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Dissociation bounds from aspect ratio and coefficient imbalance

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_proof: A discrete Kraft deficiency estimate bounds a positive set in terms of its aspect ratio and the least possible nonzero coefficient sum of a relation. It gives the sharp positive bound at ratio 4/3 and the factor 3 at ratio 2.


Statement

For a finite set X⊂RX\subset\mathbb R let d(X)d(X) be the largest size of a dissociated subset, a subset whose distinct subsets have distinct sums. If PP is a nonempty finite set of positive reals with max⁡P≤2min⁡P\max P\le2\min P and n=∣P∣n=|P|, then

n<3 2d(P),henced(P)>log⁡2n−log⁡23andd(P)≥⌊log⁡2n⌋−1.n<3\,2^{d(P)}, \qquad\text{hence}\qquad d(P)>\log_2 n-\log_2 3 \quad\text{and}\quad d(P)\ge\lfloor\log_2n\rfloor-1.

Every set of two distinct positive reals is dissociated, so d(P)≥2d(P)\ge2 for n≥2n\ge2 independently of this bound.

There is also a bound that uses the possible coefficient sums of relations. Let q≥1q\ge1 be an integer and 0<θ<q0<\theta<q a rational number. Suppose max⁡P≤(1+θ)min⁡P\max P\le(1+\theta)\min P, and every ternary relation ∑x∈Pεxx=0\sum_{x\in P}\varepsilon_xx=0 has coefficient sum ∑xεx\sum_x\varepsilon_x either zero or of absolute value at least qq. Then

∣P∣<q+θ2(q−θ) 2d(P).|P|<\frac{q+\theta}{2(q-\theta)}\,2^{d(P)}.

The relation hypothesis is automatic for q=1q=1, and holds for q=2q=2 on odd integer sets. Taking q=1q=1 and θ=1/3\theta=1/3 proves ∣P∣≤2d(P)−1|P|\le2^{d(P)}-1 when max⁡P≤43min⁡P\max P\le\tfrac43\min P. Thus the exact bound asked in Problem 963 holds for real sets whose nonzero absolute values lie within a factor of 4/34/3: their symmetric reduction has ∣A∣≤2∣P∣+1≤2d(A)+1−1|A|\le2|P|+1\le2^{d(A)+1}-1. Sets with no nonzero element satisfy the target directly.

Significance

Problem 963 asks whether every nn-element real set has a dissociated subset of size ⌊log⁡2n⌋\lfloor\log_2n\rfloor. Its symmetric reduction replaces a set AA by the positive set PP of absolute values of its nonzero elements, with d(P)=d(A)d(P)=d(A) and ∣A∣≤2∣P∣+1|A|\le2|P|+1. For sets whose nonzero absolute values lie within a factor two of each other this claim gives d(A)≥⌊log⁡2((∣A∣−1)/2)⌋−1d(A)\ge\lfloor\log_2((|A|-1)/2)\rfloor-1 when ∣A∣≥3|A|\ge3, a constant loss against the asked bound. The tier-2 bound of [[theory/number_theory/L22_near_logarithmic_dissociated_subset/_index|L22, which loses log⁡2(log⁡2n+1)\log_2(\log_2n+1) for arbitrary real sets]], is the comparison: in the bounded-ratio regime the loss here does not grow with nn. The general consequence, a loss of order log⁡2log⁡2log⁡2n\log_2\log_2\log_2 n for arbitrary sets, is L213, which applies this claim block by block.

Argument

The proof, by Plasma AI, is author-recorded at tier 0. It has no independent review and no formalization. An integer model reduces the claim to positive integers. A pivot and companions at distinct difference valuations are dissociated if their total distance from the pivot is below qq times the pivot, under the stated relation hypothesis. The truncated cost of dd companions is an integral over distance scales, even at pivots with too few levels. The Kraft equality and an integer deficiency inequality bound its mean above by 1+2d/n1+2^d/n times the diameter. Failure of extraction gives a lower mean in terms of qq times the minimum divided by the diameter. Their comparison proves the quantitative bound. At q=2q=2 and ratio two it gives n<3 2d−1n<3\,2^{d-1} for odd sets; summing valuation layers proves the factor three. No computational premise is used.

The editable proof sketch of the sharp-bound attack places this claim in its route: the claim settles the bounded-ratio regime up to the constant, and the remaining gaps toward the asked bound are the combination of different dyadic scales and the last constant.

Verification and remaining obligations

Read the proof page: its steps are the integer model, the pivot lemma, the Kraft equality, the integer deficiency estimate, the upper and lower averages of the same truncated cost, and the valuation-layer sum. No mathematical gap is known in this statement. Independent review has not been commissioned; the statement does not resolve Problem 963.