OpenAI 证明 Erdős 相似性猜想后,arXiv 出现人工详细解读
The geometric case of the Erdős Similarity Conjecture. A human exposition of the OpenAI proof
OpenAI 证了 Erdős 相似性猜想,这篇解读只用基础测度论就能看懂证明思路,数学爱好者别错过。
OpenAI 于 2026 年 10 月 6 日发布了对 Erdős 相似性猜想的证明,构造出 [0,1] 中测度可任意接近 1 的紧子集,可避开 {q^n} 的所有平移与伸缩副本。arXiv 2611.02454 给出仅依赖有限概率和基础测度论的人工详细解读。证明核心是嵌套网格上的随机构造:每个副本获得多个独立相交机会,再用有限代表伸缩上的联合界覆盖所有情形。文章还追溯了 Kolountzakis 1997 年的随机格点方法,并说明条件期望法可将构造确定化。
The geometric case of the Erdős Similarity Conjecture. A human exposition of the OpenAI proof
For a fixed $q\in(0,1)$, can a set of positive Lebesgue measure avoid every translated and dilated copy of $\{q^n:n\ge1\}$? OpenAI's release of October 6, 2026 shows that it can, with a compact subset of $[0,1]$ whose measure is arbitrarily close to one. We give a detailed exposition of the proof, using only finite probability and basic measure theory. The proof removes a small open set intersecting every copy, obtained from a random construction: fair random choices, attached to the cells of nested grids, route each point down a finite tree, and at the leaf reached a further random choice removes the point with probability $p$. The removed set therefore has expected density $p$, however large the tree, while the branching gives each copy many independent chances to intersect it. Each chance depends on only part of the tree, so its outcome changes at only a controlled number of dilations, and a union bound over finitely many representative dilations then covers every dilation. We also relate the proof to the earlier random-cell arguments going back to Kolountzakis (Bull. London Math. Soc. 29 (1997), 415--424) and to recent work on two-fold sumsets (arXiv:2607.03584) and on Rajchman measures (arXiv:2609.04456), and we show how the method of conditional expectations makes the construction deterministic.