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Rests on 1 statement not proved in this library. Every dependency marked below is recorded with a citation but is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Every H-free graph has a homogeneous set of size at least 2clog2nlog2log2n

Statement

Let H be a finite graph. Then there exists a constant cH>0 such that every nonnull finite H-free graph G with n:=V(G)2 satisfies hom(G)2cHlog2nlog2log2n.

Facts & Assumptions

Given: A finite graph H, a nonnull finite H-free graph G, and n:=V(G)2.

[L1]

The homogeneous number is hom(G)=max{ω(G),α(G)} (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L2]

Bucić-Nguyen-Scott-Seymour quantitative density: there exists CH>0 such that for every real x with 0<x<1/2 there is SV(G) with S2CH(log2(1/x))2/log2log2(1/x)n and one of G[S] and G[S] has at most x(S2) edges (Bucić–Nguyen–Scott–Seymour: a log-log quantitative density theorem ).

[L3]

If a nonempty vertex set X satisfies dG(X,X)c, then some XX has XX/2 and is 4c-sparse (A set of self-density at most c has a subset of at least half its size that is 4c-sparse).

[L4]

A nonempty set X is c-sparse exactly when every vertex of G[X] has degree at most cX (A set is c-sparse exactly when the maximum degree of the graph it induces is at most c times its size).

[L5]

Every nonnull finite graph F satisfies χ(F)Δ(F)+1 (The greedy colouring bound χ(G)Δ(G)+1 for every nonnull finite graph).

[L6]

Every finite graph F satisfies V(F)χ(F)α(F) (The bounds ω(G)χ(G) and V(G)χ(G)α(G)).

[L7]

A vertex set is a clique in G if and only if it is a stable set in G (Complementation swaps cliques with stable sets, so ω(G)=α(G)).

[L8]

For nonempty X, the self-density is dG(X,X)=2E(G[X])/X2 (Edge counts and densities between nonempty vertex sets).

Proof

technique · direct
1.1

By [L2], choose a constant CH>0. Because every nonnull graph has hom(G)1, it is enough to prove the bound for all sufficiently large n; assume from now on that n is large enough that L:=log2n4. Set β:=1/(4CH) and x:=2βLlog2L. Then 0<x<1/2.

L1 L2choose
2.1

For large enough L, the inequality βLlog2LL holds, so log2log2(1/x)=log2(βLlog2L)12log2L. Therefore CH(log2(1/x))2/log2log2(1/x)2CHβ2L=L/8. Using [L2], obtain SV(G) with S2L/8n=27L/8n, and one of G[S] and G[S] has at most x(S2) edges. For that chosen graph F on vertex set S, [L8] gives dF(S,S)2x(S2)/S2=x(S1)/Sx.

step 1.1 L2L8algebra
3.1

If F=G[S], then [L3] gives XS with XS/2 and X 4x-sparse in G. By [L4], [L5], and [L6], α(G[X])X/(4xX+1).

step 2.1L3L4L5L6
3.2

If F=G[S], then the same argument inside the complement produces a stable set of G[X] of size at least X/(4xX+1) for some XS with XS/2, and [L7] turns it into a clique of G[X].

step 2.1L3L4L5L6L7
4.1

Steps 3.1 and 3.2 show that G has a homogeneous set Y with YX/(4xX+1) for some XS satisfying XS/2n/2.

step 3.1step 3.2step 2.1L1
5.1

Because xn=2L/2βLlog2L, choose a threshold NH2 so that xn1 whenever nNH. For those n, step 4.1 gives 4xX2, hence 4xX+18xX, and therefore Y1/(8x)=2βLlog2L3.

step 4.1step 1.1choosealgebra
6.1

Set cH:=β/2. For all sufficiently large n, the inequality βLlog2L3cHLlog2L holds, so step 5.1 gives Y2cHLlog2L. Shrinking cH if necessary handles the finitely many smaller values of n.

step 5.1choosealgebra
7.1

Hence every nonnull finite H-free graph G with V(G)=n2 satisfies hom(G)2cHlog2nlog2log2n.

step 6.1L1

Depends on

Used by

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