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Every finite graph has a linearly large ϵ-self-regular vertex subset

Statement

For every 0<ϵ<1 there is δ=δ(ϵ)>0 such that every finite graph G with at least one vertex has a nonempty set WV(G) with WδV(G) and (W,W) is ϵ-regular. The hypothesis that G is nonempty cannot be dropped: ϵ-self-regularity is defined only for nonempty vertex sets.

Facts & Assumptions

Given: 0<ϵ<1 and a finite graph G with at least one vertex.

[L1]

For arbitrarily small parameters and prescribed minimum part counts, every sufficiently large graph has a bounded equitable regular partition (Szemerédi regularity lemma with an equitable partition and an explicit tower-type upper bound for graphs of order at least m0).

[L2]

A set W is ϵ-self-regular when every two subsets of W of size at least ϵW have density within ϵ of d(W,W) (ϵ-regular pairs and self-regular vertex sets).

Proof

technique · direct
1.1

Choose an integer sϵ3 and divide [0,1] into q=8/ϵ intervals of length at most ϵ/8. Repeated pigeonhole selection gives an integer R=R(q,s) such that every q-colouring of the pairs of an R-set has a monochromatic s-set: select successively a vertex and a colour occurring on at least a 1/q fraction of its remaining incident pairs, and take the initial set large enough for s selections.

givenchooseinduction
2.1

Choose ρ>0 much smaller than R2ϵ3, apply [L1] at parameter ρ and minimum part count much larger than R/ϵ, and let M be the resulting upper bound on the number of parts.

step 1.1L1choose
3.1

In the graph on the partition indices whose edges are the regular cross-pairs, fewer than 2ρk2 pairs are missing by equitability. If every R-set contained a missing pair, double-counting pairs inside R-sets would force at least (k2)/(R2) missing pairs, contrary to the choice of ρ. Hence there is an R-set of indices all of whose cross-pairs are ρ-regular.

step 2.1L1algebra
4.1

Colour those regular pairs by the interval containing their density. Step 1.1 supplies s parts V1,,Vs whose cross-densities all lie in one interval; let W=V1Vs. Since the partition is equitable and has at most M parts, Wsn/(2M).

step 1.1step 3.1choosealgebra
5.1

Let A,BW have size at least ϵW, and write Ai=AVi, Bj=BVj. Pairs with i=j, or with Ai<ρVi or Bj<ρVj, contribute at most 3/(ϵ2s)+4ρ/ϵ to the normalized density comparison.

step 2.1step 4.1algebra
6.1

On every remaining pair, regularity gives d(Ai,Bj)d(Vi,Vj)ρ, while the density colour in step 4.1 makes any two cross-densities differ by at most ϵ/8. Decomposing both d(A,B) and d(W,W) over the s2 pairs and using step 5.1 therefore gives d(A,B)d(W,W)ϵ by the choices of s and ρ.

step 4.1step 5.1L2algebra
7.1

For graphs large enough for [L1], steps 4.1 and 6.1 give an ϵ-self-regular W of size at least sn/(2M). For the finitely many smaller orders n1, a singleton is 0-self-regular and hence ϵ-self-regular. Shrinking δ to the minimum of s/(2M) and the reciprocals of those orders proves the Statement for every finite graph with at least one vertex.

step 2.1step 4.1step 6.1L1L2choose

Depends on

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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