bibkey: glabkula2025kenyon
authors: Szymon Głąb and Mateusz Kula
year: 2025
title: Kenyon theorem revisited
doi: null
url: https://arxiv.org/abs/2508.19454v1
claim: Section 5.1 Problem 1 asks whether positive measure forces every residue modulo the cardinality of the Boolean block-sum set when no indexed coefficient is divisible by that cardinality; this is the alias of arXiv 2512.17285v1 Section 2 Problem 9.
strata_touched:
  - D5/S3/Topology/Measures
license: citation-only
triage: anchor

The unnormalized residue question

Section 5.1, Problem 1 of the fixed arXiv v1 source asks about a multigeometric sequence $(k_1,\dots ,k_m;1/n)$ with positive natural coefficients. Let

$$\Sigma =\left\{\sum _{i=1}^m{\epsilon }_i{k}_i:{\epsilon }_i\in \{0,1\}\right\},|\Sigma |=n.$$

Assuming no $k_i$ is divisible by $n$ and the infinite achievement set has positive Lebesgue measure, must $\Sigma$ have every remainder modulo $n$? No primitive-coefficient or gcd-one condition appears in this question. Its later appearance as Section 2, Problem 9 of the achievement-set survey is one target, not a second problem.

The module D5.S3.Topology.Measures.AchievementResidueConverseRefutation retains the entire unnormalized universal claim with weakly decreasing indexed coefficients and independent repeated positions. For $(9,3,3)$ and $n=6$, the six distinct block sums are $\{0,3,6,9,12,15\}$ and the residues are only $\{0,3\}$. The genuine achievement set, starting at denominator exponent one, is $[0,3]$. Summability, regrouping and the selection of every base-six digit sequence are proved on the original indices, using Mathlib’s infinite digit expansion. The conclusion is the literal closed negation result : ¬ claim.

Source and scope limits

The complete fixed-v1 TeX source is available; the passages used here are the multigeometric convention and Section 5.1’s definition, example and question. The paper’s normalized prime-case theorem involves a translation and scaling by a common divisor; it is not a substitute for the unnormalized question. Dividing $(9,3,3)$ by three changes the residue conclusion and is not performed in the Lean model.

The supplied literature check was bounded. Original journal sources cited by these papers have not all been read, and exhaustive novelty, current-openness clearance and official acceptance are not asserted. The result makes no claim about primitive or normalized variants, nor about the other open questions or main theorems of this paper. This note is citation and paraphrase only.

Verified locator

• Fixed primary source: https://arxiv.org/abs/2508.19454v1
• Fixed source archive: https://arxiv.org/src/2508.19454v1
• Fixed PDF: https://arxiv.org/pdf/2508.19454v1