Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Source. Proposition 3.4, pp. 5–6, with the definition of on p. 4 and Lemmas 3.1–3.3 (pp. 4–5), of David Turturean, A Negative Answer to Erdős Problem #870, preprint dated April 2026 (11 pp.), https://www.overleaf.com/read/gknkvvxrymfv; the edition read is named on the source card.
Setting
P. 4. For finite and , , so lies in it when for some or for some . and are as in Proposition 2.3.
Statement
Proposition 3.4 (pp. 5–6). There is an absolute constant with the following property. Let be any finite list of pairs of finite subsets of , each nonempty, and let be finite. There is a set such that
- ;
- is cofinite;
- for all sufficiently large ;
- ;
- for every , every and every finite exceptional set : if is cofinite, then there is such that, with , the set is still cofinite and is an additive basis of order 3.
The proof (p. 7) takes .
Proof pointer
Pp. 6–8. The Larsen–Larsen construction is run with lag 10 and a Bernoulli constant chosen in terms of , but each canary is replaced by a cluster , , around a random center , with restoration elements for old control summands . Lemma 3.2 supplies, for each robust element, many representations whose summands avoid the fixed shift differences, which rules out same-cluster accidental representations; Lemma 3.1, a hypergeometric and difference-set estimate, gives a summable Borel–Cantelli bound that rules out accidental representations across clusters and keeps the points , , out of . Lemma 3.3 makes fixed differences between canaries outside occur only finitely often, which gives the order-3 property of . The set is avoided by deleting finitely many Bernoulli variables at the outset.
Dependencies
Lemmas 3.1–3.3 and the construction of D. Larsen and M. Larsen, Robust additive bases without minimal subbases, arXiv:2601.18507 (2026), whose Lemmas 2, 6 and 7, Proposition 5, p. 8 bound and finite-incidence argument the paper cites.
Read depth: claims checked. The statement was read clause by clause on the print and the proof followed in outline; the cited Larsen–Larsen results were not read.
Bears on
- Problem 870: the input of Proposition 4.1, which gives the case of Theorem 1.1. On its own it gives no order-3 basis without a minimal subbasis.