Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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A set of self-density at most c has a subset of at least half its size that is 4c-sparse

Statement

Let G be a finite simple graph, let c≥0, and let X⊆V(G) be nonempty. If dG(X,X)≤c, then there is a subset X′⊆X with ∣X′∣≥∣X∣/2 such that X′ is 4c-sparse.

Facts & Assumptions

Given: A finite simple graph G, a real c≥0, and a nonempty set X⊆V(G) with dG(X,X)≤c.

[L2]

The self-density inequality dG(X,X)≤c is equivalent to eG(X,X)≤c∣X∣2 (Edge counts and densities between nonempty vertex sets).

[L3]

A set is 4c-sparse exactly when every vertex of the induced graph on it has degree at most 4c times its size (A set is c-sparse exactly when the maximum degree of the graph it induces is at most c times its size, c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · direct
1.1L1L2algebra

By [L1] and [L2], the average internal degree of a vertex of X is at most c∣X∣.

2.1step 1.1algebrachoose

Let B:={x∈X:deg⁡G[X](x)>2c∣X∣}. If ∣B∣≥∣X∣/2, then the sum of the nonnegative internal degrees would be strictly larger than ∣B∣ 2c∣X∣≥c∣X∣2, contradicting step 1.1. Hence X′:=X∖B has ∣X′∣>∣X∣/2, and every x∈X′ has deg⁡G[X](x)≤2c∣X∣.

3.1step 2.1L3algebra∎

For x∈X′ one has deg⁡G[X′](x)≤deg⁡G[X](x)≤2c∣X∣≤4c∣X′∣, because ∣X′∣≥∣X∣/2. Thus [L3] makes X′ 4c-sparse.

Depends on

Used by

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources