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Statement

Setting (pp. 1--2). A(x)A(x) and B(x)B(x) count the elements up to xx, condition (1.2) is A(x)B(x)/x→1A(x)B(x)/x\to1, and a∗(x)=max⁡{a∈A, a≤x}a^*(x)=\max\{a\in A,\ a\le x\}.

Theorem 1.3 (p. 2, quoted). "Let ω\omega be a function tending to infinity arbitrarily slowly. There are additive complements satisfying (1.2) such that for infinitely many values of xx we have

A(x)B(x)−x<min⁡(ω(x),ca∗(x))(1.7)A(x)B(x)-x<\min\bigl(\omega(x),ca^*(x)\bigr) \tag{1.7}

with some constant cc."

The paper introduces the theorem (p. 2) as the answer to the question, which it says Chen and Fang also formulated, whether an absolute lower bound such as A(x)B(x)−x>log⁡xA(x)B(x)-x>\log x holds: it does not. It is the example by which the abstract says the paper's lower bound, Theorem 1.2, is nearly best possible.

Source. I. Z. Ruzsa, Exact additive complements, Q. J. Math. 68 (2017), 227--235, doi:10.1093/qmath/haw029; labels and pages are those of the arXiv version arXiv:1510.00812v1 (3 October 2015), as identified on the source card: the statement on p. 2, the construction in Section 3, pp. 5--7.

Read depth. Claims checked: the statement was read clause by clause on the printed page, and the construction (pp. 5--7) was read through for structure. No step of it was independently checked, and nothing here is independently reviewed.

Proof pointer

Section 3, pp. 5--7. Take primes pkp_k with k3<pk<(k+1)3k^3<p_k<(k+1)^3 (with finitely many exceptions) and a fast-growing sequence uku_k with uk>kuk−1u_k>ku_{k-1} and pk∣ukp_k\mid u_k. Let AA be the union of blocks A1={1,…,p1}A_1=\{1,\ldots,p_1\} and Ak⊂(uk,2uk)A_k\subset(u_k,2u_k) of pk−pk−1p_k-p_{k-1} elements, chosen so that A1∪⋯∪AkA_1\cup\cdots\cup A_k is a complete residue system modulo pkp_k; Lemma 3.1 (p. 5) shows such blocks exist once uku_k exceeds a bound depending only on the primes. Let BB be the union of the sets BkB_k of multiples of pkp_k in (kuk,(k+3)uk+1)(ku_k,(k+3)u_{k+1}). Every n>3u1n>3u_1 lies in A+BA+B (p. 6), and counting gives A(x)B(x)−x=O(x/k)A(x)B(x)-x=O(x/k), so the complements are exact. At x=uk+1x=u_{k+1} one has A(x)=pkA(x)=p_k and uk<a∗(x)<2uku_k<a^*(x)<2u_k, giving A(x)B(x)−x<c3uk<c3a∗(x)A(x)B(x)-x<c_3u_k<c_3a^*(x), and c3uk<ω(x)c_3u_k<\omega(x) once uju_j grows so fast that ω(uk+1)>uk\omega(u_{k+1})>u_k (p. 7).

Dependencies

Lemma 3.1 (p. 5) and the convergence of ∑1/pi\sum1/p_i over the chosen primes; no other result of the paper.

Bears on

  • Problem 785: the sets constructed are infinite, A+BA+B contains every integer above 3u13u_1, and A(x)B(x)∼xA(x)B(x)\sim x, so they satisfy the problem's hypotheses (as sets of positive integers; this matching is an observation on this page, not the paper's). Along infinitely many xx their excess A(x)B(x)−xA(x)B(x)-x stays below ω(x)\omega(x) for a prescribed ω\omega tending to infinity arbitrarily slowly. So the excess in the problem's conclusion A(x)B(x)−x→∞A(x)B(x)-x\to\infty admits no absolute lower bound such as log⁡x\log x, as the paper says (p. 2); the theorem does not contradict the conclusion.