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Statement

Setting (p. 1). For a set AA of positive integers, A(x)A(x) counts its elements up to xx. Condition (1.2) is A(x)B(x)/x→1A(x)B(x)/x\to1; additive complements satisfying it are called exact.

Theorem 1.1 (Narkiewicz's dichotomy; p. 1). Let A,BA,B be infinite sets of positive integers, and let r(x)r(x), the number of integers up to xx that do not lie in A+BA+B, satisfy r(x)=o(x)r(x)=o(x). If (1.2) holds, then

A(2x)A(x)→1,B(2x)B(x)→2(1.3)\frac{A(2x)}{A(x)}\to1,\qquad\frac{B(2x)}{B(x)}\to2 \tag{1.3}

or (1.3) holds with the roles of AA and BB exchanged. If (1.3) holds, then for every ε>0\varepsilon>0 and x>x0(ε)x>x_0(\varepsilon),

A(x)<xε,B(x)>x1−ε.(1.4)A(x)<x^{\varepsilon},\qquad B(x)>x^{1-\varepsilon}. \tag{1.4}

The print's statement reads "infitite" [sic] for "infinite". The paper assumes (1.3) from then on, so that AA is the small set and BB the large one (p. 2), and adds that (1.4) shows polynomial sequences have no exact complement (p. 2).

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.

Read depth. Claims checked: the statement was read clause by clause on the printed page. The theorem is quoted from Narkiewicz and is not proved in this paper, and Narkiewicz's paper was not read. Nothing here is independently reviewed.

Proof pointer

None in this paper: the result is attributed to W. Narkiewicz, Remarks on a conjecture of Hanani in additive number theory, Colloq. Math. 7 (1959/60), 161--165 (the paper's reference [4]).

Dependencies

External: Narkiewicz's paper, as above. Used by Theorem 1.2, whose normalization is (1.3) and whose comparison with Chen and Fang's bound uses (1.4).

Bears on

  • Problem 785: context only. The dichotomy fixes which of the two sets is small, the normalization under which the paper states its lower bound for the excess A(x)B(x)−xA(x)B(x)-x; on its own it says nothing about that excess.