1 Introduction
Two sets $A,B$ of positive integers are called additive complements, if $A+B$ contains all sufficiently large integers. Let $A=\{a_1<a_2<\cdots \}$ be a set of positive integers. Denote by $A(x)$ the counting function of A and by $a^*(x)$ the largest element in $A\bigcap [1,x]$ . If additive complements $A,B$ satisfy
then we call such $A,B$ exact additive complements. In 2001, Ruzsa [Reference Ruzsa2] introduced the following notation which is powerful for the proof of additive complements. Let ${m>a_1}$ be an integer and let $k=A(m)$ . Denote by $L(m)$ the smallest number l for which there are integers $b_1,\ldots ,b_l$ such that the numbers $a_i+b_j,1\leqslant i\leqslant k, 1\leqslant j\leqslant l$ , contain every residue modulo m. Obviously, $L(m)\geqslant m/k$ .
Theorem 1.1 (Ruzsa [Reference Ruzsa2]).
If
then A has an exact complement.
Theorem 1.2 (Ruzsa [Reference Ruzsa2]).
Let A be a set satisfying ${A(2x)}/{A(x)}\rightarrow 1$ . The following are equivalent.
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(a) A has an exact complement.
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(b) $A(m)L(m)/m\rightarrow 1$ .
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(c) There is a sequence $m_1\kern1.5pt{<}\kern1.5pt m_2\kern1.5pt{<}\kern1.2pt\cdots $ of positive integers such that ${A(m_{i+1})/ A(m_i)\kern1.2pt{\rightarrow}\kern1.2pt 1}$ and $A(m_i)L(m_i)/m_i\rightarrow 1$ .
Theorem 1.3 (Ruzsa [Reference Ruzsa3]).
For exact additive complements $A,B$ with ${A(2x)}/ {A(x)}\rightarrow 1$ ,
In 2019, Chen and Fang [Reference Chen and Fang1] improved Theorem 1.3 by removing the exact condition. Chen and Fang also showed in [Reference Chen and Fang1] that Theorem 1.3 is the best possible.
Theorem 1.4 (Chen and Fang [Reference Chen and Fang1]).
There exist exact additive complements $A,B$ with ${A(2x)}/{A(x)}\rightarrow 1$ such that
for infinitely many positive integers x.
In this paper, under condition (1.1) from [Reference Ruzsa2], we obtain the following result.
Theorem 1.5. For exact additive complements $A,B$ with (1.1),
Moreover, we also show that ${a^*(x)}/{A(x)^2}$ is the best possible.
Theorem 1.6. There exist exact additive complements $A,B$ with (1.1) such that
2 Proofs of the main results
Let
and
The ideas used in the proofs of the main results come from [Reference Chen and Fang1–Reference Ruzsa3]. We use the following lemma of Ruzsa in the proof of Theorem 1.5.
Lemma 2.1 [Reference Ruzsa3, Lemma 2.1].
Let U and V be finite sets of integers and let
and
Then
Proof of Theorem 1.5.
Assume the contrary. Suppose that (1.2) does not hold. Then there exist a positive number $\delta _0(<1)$ and a sequence $x_1<x_2<\cdots <x_k<\cdots $ such that
We know that
Since $a^*(x_k)\in A$ and $a^*(x_k)+b>x_k$ for all $b\in B$ with $x_k-a^*(x_k)<b\leqslant x_k$ ,
Thus,
From Ruzsa’s Lemma 2.1,
Let
Then
Now we need a lower bound for $|D|$ . We consider the following two cases.
Case 1: $a^*(x_k)>\frac 12x_k$ for infinitely many k. By (1.1),
Thus, in this case, by Theorem 1.3 and $A(x)B(x)/x\rightarrow 1$ ,
for sufficiently large k. It follows from (2.2) and (2.3) that
for sufficiently large k, which is in contradiction with (2.1).
Case 2: $a^*(x_k)\leqslant \tfrac 12x_k$ for infinitely many k. By (1.1),
Thus, in this case, by Theorem 1.3 and $A(x)B(x)/x\rightarrow 1$ ,
for sufficiently large k. It follows from (2.2) and (2.3) that
for sufficiently large k, which is in contradiction with (2.1).
This completes the proof of Theorem 1.5.
Proof of Theorem 1.6.
Let $a_1=1$ and $a_2=4$ . We construct the sequence $a_3,a_4,\dots $ with
and a sequence $n_1,n_2,\dots $ such that $a_1,a_2,\dots ,a_{n_k}$ form a complete residue system modulo $n_k$ and $n_k\mid a_{n_k}$ . We get such a sequence by a greedy algorithm: let $n_1=2$ , and if $n_1,n_2,\dots ,n_k$ are already defined, then let $n_{k+1}=a_{n_k}$ . Since $a_1,\dots ,a_{n_k}$ are distinct residues modulo $a_{n_k}$ , we can choose $a_{n_k+1},\dots ,a_{n_{k+1}}$ such that $a_{m+1}\gg m^4a_m$ for ${m=n_k,\ldots , n_{k+1}-1}$ , $a_{n_k}\mid a_{a_{n_k}}$ and $a_1,\dots ,a_{n_{k+1}}$ form a complete residue system modulo $n_{k+1}$ .
For every positive integer k, we further take
Then $a_i+b_j$ for $1\leqslant i\leqslant p, 1\leqslant j\leqslant {a_{n_k}}/{n_k}$ , form a complete residue system modulo $a_{n_k}$ . From the definition of $L(a_{n_k})$ ,
For the set $A=\{a_k\}_{k=1}^{\infty }$ and every positive integer k, define $q_k$ by
Define the same sets $A,B$ as in [Reference Ruzsa2, Theorem 3] (replacing $m_k$ by $a_k$ ) and write ${A_k=A\bigcap [1,a_k]}$ . Take $U_k\subseteq [1,a_k]$ so that $|U_k|=L(a_k)$ and $A_k+U_k$ contains every residue module $a_k$ . Let
Let $q_ka_k\leqslant x\leqslant q_{k+1}a_{k+1}$ . The sequence $\{q_k\}_{k=1}^{\infty }$ defined in (2.6) is increasing to infinity by (2.4) and $A(q_ka_k)\sim A(a_k)$ . (In fact, $A(q_ka_k)=k=A(a_k)$ by (2.6).) By the same proof as in [Reference Ruzsa2, Theorem 3], $A,B$ are additive complements and $A(x)B(x)\sim x$ . Thus, the set A with (2.4) has an exact complement B. Obviously, such an A with (2.4) satisfies (1.1).
Finally, we prove that (1.3) holds for infinitely many $x_k$ . For x with $q_ka_k\leqslant x<(q_{k+1}-1)a_{k+1}$ , we have $k\leqslant A(x)\leqslant k+1$ and
By Theorem 1.2(b), $L(a_{k-1})\leqslant {2a_{k-1}}/{(k-1)} \mbox { for large } k$ . From (2.6),
It is easy to see that, for large k,
It follows from (2.7) that
Choose $x_k=a_{n_k+1}$ , where $n_k$ is the index satisfying (2.5). Then by (2.8),
This completes the proof of Theorem 1.6.