ProbXiv
sign in
Problem archiveProblem record

Wegner's Piercing Conjecture for Rectangles

Geometry & topology · posed by Gerd Wegner, 1965 · disproved

2 comments

Statement

Wegner conjectured in 1965 that every finite family R\mathcal{R} of axis-parallel rectangles satisfies τ(R)≤2ν(R)−1\tau(\mathcal{R}) \le 2\nu(\mathcal{R}) - 1, where τ\tau is the minimum number of piercing points and ν\nu the largest pairwise-disjoint subfamily. False, by an explicit triangle-free counterexample.

Record

Source
Added

Comments

No person has examined this. Nothing here has been checked at all. say whether it holds →

  1. construction · #1

    Deepak Ajwani, Rishikesh Gajjala and Rajiv Raman, using GPT-5.5 Pro, Codex

    That credit came with the record as it was imported. No ProbXiv account is credited for this work, and nobody has answered for it here.

    AI involvement
    ai co developed
    — a person and a model developed the result together.

    The authors say the construction of the initial counterexamples relied heavily on trial and error, and that GPT-5.5 Pro was used extensively to search for suitable constructions. Codex drew the figures and drafted portions of the text. They note the correctness of the proofs does not rest on the auxiliary machine verifications.

  2. Recorded elsewhere on #1 · not checked here

    recorded: correctVibeMathed site check

    scope Reproduction by the VibeMathed site

    The refutation is an explicit finite family of rectangles, so it reduces to a finite piercing computation. arXiv preprint, not peer-reviewed.

    Repeated from the source; nothing was checked here.

Sign in with an institutional address to take part in the discussion. Reading every thread stays open to everyone.

Sign in

Solve with an agent

Open the statement in a chat, with the problem and the ground rules already written into the prompt.

open in Claude →open in ChatGPT →

This opens a third-party site. Nothing is posted back to ProbXiv and nothing you write there is recorded here — what a model gives you is an attempt, which a person still has to check.