Recognition: 2 theorem links
· Lean TheoremA solution to a strengthened conjecture of Bukh, van Hintum and Keevash on additive bases
Pith reviewed 2026-05-12 04:23 UTC · model grok-4.3
The pith
If S is any basis of R^n and S+S sits inside A+B with |A| at most n-t, then |B| must be at least n plus binom(t+1,2).
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
We prove the full strengthened statement over R^n: if S+S ⊆ A+B and |A| ≤ n-t with 0 ≤ t ≤ n-1, then |B| ≥ n + binom(t+1,2), which is sharp for every basis S and every 0 ≤ t ≤ n-1. The proof is short, using edge contractions in a graph-theoretical framework and a new coloring lemma over F_2^n.
What carries the argument
Graph modeling of the containment S+S ⊆ A+B, reduced via edge contractions to a new coloring lemma on F_2^n.
If this is right
- The bound is attained for explicit choices of A and B for every basis S.
- The original conjecture over Q^n follows immediately as a special case.
- When t=0 the result forces |A|+|B| ≥ 2n whenever A+B covers S+S.
- The same minimal sizes are forced for every choice of basis S.
Where Pith is reading between the lines
- The reduction to an F_2^n coloring problem may adapt to sumset questions in other abelian groups.
- Similar strengthened bounds could be sought for higher-order sumsets such as S+S+S.
- The sharpness constructions may supply minimal covering sets in other vector-space problems.
Load-bearing premise
S must be a linearly independent spanning set for R^n, since this property is required for the sumset condition to translate into a graph on which the F_2^n coloring lemma applies.
What would settle it
A concrete counterexample would be any basis S of R^n, any set A of size n-t, and any set B with fewer than n + binom(t+1,2) elements such that S+S is still contained in A+B.
read the original abstract
Motivated by the change-of-domain problem for additive bases, Bukh, van Hintum and Keevash conjectured that if \(A,B\subseteq \mathbb{Q}^{n}\) and \(\{\boldsymbol{e}_i+\boldsymbol{e}_j:1\le i\le j\le n\}\subseteq A+B,\) then \(|A|+|B|\ge 2n\). They further proposed the strengthened conjecture: if \(|A|=n-t\), then \(|B|\ge n+\binom{t+1}{2}.\) Bukh also explicitly asked whether the same bounds hold for \(A,B\subseteq \mathbb{R}^{n}\) and an arbitrary basis \(S\) of \(\mathbb{R}^{n}\), under the assumption \(S+S\subseteq A+B\). We prove the full strengthened statement over \(\mathbb{R}^{n}\): if \(S+S\subseteq A+B\) and \(|A|\le n-t\) with \(0\le t\le n-1\), then \(|B|\ge n+\binom{t+1}{2},\) which is sharp for every basis \(S\) and every \(0\le t\le n-1.\) The proof is short, using edge contractions in a graph-theoretical framework and a new coloring lemma over \(\mathbb F_2^n\).
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript proves the strengthened form of the Bukh-van Hintum-Keevash conjecture over the reals: for any basis S of R^n and subsets A,B of R^n satisfying S+S ⊆ A+B with |A| ≤ n-t (0 ≤ t ≤ n-1), one has |B| ≥ n + binom(t+1,2). The bound is shown to be sharp by explicit constructions that work for every basis S. The argument models the sumset condition as a graph on the basis vectors, performs edge contractions, and invokes a new coloring lemma over F_2^n.
Significance. If the proof is correct, the result fully resolves the strengthened conjecture in the general real-vector-space setting with arbitrary bases, extending the original rational case. The short, direct derivation (no free parameters, no circular reductions) and the matching sharpness examples for all admissible t constitute a clean contribution to additive combinatorics. The new F_2^n coloring lemma is a reusable tool whose introduction strengthens the paper.
minor comments (1)
- A one-sentence informal statement of the new coloring lemma in the abstract or introduction would improve readability for readers who do not immediately consult the body.
Simulated Author's Rebuttal
We thank the referee for the positive assessment of our manuscript and for recommending acceptance. We appreciate the recognition of the result's significance, the shortness of the proof, and the utility of the new coloring lemma.
Circularity Check
No significant circularity; derivation self-contained
full rationale
The paper supplies an explicit short proof of the strengthened bound for arbitrary bases S of R^n. It models the sumset condition S+S ⊆ A+B as a graph on basis elements, performs edge contractions, and invokes a new coloring lemma over F_2^n that is stated and proved within the manuscript. No load-bearing step reduces by definition to the target bound, no parameter is fitted and renamed as a prediction, and no self-citation chain is used to justify the central claim. The sharpness constructions are explicit and independent of the proof. This matches the default expectation for a self-contained combinatorial proof paper.
Axiom & Free-Parameter Ledger
axioms (2)
- standard math R^n forms a vector space over the reals under componentwise addition and scalar multiplication
- domain assumption Standard notions of graphs, edge contraction, and vertex coloring apply to the auxiliary graph constructed from the sumset condition
invented entities (1)
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New coloring lemma over F_2^n
no independent evidence
Lean theorems connected to this paper
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IndisputableMonolith/Foundation/AbsoluteFloorClosure.leanreality_from_one_distinction unclearWe construct a bipartite graph G ... contract all diagonal edges εii ... color the vertices of H by elements of the quotient group R^n / 2Z^n ... Lemma 2.2 ... Dn ⊆ X−C
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IndisputableMonolith/Cost/FunctionalEquation.leanwashburn_uniqueness_aczel unclearif S+S ⊆ A+B and |A| ≤ n−t ... |B| ≥ n + binom(t+1,2)
Reference graph
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discussion (0)
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