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Source. Published pp. 284–285, Section 11 and Theorem 11.1 (PDF).
Let be the least size of a family of -subsets of such that every -set has for some family member .
Statement. There is an absolute such that for every positive odd integer . Also for every positive integer .
Proof. Let be the span of the characteristic vectors of a qualifying family . When is odd, every weight- vector has nonzero inner product with at least one generator, so contains no vector of weight . Since it is a linear subspace, it contains no pair at Hamming distance either: their difference would have that weight. Theorem 1.10 at alphabet size two, length , and distance gives for an absolute . Taking base-two logarithms and using orthogonal-complement dimensions,
If a theorem threshold was imposed before its finite-dimensional extension, decrease the final constant to include the finitely many smaller positive odd .
For the upper bound, cyclically order and let be the consecutive window of length starting at , for . Fix a -set and also consider . The integers satisfy and . If we are done; otherwise the endpoints are on opposite sides of , so an integer intermediate value equals . If the last endpoint equals , the first does too. Thus one of the first windows works for every , proving the upper bound.
Source precision and historical scope. The printed interval in the upper construction has the wrong size for a family of -sets; the proof uses length as above. The paper's later Conjecture 11.2 is explicitly reported as proved in its own added-in-proof note. No current open-problem or optimal-constant claim is inferred from the earlier conjecture paragraph.
Dependencies. theorem_1_10.