$ depending only on $Q$:
$$
|(A_j\cdot A_j)(n)|\geq \alpha n \text{ for every $n\geq n_0$.}\leqno(*)
$$

Since $d(Q\cdot Q)=\beta>\alpha$, a threshold $n_0$ can be chosen in such a way that $(*)$ holds for $A_1=Q$ with this choice of $n_0$.
Now, assume that $n_j$ and $A_j$ are already defined for some $j$.
We continue in the following way depending on the parity of $j$:
\begin{itemize}
\item[Case I:] $j$ is odd.



Let $n_j<s$ be the smallest integer such that
$$|(A_j\setminus [n_j+1,s])\cdot (A_j\setminus [n_j+1,s])(n)|<\alpha n$$
for some $n\geq n_0$. We claim that such an $s$ exists, indeed it is at most $\lfloor n_j^2/\alpha \rfloor +1$.
For $s'=\lfloor n_j^2/\alpha \rfloor +1$ we have
$$|(A_j\setminus [n_j+1,s'])\cdot (A_j\setminus [n_j+1,s'])(s')|\leq n_j^2<\alpha s'.$$
Hence, $s$ is well-defined (and $s\leq s'$). Let $n_{j+1}:=s-1$ and $A_{j+1}:=A_j\setminus [n_j+1,s-1]$. (Specially, it can happen that $n_{j+1}=n_j$ and $A_{j+1}=A_j$.) Note that $A_{j+1}$ satisfies $(*)$.\\



\item[Case II:] $j$ is even.

Now, let $n_j< s$ be the smallest index for which $|(A_{j}\cdot A_{j})(s)|>(\beta-1/j)s$.

We have $\d(Q\cdot Q)=\beta$ and $A_{j}$ is obtained from $Q$ by deleting finitely many elements of it: $A_{j}=Q\setminus R$, where $R\subseteq [n_j]$. As $\d(Q)=0$, we have that
$$|((Q\cdot Q)\setminus (Q\setminus R)\cdot (Q\setminus R))(n)|\leq |R|^2+\sum\limits_{r\in  R} |Q(n/r)|=o(n),$$
 therefore, $\d(A_{j}\cdot A_{j})=\beta$. So for some $n>n_j$ we have that $(A_{j}\cdot A_{j})(n)>(\beta-1/j)n$, that is, $s$ is well-defined. Let $n_{j+1}:=s$ and $A_{j+1}=A_j$. Clearly, $A_{j+1}$ satisfies $(*)$.

\end{itemize}

This way an increasing sequence $(n_j)_{j=1}^\infty$ and sets $A_j (j\in \mathbb N)$ are defined, these satisfy conditions (i)-(iii). Finally, let us set $A:=\bigcap\limits_{j=1}^\infty A_j$. Note that $A(n_j)=A_j(n_j)$.

We have already seen that $A\subseteq Q$ implies that  $\d(A)=0$ and $\overline{\d}(A\cdot A)\leq \beta$. At  first we show that $\underline{\d}(A\cdot A)\geq \alpha$. Let $n\geq n_0$ be arbitrary. If $j$ is large enough, then $n_j>n$. As $A_j$ satisfies $(*)$ and $(A\cdot A)(n)=(A_j\cdot A_j)(n)$ we obtain that
$$|(A\cdot A)(n)|=|(A_j\cdot A_j)(n)|\geq \alpha n.$$
This holds for every $n\geq n_0$, therefore, $\underline{\d}(A\cdot A)\geq \alpha$.

As a next step, we show that $\underline{\d}(A\cdot A)=\alpha$. Let $j$ be odd. According to the definition of $n_{j+1}$ and $A_{j+1}$ there exists some $n\geq n_0$ such that
$$|((A_j\setminus \{n_{j+1}+1\})\cdot (A_j\setminus \{n_{j+1}+1\}))(n)|<\alpha n. $$
For brevity, let $s:=n_{j+1}+1$. As $A\subseteq A_j$ we get that $|(A\setminus \{s\})\cdot (A\setminus \{s\})(n)|<\alpha n$. Also, $$|(A\cdot A)\setminus ((A\setminus\{s\})\cdot (A\setminus\{s\})(n))|\leq 1+|A(n/s)| \leq 1+|Q(n/s)|,$$
since $A\subseteq Q$.  Thus $|(A\cdot A)(n)|\leq \alpha n+1+|Q(n/s)|\leq n(\alpha+1/n+1/s)$.
Clearly $s=n_{j+1}+1\leq n$, and as $j\to \infty$ we have $n_{j+1}\to \infty$, therefore $\underline{\d}(A\cdot A)=\alpha$.

Finally, we prove that $\overline{\d}(A\cdot A)=\beta$. Let $j$ be even. According to the definition of $A_{j+1}$ and $n_{j+1}$, we have $|(A_{j+1}\cdot A_{j+1})(n_{j+1})|>(\beta-1/j)n_{j+1}$. However, $(A\cdot A)(n_{j+1})=(A_{j+1}\cdot A_{j+1})(n_{j+1})$, therefore $\overline{\d}(A\cdot A)\geq \lim (\beta-1/j)=\beta$, thus $\overline{\d}(A\cdot A)=\beta$ as it was claimed.



\end{proof}




