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CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31
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An H-free graph has a linearly large induced subgraph whose graph or complement has bounded maximum degree

Statement

For every finite graph H and every real ϵ>0, there exists δ>0 such that every nonempty H-free graph G contains a set X⊆V(G) with ∣X∣≥δ∣V(G)∣ for which one of G[X] or G‾[X] has maximum degree at most ϵδ∣V(G)∣.

Facts & Assumptions

Given: A finite graph H, a real ϵ>0, and a nonempty H-free graph G.

[L1]

For every graph H and every real η∈(0,12), there exists δ0>0 such that every nonempty H-free graph contains a set Z⊆V(G) with ∣Z∣≥δ0∣V(G)∣ and either dG(Z,Z)≤η or dG(Z,Z)≥1−η (The edge-density form of Rödl's theorem: every nonempty H-free graph has a linearly large set of self-density at most ϵ or at least 1−ϵ).

[L2]

If a graph on n vertices has at most λn(n−1)/2 edges, then for every integer m with 1≤m≤(n+1)/2 it has an m-vertex induced subgraph of maximum degree at most 4λ(m−1) (A sparse graph has a prescribed-size induced subgraph of bounded maximum degree).

Proof

technique · direct
1.1givenL1choosecases

Put η:=ϵ/8. If η≥1/2, then ϵ≥4. The conclusion is trivial with δ:=1, because every induced subgraph has maximum degree at most ∣V(G)∣−1<∣V(G)∣≤ϵ∣V(G)∣=ϵδ∣V(G)∣. So we may assume η<1/2, and [L1] gives δ0>0 for the parameter η. Set δ:=min⁡{δ0,1}/2.

2.1step 1.1L1choosealgebra

Apply [L1] to the given nonempty H-free graph G. Then there is Z⊆V(G) with ∣Z∣≥δ0∣V(G)∣≥2δ∣V(G)∣ and either dG(Z,Z)≤η or dG(Z,Z)≥1−η. Let m:=⌈δ∣V(G)∣⌉. Since ∣Z∣ is an integer and ∣Z∣≥2δ∣V(G)∣, we have ∣Z∣≥⌈2δ∣V(G)∣⌉≥2m−1, so m≤(∣Z∣+1)/2.

3.1step 2.1L2algebracases

First suppose dG(Z,Z)≤η. Writing n:=∣Z∣, the definition of density gives 2∣E(G[Z])∣/n2≤η. If n=1, then m=1 and Z itself already has maximum degree 0≤ϵδ∣V(G)∣. If n≥2, then n2≤2n(n−1), so ∣E(G[Z])∣≤ηn2/2≤(ϵ/4)n(n−1)/2. Applying [L2] with λ=ϵ/4 gives a set X⊆Z with ∣X∣=m and maximum degree at most 4(ϵ/4)(m−1)=ϵ(m−1)≤ϵδ∣V(G)∣.

4.1step 2.1step 3.1algebra

Now suppose instead that dG(Z,Z)≥1−η. If n:=∣Z∣, then dG‾(Z,Z)=1−1n−dG(Z,Z)≤η−1n≤η. Applying step 3.1 to the complement graph on the same vertex set yields X⊆Z with ∣X∣=m such that G‾[X] has maximum degree at most ϵδ∣V(G)∣.

5.1step 3.1step 4.1cases-exhaustive∎

In either case there is a set X with ∣X∣=m≥δ∣V(G)∣ such that one of G[X] or G‾[X] has maximum degree at most ϵδ∣V(G)∣.

Depends on

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