On the density of sumsets and product sets

TL;DR

This paper explores the relationships between the density of integer sets and the densities of their sumsets, product sets, and subset sums, revealing new insights into their combinatorial structure.

Contribution

It introduces novel links between the densities of sets and their sumsets, product sets, and subset sums, advancing understanding in additive and multiplicative combinatorics.

Findings

Established bounds relating set density to sumset density

Identified conditions under which product sets have increased density

Connected the density of subset sums to original set properties

Abstract

In this paper some links between the density of a set of integers and the density of its sumset, product set and set of subset sums are presented.

Equations

$$\underline{\mathbf{d}}A=\underset{n\to \infty }{\mathrm{liminf}}\frac{|A\cap [1,n]|}{n},\overline{\mathbf{d}}A=\underset{n\to \infty }{\mathrm{limsup}}\frac{|A\cap [1,n]|}{n}$$

$$\underline{\mathbf{d}}A$$

$$\overline{\mathbf{d}}A$$

$$\underline{\mathbf{d}}A$$

$$\overline{\mathbf{d}}A$$

$$\mathbf{d}(A+A)$$

$$({\alpha }_A,{\beta }_A,{\gamma }_A)$$

$${\alpha }_A\le {\beta }_A\le {\gamma }_A.$$

$$\gamma_{A}\geq\frac{\alpha_{A}}{2}+\min\Big{(}\alpha_{A},\frac{1}{2}\Big{)}.$$

$$U+(U\cup V)=\{0,1,2,3,4,5,6\}V+(U\cup V)=\{0,1,2,3,4,5\}.$$

$$A=(U\cup V)\cup\bigcup_{k\geq 1}\Big{(}(U+7\mathbb{Z})\cap[7N_{2k},7N_{2k+1}]\cup(V+7\mathbb{Z})\cap[7N_{2k+1},7N_{2k+2}]\Big{)}.$$

$$[7N_{2k},7N_{2k+1}]\subset A+A,$$

$$(A+A)\cap [14N_{2k-1},7N_{2k}]=(\{0,1,2,3,4,5\}+7\mathbb{N})\cap [14N_{2k-1},7N_{2k}],$$

$$\mathbf{d}A$$

$$\underline{\mathbf{d}}(A+A)$$

$$\overline{\mathbf{d}}(A+A)$$

$$({\alpha }_A,{\underline{\gamma }}_A,{\overline{\gamma }}_A)$$

$${\alpha }_A\le {\underline{\gamma }}_A\le {\overline{\gamma }}_A.$$

$$\underline{\mathbf{d}}(A+A)\ge c\cdot (\overline{\mathbf{d}}(A+A){)}^{1-\nu }(\mathbf{d}A{)}^{\nu }?$$

$$P(A):=\Big{\{}\sum_{i=1}^{k}\varepsilon_{i}a_{i}\,:\,k\geq 0,\ \varepsilon_{i}\in\{0,1\}\ (1\leq i\leq k)\Big{\}}.$$

$$|a_n-s_{n-1}|=\theta (s_{n-1})\text{ for every }n,$$

$$|P(A)(s_{n})|=\big{(}\delta+o(1)\big{)}s_{n}\quad\text{as $n\to\infty$.}$$

$$P(A)\cap[1,s_{n}]=\Big{(}P(A)\cap[1,s_{n-1}]\Big{)}\cup\Big{(}P(A)\cap(s_{n-1},s_{n}-\theta(s_{n-1}))\Big{)}.$$

$$\Big{|}P(A)\cap[1,s_{n}]\Big{|}\geq 2\Big{|}P(A)\cap[1,s_{n-1}]\Big{|}-2\theta(s_{n-1})-1.$$

$$P(A)\cap [1,s_n]\subseteq \left(P(A)\cap [1,s_{n-1}]\right)\cup \left(a_n+P(A)\cap [1,s_{n-1}]\right)\cup [s_n-\theta (s_n),s_n],$$

$$\Big{|}P(A)\cap[1,s_{n}]\Big{|}\leq 2\Big{|}P(A)\cap[1,s_{n-1}]\Big{|}+\theta(s_{n})+1.$$

$$\delta_{n}=\frac{\Big{|}P(A)\cap[1,s_{n}]\Big{|}}{s_{n}},$$

$${\delta }_n-{\delta }_{n-1}=O\left(\frac{\theta (s_n)}{s_n}\right).$$

$$s_n=s_{n-1}+a_n\ge 2s_{n-1}-\theta (s_{n-1})=s_{n-1}\left(2-\frac{\theta (s_{n-1})}{s_{n-1}}\right),$$

$$s_n\ge s_K\prod _{i=K+1}^n\left(2-\frac{c}{i^2}\right)\ge s_K\left[2^{n-k}-2^{n-k-1}\sum _{i=K+1}^n\frac{c}{i^2}\right]\gg 2^n,$$

$${\delta }_n-{\delta }_{n-1}=O(n^{-2}).$$

$${\delta }_n=\delta +O(1/n).$$

$$x=a_{n_1+1}+a_{n_2+1}+\cdots +a_{n_j+1}+z,$$

Full text

1 Supported by the National Research, Development and Innovation Office of Hungary (Grant Nr. NKFIH (OTKA) PD115978 and NKFIH (OTKA) K124171) and the János Bolyai Research Scholarship of the Hungarian Academy of Sciences.

1. 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 [9]). 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 $\underline{\mathbf{d}}A$, the upper asymptotic density $\overline{\mathbf{d}}A$ and the asymptotic density $\mathbf{d}A$ in the natural way:

[TABLE]

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 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 Erdős 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.

2. 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:

[TABLE]

We shall assume that our sets $A$ are normalized in the sense that $A$ contains [math] 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 $A=\{0\}\cup\bigcup_{n\geq 0}[2^{2n},2^{2n+1}]$ 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

[TABLE]

and we have

[TABLE]

The first question arising from this is to decide whether or not for any $p=(\alpha,\beta,\gamma)$ such that $0\leq\alpha\leq\beta\leq\gamma\leq 1$ 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.

Proposition 2.1**.**

Let $0\leq\alpha\leq 1$. There exists a normalized set $A\subset\mathbb{N}$ such that $\mathbf{d}A=\alpha$ and $\mathbf{d}(A+A)=1$.

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 $A=B\cup\{\lfloor n/\alpha\rfloor,\ n\geq 1\}$. Then $A$ is a normalized set satisfying $A+A=\mathbb{N}_{0}$ and $\mathbf{d}A=\alpha$.

(Note that $B=\{0,1,2,\dots,\lfloor 1/\alpha\rfloor\}$ is also an appropriate choice for $B$.) ∎

Remark**.**

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

After a conjecture stated by Pichorides, the related question about the characterisation of the two-dimensional domains $\{(\underline{\mathbf{d}}B,\overline{\mathbf{d}}B)\,:\,B\subset A\}$ has been solved (see [3] and [6]).

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. [2]). Namely, every normalized set $A$ satisfies

[TABLE]

A related but more surprising statement is the following:

Proposition 2.2**.**

There is a set of positive integers for which $\mathbf{d}(A)$ does exist and $\mathbf{d}(A+A)$ does not exist.

Proof.

Let us take $U=\{0,2,3\}$ and $V=\{0,1,2\}$, then observe that

[TABLE]

Let $(N_{k})_{k\geq 0}$ be a sufficiently quickly increasing sequence of integers with $N_{0}=0$, $N_{1}=1$, and define $A$ by

[TABLE]

Then $A$ has density $3/7$. Moreover, for any $k\geq 0$

[TABLE]

thus $\overline{\mathbf{d}}(A+A)=1$, if we assume $\lim_{k\to\infty}N_{k+1}/N_{k}=\infty$.

We also have

[TABLE]

hence $\underline{\mathbf{d}}(A+A)=6/7$ using again the assumption that $\lim_{k\to\infty}N_{k+1}/N_{k}=\infty$. ∎

For any set $A$ having a density, let

[TABLE]

then we have

[TABLE]

A question similar to the one asked for $p_{A}$ can be stated as follows: given $q=(\alpha,\underline{\gamma},\overline{\gamma})$ such that $0\leq\alpha\leq\underline{\gamma}\leq\overline{\gamma}\leq 1$, does there exist a set $A$ such that $q=q_{A}$ ?

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,

[TABLE]

Ruzsa proved (unpublished) that in case of an affirmative answer, we necessarily have $\nu\geq 1/2$.

3. Density of subset sums

Let $A=\{a_{1}<a_{2}<\cdots\}$ be a sequence of positive integers. Denote the set of all subset sums of $A$ by

[TABLE]

Zannier conjectured and Ruzsa proved that the condition $a_{n}\leq 2a_{n-1}$ implies that the density $\mathbf{d}(P(A))$ exists (see [8]). Ruzsa also asked the following questions:

• 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 [5].

• 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 ?

We shall prove the following statement.

Proposition 3.1**.**

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

[TABLE]

*where $s_{n-1}:=a_{1}+a_{2}+\dots+a_{n-1}$.

Then $\mathbf{d}(P(A))$ exists.*

Proof.

We first prove that there exists a real number $\delta$ such that

[TABLE]

Let $n\geq 2$ be large enough. Then

[TABLE]

Since $a_{n}\geq s_{n-1}-\theta(s_{n-1})$, we have $P(A)\cap(s_{n-1},s_{n}]\supseteq a_{n}+P(A)\cap(\theta(s_{n-1}),s_{n-1}]$, thus

[TABLE]

On the other hand,

[TABLE]

since $a_{n+1}\geq s_{n}-\theta(s_{n})$. Therefore,

[TABLE]

Observe that $s_{n}=a_{n}+s_{n-1}\leq 2s_{n-1}+\theta(s_{n-1})$, hence letting

[TABLE]

we obtain from (1) and (2) that

[TABLE]

Now, we show that $s_{n}\gg 2^{n}$. Since

[TABLE]

the condition $\theta(k)\ll\frac{k}{(\log k)^{2}}$ implies that from (4) 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

[TABLE]

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 (3) yields that

[TABLE]

Therefore, the sequence $\delta_{n}$ has a limit which we denote by $\delta$. Furthermore, observe that

[TABLE]

The next step is to consider an arbitrary sufficiently large positive integer $x$ and decompose it as

[TABLE]

where $n_{1}>n_{2}>\cdots>n_{j}>k$ and $0\leq z$ 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 $a_{n_{1}+1}\leq x<a_{n_{1}+2}$. 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

[TABLE]

Furthermore, let

[TABLE]

(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

[TABLE]

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

[TABLE]

holds for every $0\leq i\leq j-2$.

The set of those elements of $[1,x]$ that are not covered by $X$ is:

[TABLE]

Therefore,

[TABLE]

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 (6), 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,

[TABLE]

Therefore,

[TABLE]

and

[TABLE]

Now, observe that

• •

$|z|=o(x)$ by (6),

• •

$\sum\limits_{i=0}^{j-1}|s_{n_{i+1}}-a_{n_{i+1}+1}|=o(x)$ by using $|s_{n_{i+1}}-a_{n_{i+1}+1}|=\theta(s_{n_{i+1}})$ and $\sum\limits_{i=0}^{j-1}a_{n_{i+1}+1}\leq x$,

• •

$\sum\limits_{i=0}^{j-1}(\delta_{n_{i+1}}-\delta)s_{n_{i+1}}\ll x/k$ by using (5). Letting $k\to\infty$ this term is also of size $o(x)$.

Hence, we obtain from (7) and (8) that $|P(A)\cap[x]|=\delta x+o(x)$.

∎

4. Density of product sets

For any semigroup $G$ and any subset $A\subseteq G$, we denote by $A^{2}$ the product set

[TABLE]

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 $\mathbf{d}A^{2}=\frac{1}{(\gcd(A))^{2}}\mathbf{d}(B^{2})$.

Examples**.**

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

[TABLE]

ii) However, while $\gcd(A_{\mathrm{nsf}})=1$, we have

[TABLE]

iii) Furthermore, the set $A_{\mathrm{sf}}$ of all squarefree integers satisfies

[TABLE]

since $A_{\mathrm{sf}}^{2}$ consists of all cubefree integers.

iv) Given a positive integer $k$, the set A_{k}=\big{\{}n\in\mathbb{N}\,:\,\gcd(n,k)=1\big{\}} satisfies

[TABLE]

where $\phi$ is Euler’s totient function.

We have the following result:

Proposition 4.1**.**

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

Proof.

Assume first that $\alpha<1/2$.

For $k\geq 1$ let $A_{k}=k\mathbb{N}=\{kn,\ n\geq 1\}$, 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 $1/(k+1)\leq\alpha<1/k$, 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 $\alpha\in(0,1/2).$

Assume now that $1>\alpha\geq 1/2$.

Let $p_{1}<p_{2}<\cdots$ be the increasing sequence of prime numbers and

[TABLE]

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

[TABLE]

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

[TABLE]

Note that

[TABLE]

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 (9) 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.

∎

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

Question**.**

If $1\in A$ and $\mathbf{d}A=1$, then $\mathbf{d}(A^{2})=1$, too. Given two integers $k,\ell$, the set

[TABLE]

is multiplicatively stable. What are the sets $A$ of positive integers such that $A^{2}=A$ or less restrictively

[TABLE]

Question**.**

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

[TABLE]

Is it true that $f(\alpha)=0$ for any $\alpha$ or at least for $\alpha<\alpha_{0}$ ?

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

Proposition 4.2**.**

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 that

[TABLE]

Then the product set $A^{2}$ also has density $\mathbf{d}(A^{2})=1$.

Proof.

Let $\varepsilon>0$ be arbitrary and choose a large enough $k$ such that

[TABLE]

Let $x$ be a large integer. For any $i=1,\dots,k$, the set $A^{2}(x)$ contains all the products $a_{i}a$ with $a\leq x/a_{i}$. We shall use a sieve argument. Let $A^{\prime}$ be a finite subset of $A$ and $X=[1,x]\cap\mathbb{N}$ for some $x>\max(A^{\prime})$. For any $a\in A^{\prime}$, let

[TABLE]

Observe that

[TABLE]

Then

[TABLE]

By the inclusion-exclusion principle we obtain

[TABLE]

whence

[TABLE]

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

[TABLE]

By the assumption $\mathbf{d}A=1$ we immediately get

[TABLE]

Plugging this into (11):

[TABLE]

Since the elements of $A^{\prime}=\{a_{1},a_{2},\dots,a_{k}\}$ are mutually coprime,

[TABLE]

(Note that for $j=0$ the empty product is defined to be 1, as usual.) Since $1-u\leq\exp(-u)$ we get

[TABLE]

by our assumption (10). Thus finally

[TABLE]

This ends the proof. ∎

Remark**.**

Specially, the preceding result applies when $A$ contains a sequence of prime numbers $p_{1}<p_{2}<\cdots$ such that $\sum_{i\geq 1}1/p_{i}=\infty$. For this it is enough to assume that

[TABLE]

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:

Question**.**

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

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

[TABLE]

where $\Omega(n)$ denotes the number of prime factors (with multiplicity) of $n$. An appropriate generalisation of the Hardy-Ramanujan theorem (cf. [4] and [10]) shows that the normal order of $\Omega(n)$ is $\log\log n$ and the Erdős-Kac theorem asserts that

[TABLE]

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 $\omega(n)\leq 1.2\log\log n$.

For $n\in\mathbb{N}$ let

[TABLE]

Let us consider first the density of the integers $n$ such that

[TABLE]

Let $x$ be a large number and write

[TABLE]

By a theorem of Hildebrand (cf. [7]) 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 (12). 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

[TABLE]

Our goal is to find a decomposition $n=n_{1}n_{2}$ with $\Omega(n_{i})\leq 0.75\log\log n_{i}+1$, $i=1,2$.

Let

[TABLE]

where $t=\Omega(n)$. We also assume that $p_{1}\leq p_{2}\leq\cdots\leq p_{t-1}\leq P_{+}(n)$. Let $m=\frac{n}{P_{+}(n)}$. Then

[TABLE]

Let

[TABLE]

where $u=\lfloor(t-1)/2\rfloor$. Then $n_{2}\geq\sqrt{m}$, which yields

[TABLE]

On the other hand,

[TABLE]

Now $n_{1}\geq P_{+}(n)\geq\exp((\log n)^{4/5})$, hence

[TABLE]

and

[TABLE]

Therefore, the following statement is obtained:

Proposition 4.3**.**

The set

[TABLE]

has density [math] and its product set $A^{2}$ has density $1$.

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

Theorem 4.4**.**

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

Proof.

We start with defining a set $Q$ such that $\mathbf{d}(Q)=0$ and $\mathbf{d}(Q\cdot Q)=\beta$. 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

[TABLE]

[TABLE]

Furthermore, let

[TABLE]

and

[TABLE]

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 $\mathbf{d}(Q\cdot Q)=\beta$. 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 $\prod\limits_{p<x,p\in P_{i}}(1-1/p)\leq\frac{1}{\beta}\prod\limits_{p<x,p\in P_{i}\cup P_{0}}(1-1/p)\leq\frac{1}{\beta}\exp\left\{-\sum\limits_{\begin{subarray}{c}j:\ p_{j}<x,\\ j\equiv i\pmod{2}\end{subarray}}\frac{1}{p_{j}}\right\}=O\left(\frac{1}{\beta\sqrt{\log x}}\right)$. Therefore, $\mathbf{d}(Q_{1})=\mathbf{d}(Q_{2})=0$, and consequently $\mathbf{d}(Q)=0$ 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 $\underline{\mathbf{d}}(A\cdot A)=\alpha$ and $\overline{\mathbf{d}}(A\cdot A)=\beta$. As $A\subseteq Q$ we will have $\mathbf{d}(A)=0$ and $\overline{\mathbf{d}}(A\cdot A)\leq\beta$. 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):

• (i)

$A_{j}\subseteq A_{j-1}$,

• (ii)

$A_{j}\cap[1,n_{j-1}]=A_{j-1}\cap[1,n_{j-1}]$,

• (iii)

$A_{j}\cap[n_{j}+1,\infty]=Q\cap[n_{j}+1,\infty]$.

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 $Q$:

[TABLE]

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 $j$:

• Case I:

$j$ is odd.

Let $n_{j}<s$ be the smallest integer such that

[TABLE]

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 $s^{\prime}=\lfloor n_{j}^{2}/\alpha\rfloor+1$ we have

[TABLE]

Hence, $s$ is well-defined (and $s\leq s^{\prime}$). 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 $(*)$.

• 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 $\mathbf{d}(Q\cdot Q)=\beta$ 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 $\mathbf{d}(Q)=0$, we have that

[TABLE]

therefore, $\mathbf{d}(A_{j}\cdot A_{j})=\beta$. 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 $(*)$.

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 $\mathbf{d}(A)=0$ and $\overline{\mathbf{d}}(A\cdot A)\leq\beta$. At first we show that $\underline{\mathbf{d}}(A\cdot A)\geq\alpha$. 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

[TABLE]

This holds for every $n\geq n_{0}$, therefore, $\underline{\mathbf{d}}(A\cdot A)\geq\alpha$.

As a next step, we show that $\underline{\mathbf{d}}(A\cdot A)=\alpha$. 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

[TABLE]

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,

[TABLE]

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 $\underline{\mathbf{d}}(A\cdot A)=\alpha$.

Finally, we prove that $\overline{\mathbf{d}}(A\cdot A)=\beta$. 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 $\overline{\mathbf{d}}(A\cdot A)\geq\lim(\beta-1/j)=\beta$, thus $\overline{\mathbf{d}}(A\cdot A)=\beta$ as it was claimed.

∎