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Statement

Let $m$ be Lebesgue measure on $\mathbb R$ and let $D=\{2^{-n}:n\in\mathbb N,\ n\ge1\}$. A nontrivial affine copy of a set $A$ is $x+sA$ with $x\in\mathbb R$ and $s\in\mathbb R\setminus\{0\}$; $A$ is measure universal when every measurable set of positive measure contains such a copy.

Theorem 1.1. For every $\eta\in(0,1)$ there is a compact set $E_\eta\subseteq[0,1]$ with $m(E_\eta)>1-\eta$ such that for every $x\in\mathbb R$ and every $s\in\mathbb R\setminus\{0\}$,

$$x+sD\nsubseteq E_\eta,$$

that is, some integer $n\ge1$ has $x+s2^{-n}\notin E_\eta$.

The quantifiers are as the manuscript prints them: the measure deficit $\eta$ is fixed first, the set depends on it, and the conclusion runs over every real translation and every nonzero dilation of either sign. The manuscript states the consequence as "the dyadic sequence is not measure universal" and adds that the theorem "concerns one infinite pattern and all of its signed affine copies; the conjecture for arbitrary infinite sets is not addressed" (Section 1, p. 2).

Source. OpenAI, The dyadic case of the Erdős similarity conjecture, release folder preprints/The-dyadic-case-of-the-Erdos-similarity-conjecture-September-25-2026; TeX sections/introduction.tex, environment thm:main (lines 24--33), PDF p. 2; proof in sections/global.tex lines 7--76, PDF pp. 14--15; read. The card records the provenance and attestations.

Read depth. Claims checked: the statement, the definitions of affine copy and measure universality, and the two qualifying sentences above were read clause by clause in the TeX source and located in the PDF. The deduction from Lemma 2.1 and the four-section proof of that lemma were read for their structure only (below); no step was checked. Nothing here is independently reviewed.

Proof pointer

Section 6 deduces the theorem from Lemma 2.1, an open $1$-periodic set $H$ of density at most $6p$ that meets $x+tD$ for every real $x$ and every normalized dilation $t\in[1,2]$. Given the lemma, the deduction is short: take $H_j$ with $p_j=\eta2^{-j}/24$ for $j=1,2,\dots$, let $C=\bigcup_{j\ge1}(2^{-j}H_j\cup(-2^{-j}H_j))$ and $E_\eta=[0,1]\setminus C$. The set $C$ is open and symmetric, so $E_\eta$ is compact; each $\pm2^{-j}H_j$ has measure $\rho(H_j)$ in $[0,1]$ because $[0,2^j]$ holds exactly $2^j$ periods, so $m(C\cap[0,1])\le12\sum_jp_j=\eta/2$. For $s>0$ write $t=2^ks\in[1,2)$, put $j=\max\{1,k\}$ and apply the lemma to $H_j$ at the center $2^jx$: the hit $2^jx+t2^{-n}\in H_j$ rescales to $x+s2^{-(n+j-k)}\in C$ with $n+j-k\ge n\ge1$ because $j\ge k$. For $s<0$ apply the positive case to $(-x,-s)$ and use $-C=C$.

The manuscript proves Lemma 2.1 in Sections 3--5 by a random construction; its structure is summarized on the lemma's page. The hypothesis $\eta<1$ only fixes the measure budget; the dyadic ratio enters through the normalization $t=2^ks\in[1,2)$ and through the dyadic grids $J_b(z)=\lfloor2^{b+2}\{z\}\rfloor$ on which the random tables are evaluated, so that an index window of the sequence and a grid resolution are the same kind of object.

Dependencies

None at statement level. The proof is presented as self-contained, using a finite product probability space, nested dyadic grids, and elementary measure theory on the circle (continuity of measure from above; a projection from a compact product is closed). The works cited in Sections 1 and 5 (Kolountzakis 1997; Chlebík 2015; Kolountzakis and Papageorgiou 2025; Iosevich, Kulkarni, Mora Cuéllar, Rojas Aravena and Yavicoli 2026) are named as precedents for the method, not invoked as premises. None was checked here.

Bears on

• Problem 120: claimed partial answer, the case $A=\{2^{-n}:n\ge1\}$ of the question, with the avoiding set compact in $[0,1]$, of measure above $1-\eta$, and avoiding dilations of both signs. The general question for an arbitrary infinite $A$ is not addressed. The claim is unverified here; the page's status rests on its acceptance evidence.
• [[analysis/openai_2026_geometric_case_erdos_similarity_conjecture/theorem_1_1|The companion's Theorem 1.1]]: the geometric-case manuscript of the same release family claims the same conclusion for $\{q^n:n\ge1\}$ with every fixed $q\in(0,1)$, of which this theorem is the case $q=1/2$; neither claim is verified here.