Abstract

For α,β,δ[0,1],α+β=1α, β, δ\in [0,1], α+β= 1 we consider sets BAD(α,β;δ)={ξ=(ξ1,ξ2)[0,1]2:,infpNmax{(plog(p+1))αpξ1,(plog(p+1))βpξ2}δ}. {\rm BAD}^* (α, β;δ) = \left\{ξ= (ξ_1,ξ_2) \in [0,1]^2: ,\inf_{p\in \mathbb{N}} \max \{(p\log(p+1))^α||pξ_1||, (p\log (p+1))^β||pξ_2||\} \ge δ\right\}. We prove that for different (α1,β1),(α2,β2),α1+β1=α2+β2=1(α_1,β_1), (α_2,β_2), α_1 +β_1 = α_2 +β_2 = 1 and δδ small enough BAD(α1,β1;δ)BAD(α2,β2;δ). {\rm BAD}^* (α_1, β_1 ;δ) \bigcap {\rm BAD}^* (α_2, β_2 ;δ) \neq \varnothing . Our result is based on A. Khintchine's construction and an original method due to Y. Peres and W. Schlag.

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Audit summary

Audited against arXiv v2

Not a correctness certificate. A “Correct” result may include yellow typos or minor formal corrections that do not affect substantive soundness. It means this audit found no unresolved substantive error under the stated criteria; it does not replace expert scrutiny or formal verification.

Current report

Detailed mathematical audit

Generated August 20, 2026
01Statements1 reported findingCorrect

The nonempty-intersection theorem remains valid after verified enlargement of two dyadic-cover constants and a positive starting index.

Theorem 1Correct

Intersection of the two weighted badly approximable sets

Pages 1–8 · Theorem 1 and Lemmas 1–5 · arXiv:0712.2423v2

The dangerous-set count, reciprocal-sum estimates, and nested positive-measure construction prove the theorem after the local constant corrections below. Doubling the first covering constant and using the resulting combined coefficient still leaves a bound below one half under δ220\delta\leq2^{-20}, so the induction and compactness conclusion are unchanged.

02Proofs3 reported findingsContains incorrect or incomplete proofs

The dyadic floor is used in the wrong direction and the printed recursion starts at an undefined logarithm. Both defects have verified local repairs that preserve the theorem.

Lemmas 1–3Correct

Counting and reciprocal-sum inputs

Pages 3–6 · Lemmas 1–3 · arXiv:0712.2423v2

Bad approximability separates the relevant lattice points in each dyadic rectangle. The resulting cardinality bound controls the two weighted reciprocal sums over every denominator block, and Corollary 3 supplies exactly the logarithmic estimate used by Lemmas 4 and 5. These inputs are unaffected by the later factor-two correction.

Lemma 4 and Lemma 5Incorrect as written · verified repair

The dyadic scale gives a factor four, not a factor two

Pages 6–8 · proof of Lemma 4 and final estimate in Lemma 5 · arXiv:0712.2423v2

With l=log2(x1+α(log(x+1))α/(2δ))l=\lfloor\log_2(x^{1+\alpha}(\log(x+1))^\alpha/(2\delta))\rfloor, one has x/2l<4δ/(xlog(x+1))αx/2^l<4\delta/(x\log(x+1))^\alpha, not the printed upper bound with 2δ2\delta. Consequently the first term in Lemma 4 doubles. Carrying the corrected factor through both sums in Lemma 5 gives an upper loss bounded by 211δ(1+log(1/δ))2^{11}\delta(1+\log(1/\delta)), which is still below one half for 0<δ2200<\delta\leq2^{-20}. Thus the same measure-halving induction remains valid.

Final inductionIncomplete as written · verified repair

The recursion cannot start from q0=0q_0=0

Page 8 · last paragraph of the proof of Theorem 1 · arXiv:0712.2423v2

The definition qν+1=(qν2/δ)log(qν2/δ)+1q_{\nu+1}=\lfloor(q_\nu^2/\delta)\log(q_\nu^2/\delta)\rfloor+1 is undefined at the printed choice q0=0q_0=0. Start instead with any sufficiently large positive q0q_0 for which the finite initial deletion leaves positive measure; the direct union bound used earlier supplies such a base. Lemma 5 then applies to every subsequent pair, and compactness of the nested closed sets gives the required point.

03Novelty0 reported findingsNo non-novelty findings

No non-novelty findings.

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Paper
arXiv:0712.2423v2
Authors listed
Nikolay G. Moshchevitin
Audit date
August 20, 2026
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