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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. N. Alon and D. J. Kleitman, Sum-free subsets, in: A Tribute to Paul Erdős (A. Baker, B. Bollobás and A. Hajnal, eds.), Cambridge Univ. Press (1990), 13--26, Proposition 1.1, printed pp. 13--14: "Any set BB of nn non-zero integers contains a sum-free subset AA of cardinality ∣A∣>13n|A|>\frac13n", where sum-free means (A+A)∩A=∅(A+A)\cap A=\emptyset, a=ba=b included (p. 13), the convention of Problem 792. Since ∣A∣|A| is an integer, ∣A∣≥(n+1)/3|A|\ge(n+1)/3. Proposition 1.2 (p. 14) extends the bound to sequences of nonzero integers. The site's f(n)f(n) is also taken over sets containing 00 and negative integers; 00 lies in no sum-free set, so for a set of nn integers containing 00 the bound applies to its n−1n-1 nonzero elements and gives more than (n−1)/3(n-1)/3, that is at least n/3n/3. Hence f(n)≥n/3f(n)\ge n/3 for every n≥2n\ge2 on the site's domain, while A={0}A=\{0\} gives f(1)=0f(1)=0. The proof (p. 15) multiplies the elements by a random residue modulo a prime p=3k+2p=3k+2 and keeps those landing in the sum-free interval {k+1,…,2k+1}\{k+1,\dots,2k+1\}, each with probability above 1/31/3.

Covers. The lower bound (n+1)/3(n+1)/3 for sets of nn nonzero integers, negative elements included, and with it f(n)≥n/3f(n)\ge n/3 for n≥2n\ge2. Not covered: the second-order term and the upper bound.

Depends on.

Standing. Claimed. The paper is a chapter in a tribute volume, and no evidence that the volume was refereed is on record, so refereed is not listed. The site's curator, Thomas F. Bloom, credits the improvement to (n+1)/3(n+1)/3 to Alon and Kleitman in the problem page's commentary (label OPEN), so the credit is not listed as reviewed. On sets of positive integers the bound is improved by the refereed Bourgain's Proposition 1.3, which does not apply to sets with negative elements. The proof is not checked in this corpus.