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Source. Published p. 264, Theorem 1.14, and pp. 277–279, Propositions 7.1–7.3 and their concluding reduction (PDF). Their complement, containing-set, and double-counting deductions are included in this proof.
Statement. Given , there is such that for all integer with
families and of density product at least satisfy
In particular this proves the source's fixed-proportion version. Uniformity in the four buffered cell sizes will be used in the partition induction.
Proof. We give the order of reductions explicitly. All the smaller ambient sets below have size at least a fixed positive multiple of by (1). An input loss can therefore be made smaller than any prescribed input loss for the new ambient size by choosing sufficiently small. The two independent output tolerances in Proposition 7.2 permit this choice while keeping the counting loss below any prescribed part of .
First consider equal sizes . For , Theorem 6.1 is exactly the assertion after rescaling its constants. For , apply Proposition 7.2 with . It gives at least containing sets , and in each one both -set families have density at least . Inside the four target cells have sizes , still bounded below by . Thus Theorem 6.1 gives at least pairs in each . Each pair with intersection is counted exactly times among all possible . The exact identity
follows by counting a pair and a containing -set in the two orders. It yields (2) if are small enough and the input tolerance is then chosen for Proposition 7.2 and the middle-layer theorem.
Next suppose . Interchange the two families if needed so , and complement the -sets. Both families now consist of -sets, and the target intersection becomes . This bijection preserves both density product and target-pair count. The four cell sizes in (1) are merely permuted, so the already proved equal-size case applies.
For , apply Proposition 7.2 with this . Within each useful , the four target cells are , and the two sizes sum to . Apply the preceding sum-equals-ambient case. A target pair has union size and lies in exactly sets of size . The identity
then proves (2), with output losses adding at most . Choose these less than , and choose the input losses in the stated order.
Finally, if , assume and complement the second family. The new sizes are , whose sum is at most , and the new intersection is . The new four cells are , all satisfying (1). Apply the proved case and undo the complement. This is Proposition 7.1's reduction, including the exact feasibility inequalities and the preservation of the full count.
Every tolerance used above is uniform under (1): all ambient sizes and nonempty relevant cells are bounded below proportionally, Theorem 6.1 is uniform on that compact range, and Proposition 7.2 is uniform up to . Thus a single works for all large . For the finitely many remaining , decrease it so the hypothesis forces full families; then (2) holds directly.
Source precision. The order here first proves the equal-size case, then the complementary-size case, and then the general case. It avoids reading the source's successive “sufficient to prove” reductions as a circular dependence. The containing-set argument uses the explicitly proved two-tolerance Proposition 7.2, not its unverified same-tolerance wording.
Dependencies. theorem_6_1, proposition_7_2, definitions.