Gil Kalai’s Helly-Type Question for Two-Component Convex Sets

For integers d2d\ge 2 and nd+2n\ge d+2, let F1,,FnRdF_1,\ldots,F_n\subseteq\mathbb{R}^d be sets, each of which is the union of exactly two disjoint, nonempty, closed convex sets. If, for every I{1,,n}I\subseteq\{1,\ldots,n\} with I{1,2,3,d+1}\lvert I\rvert\in\{1,2,3,d+1\}, the intersection iIFi\bigcap_{i\in I}F_i consists of exactly two nonempty closed convex components, must the total intersection i=1nFi\bigcap_{i=1}^nF_i also consist of exactly two nonempty closed convex components?

Progress summary

Solved

A 2026 preprint claims to settle the question positively in dimensions two and higher, while exhibiting a one-dimensional counterexample.

Gil Kalai posed this question in the 20202020 Discrete Geometry meeting’s open-problems collection: does the required two-piece structure for sufficiently small intersections force the whole intersection to be nonempty? The claimed theorem gives a stronger conclusion, preserving exactly two pieces globally.

2026 claimed proof

Menara’s preprint claims that for d2d\geq 2 and nd+2n\geq d+2, it suffices to check intersections of sizes 11, 22, 33, and d+1d+1; the full intersection then has exactly two nonempty closed convex components. The proof labels components over F2\mathbb{F}_2, uses triple compatibility, and applies Helly’s theorem. It also gives four subsets of R\mathbb{R} showing that d=1d=1 fails. No independent verification, objection, or retraction was found, so the claim remains unverified.

Current status (as of August 2026): The positive result is claimed in an arXiv preprint for d2d\geq 2, with d=1d=1 disproved by an explicit example; formal acceptance and independent verification remain open.

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arXiv

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