Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Source. Published p. 283, Theorem 10.2 (PDF).
Statement. If every cross intersection of has parity , then for and at most for . In the even case, equality forces both sets of characteristic vectors to be full mutually orthogonal linear subspaces over .
Proof. Empty families are immediate. In the even case their linear spans are orthogonal, so . Consequently . Equality forces equality in both inclusions, giving the stated full subspaces. In particular both families contain the empty set.
In the odd case append a coordinate one to every characteristic vector on both sides. The new spans in are orthogonal. The last-coordinate functional is nonzero on each span, so its level one contains exactly half the span. The two families lie in these two halves, yielding .