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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 η,γ>0\eta,\gamma>0, there is ϵ>0\epsilon>0 such that for all integer n,k,h,ln,k,h,l with

l, k−l, h−l, n−k−h+l≥ηn,(1)l,\ k-l,\ h-l,\ n-k-h+l\ge\eta n, \tag{1}

families A⊆Ω([n];k)\mathcal A\subseteq\Omega([n];k) and B⊆Ω([n];h)\mathcal B\subseteq\Omega([n];h) of density product at least e−ϵne^{-\epsilon n} satisfy

il(A,B)≥N(n;k,h,l)e−γn,N(n;k,h,l)=(nk)(kl)(n−kh−l).(2)i_l(\mathcal A,\mathcal B)\ge N(n;k,h,l)e^{-\gamma n},\qquad N(n;k,h,l)=\binom nk\binom kl\binom{n-k}{h-l}. \tag{2}

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 nn by (1). An input loss e−τne^{-\tau n} can therefore be made smaller than any prescribed input loss for the new ambient size by choosing τ\tau sufficiently small. The two independent output tolerances in Proposition 7.2 permit this choice while keeping the counting loss below any prescribed part of γ\gamma.

First consider equal sizes k=h≤n/2k=h\le n/2. For 2k=n2k=n, Theorem 6.1 is exactly the assertion after rescaling its constants. For 2k<n2k<n, apply Proposition 7.2 with N=2kN=2k. It gives at least (n2k)e−ρn\binom n{2k}e^{-\rho n} containing sets CC, and in each one both kk-set families have density at least e−τne^{-\tau n}. Inside CC the four target cells have sizes l,k−l,k−l,ll,k-l,k-l,l, still bounded below by ηn\eta n. Thus Theorem 6.1 gives at least N(2k;k,k,l)e−γ′2kN(2k;k,k,l)e^{-\gamma'2k} pairs in each CC. Each pair with intersection ll is counted exactly (n−2k+ll)\binom{n-2k+l}l times among all possible CC. The exact identity

(n2k)N(2k;k,k,l)=N(n;k,k,l)(n−2k+ll)(3)\binom n{2k}N(2k;k,k,l) =N(n;k,k,l)\binom{n-2k+l}l \tag{3}

follows by counting a pair and a containing 2k2k-set in the two orders. It yields (2) if ρ,γ′\rho,\gamma' are small enough and the input tolerance is then chosen for Proposition 7.2 and the middle-layer theorem.

Next suppose k+h=nk+h=n. Interchange the two families if needed so k≤hk\le h, and complement the hh-sets. Both families now consist of kk-sets, and the target intersection becomes k−lk-l. 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 k+h=N<nk+h=N<n, apply Proposition 7.2 with this NN. Within each useful CC, the four target cells are l,k−l,h−l,ll,k-l,h-l,l, and the two sizes sum to ∣C∣|C|. Apply the preceding sum-equals-ambient case. A target pair has union size N−lN-l and lies in exactly (n−N+ll)\binom{n-N+l}l sets of size NN. The identity

(nN)N(N;k,h,l)=N(n;k,h,l)(n−N+ll)(4)\binom nN N(N;k,h,l)=N(n;k,h,l)\binom{n-N+l}l \tag{4}

then proves (2), with output losses adding at most ρn+γ′N\rho n+\gamma'N. Choose these less than γn\gamma n, and choose the input losses in the stated order.

Finally, if k+h>nk+h>n, assume k≤hk\le h and complement the second family. The new sizes are k,n−hk,n-h, whose sum is at most nn, and the new intersection is k−lk-l. The new four cells are k−l,l,n−k−h+l,h−lk-l,l,n-k-h+l,h-l, 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 N=nN=n. Thus a single ϵ(η,γ)\epsilon(\eta,\gamma) works for all large nn. For the finitely many remaining nn, decrease it so the hypothesis forces full families; then (2) holds directly. □\square

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.