For α,β,δ∈[0,1],α+β=1 we consider sets BAD∗(α,β;δ)={ξ=(ξ1,ξ2)∈[0,1]2:,p∈Ninfmax{(plog(p+1))α∣∣pξ1∣∣,(plog(p+1))β∣∣pξ2∣∣}≥δ}. We prove that for different (α1,β1),(α2,β2),α1+β1=α2+β2=1 and δ small enough BAD∗(α1,β1;δ)⋂BAD∗(α2,β2;δ)=∅. Our result is based on A. Khintchine's construction and an original method due to Y. Peres and W. Schlag.
AI-generated audit
Audit summary
Audited against arXiv v2
i
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 finding✓Correct⌄
The nonempty-intersection theorem remains valid after verified enlargement of two dyadic-cover constants and a positive starting index.
Theorem 1✓Correct
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 δ≤2−20, so the induction and compactness conclusion are unchanged.
02Proofs3 reported findings×Contains 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 1–3✓Correct
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 5×Incorrect 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))⌋, one has x/2l<4δ/(xlog(x+1))α, not the printed upper bound with 2δ. 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)), which is still below one half for 0<δ≤2−20. Thus the same measure-halving induction remains valid.
Final induction×Incomplete as written · verified repair
The recursion cannot start from q0=0
Page 8 · last paragraph of the proof of Theorem 1 · arXiv:0712.2423v2
The definition qν+1=⌊(qν2/δ)log(qν2/δ)⌋+1 is undefined at the printed choice q0=0. Start instead with any sufficiently large positive q0 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.
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Paper
arXiv:0712.2423v2
Authors listed
Nikolay G. Moshchevitin
Audit date
August 20, 2026
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