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Claim. In the notation of Problem 787,

$$g(n)\ll n^{2/5+o(1)}:$$

some $n$ numbers have no subset larger than $n^{2/5+o(1)}$ in which no sum of two distinct elements lies in the whole set. S. L. G. Choi, On a combinatorial problem in number theory, Proc. London Math. Soc. (3) 23 (1971), no. 4, 629--642, cited as [Ch71] on the problem page. The paper is not held, and the bound is stated as the sources that cite it give it: the zbMATH review of Ruzsa's 2005 paper (Zbl 1155.11308) states that Choi proved $l(n)\ll n^{2/5+o(1)}$; Erdős's 1973 survey (display (9.2), printed p. 130) gives $g(n)<n^{2/5+\varepsilon}$ and credits the upper bound to Choi; Baltz, Schoen and Srivastav (Colloq. Math. 86 (2000), p. 171) give $g(n)=O(n^{2/5+\varepsilon})$; Sanders (2021, p. 1) cites it as display (2) of the paper, $M(A)\le|A|^{2/5+o(1)}$; and Beker (2025, p. 1) as $\phi(n)\le n^{2/5+o(1)}$. The same sources attest two further results of the paper that are not sub-claims: Choi's observation that the problem for real numbers reduces to sets of integers (Baltz, Schoen and Srivastav, p. 171, and Beker's footnote 1), and his greedy lower bound $\phi(n)\ge\log_2n$ (Beker, p. 1), the first published proof of a logarithmic lower bound.

Covers. The upper bound $g(n)\ll n^{2/5+o(1)}$, refined by Baltz, Schoen and Srivastav to $O(n^{2/5}(\log n)^{2/5})$ and superseded by Ruzsa's $\exp(O(\sqrt{\log n}))$. Not covered: the order of growth of $g(n)$.

Depends on. No page of this wiki.

Acceptance. Refereed: the paper is the publisher's version of record in the Proceedings of the London Mathematical Society (the Crossref record gives the issue of December 1971, with no day, so this page is named by the first of the month). The site's curator, Thomas F. Bloom, credits the bound $g(n)\ll n^{2/5+o(1)}$ to Choi in the problem page's commentary (label OPEN, page last edited 23 January 2026); the problem is not marked settled there, so the credit is recorded here and is not listed as reviewed. The paper itself is not held; the statement rests on the citing sources above.