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Large induced subgraphs in the property (*) four-outcome theorem contain a pure or x-sparse polynomial blockade

Statement

Let F have property (∗) and be leaf-reducible, and let c1′>0, c4′≥4, and d≥58c4′ be the constants from Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem. Fix ϵ∈(0,12), put x:=ϵ5d, and let G be an F‾-free graph with ∣G∣≥ϵ−10d2 such that

  1. G has no clique or stable set of size at least (ϵ156c4′d∣G∣)c1′;
  2. G has no complete or anticomplete (k,∣G∣/k2d)-blockade with k≥ϵ−5d;
  3. G has no ϵ5d-restricted induced subgraph with at least ϵ116c4′d∣G∣ vertices.

Then every induced subgraph F of G with ∣F∣≥ϵd∣G∣ has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1].

Facts & Assumptions

Given: The data and hypotheses in the Statement, together with an induced subgraph F of G satisfying ∣F∣≥ϵd∣G∣.

[L1]

The previous lemma says that every F‾-free graph of size at least x−d satisfies one of four outcomes: an x-restricted induced subgraph of size at least x23c4′ times the ambient order, a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1], a clique or stable set of size at least (x31c4′∣F∣)c1′, or a complete or anticomplete polynomial blockade (Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem).

[L2]

If ∣G∣≥ϵ−10d2 and ∣F∣≥ϵd∣G∣, then ∣F∣≥ϵ−5d2=x−d because d≥1.

[L3]

If k≥x−1=ϵ−5d, then ϵd=x1/5≥k−1/5.

Proof

technique · apply the Rödl-initialized four-outcome theorem to $F$ and use the three standing failure hypotheses to rule out every branch except the pure-or-sparse blockade branch
1.1givenL1algebra

The size hypothesis on G and the bound ∣F∣≥ϵd∣G∣ imply ∣F∣≥ϵdϵ−10d2=ϵ−10d2+d≥ϵ−5d2=x−d, because d≥1. Thus [L1] applies to F.

2.1step 1.1L1

Apply [L1] to the induced subgraph F. One of its four outcomes holds.

3.1step 2.1algebra

If [L1] yields an x-restricted induced subgraph S of F with at least x23c4′∣F∣ vertices, then ∣S∣≥x23c4′ϵd∣G∣=ϵ115c4′d+d∣G∣≥ϵ116c4′d∣G∣, because c4′≥1. This contradicts standing hypothesis 3.

3.2step 2.1algebra

If [L1] yields a clique or stable set of size at least (x31c4′∣F∣)c1′, then x31c4′∣F∣≥x31c4′ϵd∣G∣=ϵ155c4′d+d∣G∣≥ϵ156c4′d∣G∣, so standing hypothesis 1 is contradicted.

3.3step 2.1L3algebra

If [L1] yields a complete or anticomplete (k,∣F∣/kd)-blockade with k≥x−1, then [L3] gives ϵd≥k−1/5, and therefore ∣F∣kd≥ϵd∣G∣kd≥∣G∣k2d. Since k≥x−1=ϵ−5d, this contradicts standing hypothesis 2.

4.1step 3.1step 3.2step 3.3∎

The first three branches are impossible, so the remaining branch of [L1] must hold: F has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1]. This is exactly the desired conclusion.

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