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Statement

Notation (p. 14). For a subset BB of an Abelian group, s(B)s(B) is the largest size of a sum-free subset of BB, sum-free meaning that no a,b,ca,b,c of the set, not necessarily distinct, satisfy a+b=ca+b=c (p. 13). For a sequence A=(a1,…,an)A=(a_1,\dots,a_n) of elements of an Abelian group, not necessarily distinct, s(A)s(A) is the largest kk for which there are indices 1≤i1<⋯<ik≤n1\le i_1<\dots<i_k\le n such that the set {ai1,…,aik}\{a_{i_1},\dots,a_{i_k}\} is sum-free. Repeated terms are counted with multiplicity in kk, while the sum-free condition is tested on the set of their values.

Proposition 1.2 (p. 14, quoted). "For any sequence BB of non-zero integers, s(B)>13∣B∣s(B)>\frac13|B|."

For a sequence of distinct terms this is Proposition 1.1; the paper calls Proposition 1.2 "clearly stronger" (p. 14).

The upper side for sequences (p. 14, proved in Section 2, pp. 16--18). The paper constructs sequences BB with s(B)<1128∣B∣s(B)<\frac{11}{28}|B|: Corollary 2.4 (p. 18) is the sequence S=(ijl:1≤i≤2, 1≤j≤5, 1≤l≤14)S=(ijl:1\le i\le2,\ 1\le j\le5,\ 1\le l\le14) of 140140 products, with s(S)≤55s(S)\le55, so s(S)/∣S∣≤1128s(S)/|S|\le\frac{11}{28}. Corollary 2.3 (p. 17) states that for every sequence AA of nonzero integers there is a sequence BB with

s(B)∣B∣≤s(A)∣A∣−1(∣A∣−s(A)+1)! e ∣A∣,\frac{s(B)}{|B|}\le\frac{s(A)}{|A|}-\frac{1}{(|A|-s(A)+1)!\,e\,|A|},

so the infimum of s(B)/∣B∣s(B)/|B|, as BB ranges over all sequences of integers, is not attained (p. 14). Remark 2.5 (p. 18) notes that Lemma 2.2 yields sequences with smaller ratios still and omits their computation, since the method does not seem to close the gap between 13\frac13 and 1128\frac{11}{28}; on p. 26 the paper names the best-possible constants in Propositions 1.1 and 1.2 as an open question.

Source. 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, DOI 10.1017/CBO9780511983917.003, as described on the source card: the notation and the proposition on p. 14, the proof on p. 15, Corollaries 2.3 and 2.4 on pp. 17--18.

Read depth. Claims checked: the notation, the proposition, Corollaries 2.3 and 2.4 and the statements of p. 14 were read clause by clause on the page images. The proof on p. 15 was read and its steps followed; the proofs of Lemma 2.2 and Corollaries 2.3 and 2.4 were read for structure only. Nothing here is independently reviewed.

Proof pointer

P. 15. For the terms b1,…,bnb_1,\dots,b_n of BB, take a prime p=3k+2p=3k+2 with p>2max⁡i∣bi∣p>2\max_i|b_i|, so that no bib_i vanishes modulo pp, and the interval C={k+1,…,2k+1}C=\{k+1,\dots,2k+1\}, a sum-free subset of Zp\mathbb Z_p with ∣C∣/(p−1)=(k+1)/(3k+1)>13|C|/(p-1)=(k+1)/(3k+1)>\frac13. For xx uniform in {1,…,p−1}\{1,\dots,p-1\}, each xbi mod pxb_i\bmod p is uniform on the nonzero residues, so the expected number of ii with xbi mod p∈Cxb_i\bmod p\in C exceeds n/3n/3. Fix an xx doing at least as well as the expectation: more than n/3n/3 terms land in CC, and they form a sum-free subsequence, since a+b=ca+b=c among them would give xa+xb≡xc(modp)xa+xb\equiv xc\pmod p inside CC.

Dependencies

None stated. The construction side uses Schur's theorem (Theorem 2.1, p. 16, with f(2)=5f(2)=5 and f(3)=14f(3)=14) through Lemma 2.2 (pp. 16--17). Proposition 4.1' (pp. 21--22) deduces the same strict bound for sequences of nonzero reals from this proposition.

Bears on

  • Problem 792: the problem concerns sets of nn integers, where this proposition reduces to Proposition 1.1, the bound f(n)≥(n+1)/3f(n)\ge(n+1)/3 for sets of nonzero integers. The sequence constructions with ratio below 1128\frac{11}{28} allow repeated terms and give no upper bound for f(n)f(n), which is defined over sets.