Wiki
Wiki

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

Updated


Claim. The answer is yes. For every c>0c>0 there are δ=δ(c)>0\delta=\delta(c)>0 and n0(c)n_0(c) such that, whenever n≥n0(c)n\ge n_0(c) and F\mathcal F is a family of at most c2nc2^n sets of size nn with union XX, at least $\delta,2^{\lvert X\rvert}$ sets B⊂XB\subset X meet every member of F\mathcal F and contain none. Equivalently, a positive proportion of the two-colorings of XX leave no member of F\mathcal F monochromatic; one such coloring is the same as F\mathcal F having property B, the subject of Problem 901.

Argument. The proof is a random greedy partial coloring with a weight that controls the danger of each edge. Under a partial two-coloring, an edge that already carries both colors has weight 00; a monochromatic edge with at least one colored vertex and ff uncolored vertices has weight 2−f−12^{-f-1}; a wholly uncolored edge has weight 2−n2^{-n}. Each weight is half the probability that a uniformly random completion of the partial coloring leaves the edge monochromatic, so coloring any vertex of an edge at random leaves the edge's expected weight unchanged, and the total weight is a nonnegative martingale whose start is at most cc. The algorithm repeatedly colors, uniformly at random, the uncolored vertex of least weight, where a vertex's weight is the largest weight of an edge through it, and stops when the total weight exceeds 2c2c, when some edge reaches half the threshold w(2c)w(2c) of Beck's theorem, or when every vertex is colored. The optional stopping theorem bounds the probability of the first exit by 1/21/2; at the other two exits only Oc(1)O_c(1) vertices are uncolored, and the hypergraph of monochromatic edges restricted to them has bounded total weight and small individual weights, so a non-uniform form of Beck's theorem (the comment cites Theorem 1.2 of Beck's paper on property B) completes the coloring properly. The number of proper completions of the partial coloring after tt steps, multiplied by 2t2^t, is a martingale, and at the stopping time it is ≫c2∣X∣\gg_c 2^{\lvert X\rvert}, which gives the count. The comment as first posted gave a monochromatic edge with a colored vertex the weight 2−f2^{-f}, under which coloring the first vertex of an edge doubles its weight from 2−n2^{-n} to 2−n+12^{-n+1}. Stijn Cambie's comment of 24 September 2025 pointed this out, noting that the total weight was then only a nonnegative process of nondecreasing expectation, and proposed enlarging the stopping constants, from 2c2c to 4c4c. Chan amended the comment the same day and replied that it was corrected: the renormalized weight 2−f−12^{-f-1} restores the exact martingale, and the thresholds 2c2c and w(2c)/2w(2c)/2 stand.

Acceptance. Reviewed: the site's curator, Thomas Bloom, marks Problem 1027 proved and credits the proof to Koishi Chan's comment of 21 September 2025 in the site's discussion thread (problem page last edited 1 October 2025). The comments were posted by the forum account KoishiChan. The result is a forum comment, not a manuscript, and is not refereed.

Formalizations. The file in Boris Alexeev's lean-proofs collection, linked above, declares itself a formalization of a solution to the problem, names Koishi Chan as the informal author and Codex and GPT-5.6 Sol as the formal authors, and states that it completes the partial-coloring argument with Beck's non-uniform property-B theorem through the finite random greedy proof of Duraj, Gutowski and Kozik. The corpus has not built this development, so this page lists no formalized evidence. The formal-conjectures statement file the site records is a statement, not a proof; it names this development as its formal proof (the problem page links it).