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Alon–Pach–Solymosi: if H1 and H2 have the Erdős–Hajnal property, so does the graph obtained from H1 by substituting H2 for a vertex

Statement

Let H1 and H2 be finite simple graphs with the Erdős–Hajnal property, let vV(H1), and suppose the substitution H=H1[vH2] is defined (Substituting one graph for a vertex of another). Then H has the Erdős–Hajnal property.

Facts & Assumptions

Given: Finite simple graphs H1,H2 with the Erdős–Hajnal property, a vertex vV(H1), and the substitution H=H1[vH2]; write h=V(H1), so h1.

[F1]

A real ϵ>0 is an Erdős–Hajnal constant for a hereditary class C when every nonempty JC satisfies hom(J)V(J)ϵ; a finite graph K has the Erdős–Hajnal property when the class of K-free graphs has such a constant (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[F2]

hom(G)=max{ω(G),α(G)}, where ω(G) and α(G) are the largest cardinalities of a clique and of a stable set of G (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}, Cliques, stable sets, the clique number ω(G) and stability number α(G)).

[F3]

G is K-free when G has no induced copy of K, an induced copy being the image of an induced embedding (H-free and F-free graphs under the induced-subgraph convention, Induced embeddings and induced copies of a graph).

[L1]

For every family F of finite graphs, the class of F-free finite graphs is hereditary (Every class defined by forbidden induced subgraphs is hereditary).

[L2]

If ϵ>0 is an Erdős–Hajnal constant for the class of K-free graphs and WV(G) is nonempty with Wϵ>hom(G), then G[W] has an induced copy of K (If ϵ is an Erdős–Hajnal constant for H and W is a nonempty vertex set with Wϵ>hom(G), then G[W] has an induced copy of H).

[L3]

If hmn=V(G) and every m-element WV(G) has an h-element subset S with G[S]H1, then the number of h-element sets SV(G) with G[S]H1 is at least (nh+1)h/mh (If every m-element vertex set contains an induced copy of H, then at least (nh)/(mh) of the h-element vertex sets induce a copy of H).

[L4]

With Ψ the set of induced embeddings of H1v into G and Xφ the extension set of φΨ, one has indH1(G)=φΨXφ and Ψnh1 (The induced copies of H1 in G are counted by summing, over the induced embeddings of H1v, the number of vertices that extend them at v).

[L5]

If φΨ and ψ is an induced embedding of H2 into G whose image lies in Xφ, then G has an induced copy of H=H1[vH2] (An induced copy of H2 inside the extension set of an induced embedding of H1v yields an induced copy of H1 with H2 substituted for v).

[L6]

For finite sets X and Y and a relation RX×Y with row fibres Rx, there is x+X with Rx+R/X (If X is nonempty, some row fibre is at least the average size and some row fibre is at most the average size).

[F4]

indK(G) is the number of induced embeddings of K into G (The induced-embedding count indH(G)).

[F5]

G[W]=(W,E(G)[W]2), so two vertices of W are adjacent in G[W] exactly when they are adjacent in G (Subgraphs, induced subgraphs and spanning subgraphs).

[F6]

For a>0 and real x, ax=exp(xloga) (Real powers for positive bases, with the zero-base positive-exponent convention).

[L7]

For a,b>0 and real r,s: ar+s=aras, (ab)r=arbr, (a/b)r=ar/br and (ar)s=ars (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[L8]

The logarithm is continuous and strictly increasing on (0,), is onto R, satisfies log(xy)=logx+logy and log(x/y)=logxlogy, and log1=0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L9]

The exponential is continuous and strictly increasing on R (The exponential function is strictly increasing).

[F7]

For b>0 with b1 and x>0, logbx=logx/logb (The logarithm to a positive base other than one).

[L10]

Every complete ordered field is Archimedean: for every x there is a natural number k1 with x<k (Every complete ordered field is Archimedean).

[L11]

Every nonempty subset of N has a least element (The well-ordering principle).

Proof

technique · direct
1.1

Choose Erdős–Hajnal constants ϵ1>0 for the class of H1-free graphs and ϵ2>0 for the class of H2-free graphs, and set β=1/(2(h+1)), ϵ0=βmin{ϵ1,ϵ2}, n0=max{416h,2h,h2(h+1)} and ϵ=min{ϵ0,1/log2n0}. Then 0<β1/4, ϵ0>0, n064 and ϵ>0, and β(h+1)=1/2 gives 1βh=1/2+β.

givenF1F7L8choosealgebra
1.2

For a1 and st one has asat, because loga0 and both exp and log are increasing; and for 0<ab and r>0 one has arbr for the same reason.

F6L8L9
1.3

By [L1] the class of H-free graphs is hereditary, so it is a class for which [F1] can supply a constant.

L1F1
2.1

Let G be a finite simple graph with n=V(G)n0 and hom(G)<nϵ0. The set of natural numbers k with knβ is nonempty by [L10], so it has a least element m by [L11]; since nβ>0 we have m1, and m1<nβ, so nβm<nβ+12nβ, the last step because n1 gives nβ1.

step 1.1step 1.2L10L11F6choose
2.2

Turning to the small orders, let J be any nonempty H-free graph with n=V(J)<n0. If n2 then two distinct vertices of J form a clique or a stable set, so hom(J)2; and nϵ<n0ϵn01/log2n0=exp(logn0log2/logn0)=2, so hom(J)nϵ. If n=1 then hom(J)=1=1ϵ.

step 1.1step 1.2F2F6F7L8algebra
3.1

Since nh2(h+1)=h1/β, raising to the power β gives nβh, so mh; and β1/4 with n64 gives m<n1/4+1n. Hence hmn.

step 1.1step 2.1step 1.2L7algebra
4.1

Let WV(G) with W=m. Then W and Wϵ1=mϵ1(nβ)ϵ1=nβϵ1nϵ0>hom(G), using ϵ0βϵ1 and n1; so by [L2] the graph G[W] has an induced copy of H1, that is, an h-element SW with G[S]H1.

step 2.1step 3.1step 1.2L2L7F5F3
5.1

By [L3] the number g of h-element sets SV(G) with G[S]H1 is at least (nh+1)h/mh, and each such S carries at least one induced embedding of H1 into G, distinct sets carrying distinct embeddings because their images differ; so indH1(G)g(nh+1)h/mh>0.

step 3.1step 4.1L3F3F4
6.1

By [F4] and step 5.1 the set Φ of induced embeddings of H1 into G is nonempty, so the set Ψ of [L4] is nonempty as well, since each member of Φ restricts into it. Applying [L6] to the relation pairing φΨ with the members of Φ restricting to it, whose row fibres have sizes Xφ and whose total size is indH1(G) by [L4], gives φ+Ψ with Xφ+indH1(G)/Ψ.

step 5.1L4L6F4
7.1

Since Ψnh1 and nn02h gives nh+1>n/2, and m<2nβ by step 2.1, we get Xφ+(nh+1)hmhnh1>(n/2)h2hnβhnh1=n1βh4h=4hn1/2nβ.

step 1.1step 2.1step 5.1step 6.1L4L7algebra
8.1

From n416h we get n1/224h, so 4hn1/22 and step 7.1 gives Xφ+>2nβ>mnβ. In particular Xφ+ is nonempty.

step 1.1step 2.1step 7.1step 1.2L7algebra
9.1

Therefore Xφ+ϵ2(nβ)ϵ2=nβϵ2nϵ0>hom(G), so by [L2] the graph G[Xφ+] has an induced copy of H2; the corresponding induced embedding has image inside Xφ+ and, adjacency in G[Xφ+] agreeing with adjacency in G, it is an induced embedding of H2 into G.

step 2.1step 8.1step 1.2L2L7F5F3
10.1

By [L5] applied to φ+ and that embedding, the graph G of step 2.1 has an induced copy of H and so is not H-free. Hence an H-free graph G with V(G)=nn0 cannot satisfy hom(G)<nϵ0, and therefore satisfies hom(G)nϵ0nϵ, the last inequality by step 1.2 and ϵϵ0.

step 2.1step 9.1step 1.2L5F3
11.1

Steps 10.1 and 2.2 together give hom(G)V(G)ϵ for every nonempty H-free graph G, whatever its order, and the class of H-free graphs is hereditary by step 1.3; that is, ϵ is an Erdős–Hajnal constant for it and H has the Erdős–Hajnal property.

step 1.3step 10.1step 2.2F1

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