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Every H-free graph partitions into boundedly many vertex sets of self-density at most ϵ or at least 1−ϵ

Statement

Fix a graph H and a real ϵ∈(0,12). Then there exists an integer B=B(H,ϵ) such that every H-free finite simple graph G admits a partition

V(G)=X1⊔⋯⊔Xm,m≤B,

in which every part satisfies dG(Xi,Xi)≤ϵ or dG(Xi,Xi)≥1−ϵ.

Facts & Assumptions

Given: A graph H and a real ϵ∈(0,12).

[L1]

The edge-density form of Rödl's theorem supplies a constant δ0>0 such that every nonempty H-free finite simple graph J contains a set X⊆V(J) with ∣X∣≥δ0∣V(J)∣ and either dJ(X,X)≤ϵ/2 or dJ(X,X)≥1−ϵ/2 (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).

[F1]

If W⊆V(G) then G[W] has the same adjacencies on every subset X⊆W as G does, so dG[W](X,X)=dG(X,X) (Subgraphs, induced subgraphs and spanning subgraphs, Edge counts and densities between nonempty vertex sets).

Proof

technique · direct
1.1L1algebrachoose

Let δ0>0 be the constant from [L1], set ρ:=min⁡{δ0,12}, and note that 0<ρ≤12. Then every nonempty H-free graph J contains a set X⊆V(J) with ∣X∣≥ρ∣V(J)∣ and either dJ(X,X)≤ϵ/2 or dJ(X,X)≥1−ϵ/2.

2.1step 1.1L3choose

Since 0<1−ρ<1, [L3] applied to r=1−ρ yields a natural number t≥1 with (1−ρ)t<ρϵ/16.

2.2step 1.1L2F1choose

Let G be an H-free finite simple graph. If V(G)=∅, the empty partition works, so assume n:=∣V(G)∣>0. Put R0:=V(G). For each i=1,…,t, if Ri−1=∅ stop; otherwise the induced subgraph G[Ri−1] is H-free by [L2], so step 1.1 and [F1] give a nonempty set Xi⊆Ri−1 with ∣Xi∣≥ρ∣Ri−1∣ and either dG(Xi,Xi)≤ϵ/2 or dG(Xi,Xi)≥1−ϵ/2. Define Ri:=Ri−1∖Xi.

3.1step 2.2algebra

If the process stops at some stage i≤t because Ri−1=∅, then the nonempty extracted sets X1,…,Xi−1 partition V(G) and each already has self-density at most ϵ/2 or at least 1−ϵ/2, hence in particular at most ϵ or at least 1−ϵ.

3.2step 2.1step 2.2algebra

Assume now that the process does not stop before stage t. Then X1,…,Xt are pairwise disjoint nonempty sets, and an induction on i using step 2.2 gives ∣Ri∣≤(1−ρ)∣Ri−1∣ for every i≤t. Hence ∣Rt∣≤(1−ρ)tn<(ρϵ/16)n by step 2.1, while ∣X1∣≥ρn by step 2.2.

4.1step 3.2algebra

Put Y1:=X1∪Rt and Yi:=Xi for 2≤i≤t. Then Y1,…,Yt partition V(G). Writing x:=∣X1∣, r:=∣Rt∣, y:=∣Y1∣=x+r and α:=r/x, step 3.2 gives α<ϵ/16.

5.1step 4.1algebra

If dG(X1,X1)≤ϵ/2, then eG(Y1,Y1)≤eG(X1,X1)+2ry because the new ordered edges are those incident with at least one vertex of Rt. Therefore dG(Y1,Y1)≤(ϵ/2)(x/y)2+2r/y≤ϵ/2+2α<ϵ.

5.2step 4.1algebra

If instead dG(X1,X1)≥1−ϵ/2, then dG(Y1,Y1)≥dG(X1,X1)(x/y)2≥(1−ϵ/2)/(1+α)2. Since α<ϵ/16 and ϵ<1/2, one has (1+α)2<(1+ϵ/16)2≤1+ϵ/2, so dG(Y1,Y1)>(1−ϵ/2)/(1+ϵ/2)>1−ϵ.

6.1step 2.2step 4.1step 5.1step 5.2

In the situation of steps 3.2 and 4.1, step 5.1 or 5.2 handles Y1, while each Yi=Xi for i≥2 already has self-density at most ϵ/2 or at least 1−ϵ/2. Therefore every part of the partition has self-density at most ϵ or at least 1−ϵ.

7.1step 2.1step 3.1step 6.1∎

Step 3.1 settles the case where the extraction stops early, and step 6.1 settles the case where it reaches stage t. So B:=t works for every H and ϵ.

Depends on

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