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Every H-free graph has a homogeneous set of size at least 2clog2n

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)2cHlog2n. Equivalently, G has a clique or a stable set of size at least 2cHlog2n.

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]

Fox-Sudakov 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))2n and one of G[S] and G[S] has at most x(S2) edges (Fox–Sudakov: a quantitative density form of Rödl's 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. Write L:=log2n and set x:=2L/(2CH). Since n2, one has L>0, so 0<x<1/2.

L2choose
2.1

Because log2(1/x)=L/(2CH), [L2] gives a set SV(G) with S2CH(log2(1/x))2n=2L/2n=n, 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)=2E(F)/S22x(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] every vertex of G[X] has degree at most 4xX, so [L5] gives χ(G[X])4xX+1, and then [L6] yields α(G[X])X/(4xX+1).

step 2.1L3L4L5L6
3.2

If F=G[S], then [L3] gives XS with XS/2 and X 4x-sparse in G. Applying [L4], [L5], and [L6] inside the complement shows that G[X] has a stable set of size at least X/(4xX+1), and [L7] turns that stable set into a clique of the same size in 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/2L/(2CH), choose NH2 so that xn1 whenever nNH. For such n, step 4.1 gives 4xX2, hence 4xX+18xX, so Y1/(8x)=2L/(2CH)3.

step 4.1step 1.1choosealgebra
6.1

Set cH:=1/(42CH). Choose NHNH so that L/(2CH)3cHL whenever nNH. Then step 5.1 gives Y2cHL for all nNH. Shrinking cH if necessary handles the finitely many integers 2n<NH, because every nonnull graph has hom(G)1.

step 5.1L1choosealgebra
7.1

Therefore every nonnull finite H-free graph G with V(G)=n2 satisfies hom(G)2cHlog2n.

step 6.1L1

Depends on

Used by

Dependency tree · two levels

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