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Source. Published pp. 266–267, Theorem 2.1 (PDF).
Statement. If and every , satisfies , then
Proof. Empty families are immediate. Put and . Complementation shows that and are disjoint, so the smaller family has at most members. Interchange the two families so this is , and choose the integer with
The Hamming distance from to is . Thus the closed radius- neighborhood of avoids . The exact Harper input gives
If , the binomial bounds and concavity and symmetry of show that (1) is at most : among pairs of arguments a distance apart, their entropy sum is maximized when they are symmetric about . If , bound the second factor by ; then
This proves the same bound in that case.
To remove the one-unit rounding loss in , take the Cartesian powers of the two families on disjoint copies of . Their minimum cross intersection is , and their size product is . Apply the proved bound with , take th roots, and let . Continuity gives the bound with in place of . Since and decreases on , the stated inequality follows.
Source precision. The source chooses the largest integer for which all intersections exceed , then compares it directly with the real parameter . The Cartesian-power step supplies the missing rounding justification. No integral-threshold assumption is imposed.
Dependencies. external_inputs, entropy_estimates.