Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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Quantitative density theorem for ell divisive graphs

Statement

Let H be a nonempty finite graph that is ℓ-divisive for a subreciprocal function ℓ. There is CH,ℓ>0 such that, for 0<ϵ<1/2 and δ=2−CH,ℓlog⁡2(1/ϵ)2/log⁡2ℓ(ϵ), every nonempty finite graph G with ind⁡H(G)≤(δ∣G∣)∣H∣ has a nonempty S⊆V(G) of size at least δ∣G∣ with e(G[S])≤ϵ(∣S∣2) or e(G‾[S])≤ϵ(∣S∣2).

Facts & Assumptions

Given: A nonempty ℓ-divisive pattern H, a subreciprocal ℓ, and a nonempty host satisfying the copy bound for the fraction specified at each stage below.

[F1]

For the stated subreciprocal function and witnesses, the parameter construction has p>1, 0<η<1, δ<xdηt, and t the least natural with pt≥ϵ−2. (Admissible parameters for the density recursion).

[F2]

With the parameter setup of the amplification lemma, a fixed nonempty G satisfying ind⁡H(G)≤(δ∣G∣)∣H∣ and δ∣G∣>1 has βs−1(u,v)≥ηmin⁡{βs(pu,v),βs(u,pv)} for 1≤s≤t and u,v≥ϵ. (Ell divisibility amplifies through a blockade).

[F3]

From Qid finite density recursion profile: If a≥1 or b≥1, every full F qualifies and βs(a,b)=1. At s=0, the only qualifying set is V(G), so β0(a,b)=ρG(a,b).

Proof

1.1F1given

Choose divisibility witnesses c,d, put z=ℓ(c)−1/2, b=2−log⁡2(1−z)>2, and C0=20bd. First take 0<ϵ<c, and define δ0 by the displayed formula with C0. If δ0∣G∣≤1, any singleton S has the required size and zero edges in both graphs. Otherwise the parameter construction [F1] supplies x,p,η,t and δ0<xdηt<ηt; all hypotheses of [F2] hold with δ=δ0.

2.1F2step 1.1algebra

For each integer 0≤r≤t, the recurrence implies β0(ϵ,ϵ)≥ηrmin⁡0≤i≤rβr(piϵ,pr−iϵ). At r=0 this is equality. To pass from r<t to r+1, apply [F2] with s=r+1 to every pair (piϵ,pr−iϵ); both coordinates are at least ϵ because p>1. The two children have exponent pairs (i+1,r−i) and (i,r+1−i), whose union over i is exactly all pairs summing to r+1. Taking their finite minimum proves the induction step.

3.1F1F3step 2.1

At r=t, the product of the two arguments in every terminal pair is ptϵ2≥1 by the least-natural property in [F1]. At least one argument is therefore at least 1. Each terminal profile value equals 1 by [F3]. Hence ρG(ϵ,ϵ)=β0(ϵ,ϵ)≥ηt>δ0. The attained maximum defining ρG supplies a nonempty set of at least δ0∣G∣ vertices with one of the required edge bounds. Together with the singleton case, this proves the theorem on (0,c) with constant C0.

4.1step 1.1step 3.1algebra

For the full interval set a=log⁡2(1/c)>1 and CH,ℓ=a2C0. Given 0<ϵ<1/2, put ϵ′=ϵa<c and let δ′ be the small-interval fraction at ϵ′ with constant C0. Since ϵ′≤ϵ and ℓ is nonincreasing, log⁡2ℓ(ϵ′)≥log⁡2ℓ(ϵ)>0. Thus log⁡2(1/δ′)=C0a2log⁡2(1/ϵ)2/log⁡2ℓ(ϵ′)≤log⁡2(1/δ), so δ′≥δ.

5.1step 3.1step 4.1algebra∎

The original hypothesis implies ind⁡H(G)≤(δ′∣G∣)∣H∣. Apply the small-interval result to ϵ′: its set has size at least δ′∣G∣≥δ∣G∣, and its edge bound with ϵ′ implies that with ϵ. This establishes the claimed constant on the entire open interval.

Source notes

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 5.2 complete proof.

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Sources