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Statement

Announced on p. 14 and carried out on pp. 15--16. Sum-free forbids $a+b=c$ for $a,b,c$ in the set, not necessarily distinct (p. 13), and $s(\cdot)$ is the largest size of a sum-free subset.

• $B=\{1,2,3,4,5,6,10\}$ has $s(B)=3=\frac37|B|$, attained for instance by $\{1,3,10\}$ (p. 15).
• $C=B\cup7B\cup8B\cup9B\cup\{64\}$, a set of $29$ positive integers, has $s(C)\le12=\frac{12}{29}|C|$ (p. 16).
• For every positive integer $m$, the set $C_m=C\cup1000C\cup1000^2C\cup\dots\cup1000^{m-1}C$ consists of $n=29m$ positive integers and has $s(C_m)\le\frac{12}{29}|C_m|$ (p. 16).

The paper's announcement (p. 14, quoted): "We can show that the constant $\frac13$ cannot be replaced by $\frac{12}{29}$ (or any bigger constant), improving the result in [7], which asserts that the constant $\frac13$ cannot be replaced by $\frac37$." Here [7] is Erdős's 1965 paper, which credits the $\frac37$ to an example of D. Klarner (its card). The paper calls the improvement very modest and mentions it because it suggests that $\frac13$ may be the best possible constant (p. 14).

An observation made here: an exhaustive search over the subsets of the printed set $C$ confirms $|C|=29$, $s(B)=3$ and $s(C)=12$, so the bound for $C$ is attained. The search is not retained as evidence.

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 announcement on p. 14, the construction and its case analysis on pp. 15--16.

Read depth. Claims checked: the announcement, the sets $B$, $C$ and $C_m$ and the three bounds were read clause by clause on the page images. The case analysis on pp. 15--16 was read and followed; the bound for $C_m$ is stated in the paper without a separate argument. Nothing here is independently reviewed.

Proof pointer

Pp. 15--16. In $B$ a sum-free subset meets each of the pairs $\{1,2\}$, $\{3,6\}$, $\{5,10\}$ at most once, and a fourth element forces $4$, then $1$, then $6$ and $10$, against $4+6=10$; a similar case analysis shows that a three-element $A\subseteq B$ with $A\cup\{8\}$ sum-free contains $1$ and $10$. A sum-free subset of $C$ of $13$ elements would take exactly $3$ elements from each of $B,7B,8B,9B$ and contain $64$; the paper then uses $64=8\cdot8$ and the dilates $A_i'=\{a/i:a\in A\cap iB\}$, each a sum-free subset of $B$ of size $3$, to force elements whose sums lie in the set, a contradiction. For $C_m$ the paper notes only that the same estimate holds. It follows because the dilates $1000^jC$ are disjoint and a sum-free subset of $C_m$ meets each of them in a sum-free set, so $s(C_m)\le m\,s(C)\le12m$ (an observation made here).

Dependencies

None beyond the definitions.

Bears on

• Problem 792: since $C_m$ is a set of $29m$ positive integers with no sum-free subset of more than $12m$ elements, $f(29m)\le12m$ for every $m\ge1$, an upper bound with constant $\frac{12}{29}$ for the problem's $f(n)$ along $n=29m$, below the $\frac37$ of Klarner's example. It is superseded as an upper constant by the $\frac13+o(1)$ of Eberhard, Green and Manners.