Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 c0, and let XV(G) be nonempty. If dG(X,X)c, then there is a subset XX with XX/2 such that X is 4c-sparse.

Facts & Assumptions

Given: A finite simple graph G, a real c0, and a nonempty set XV(G) with dG(X,X)c.

[L2]

The self-density inequality dG(X,X)c is equivalent to eG(X,X)cX2 (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
1.1

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

L1L2algebra
2.1

Let B:={xX:degG[X](x)>2cX}. If BX/2, then the sum of the nonnegative internal degrees would be strictly larger than B2cXcX2, contradicting step 1.1. Hence X:=XB has X>X/2, and every xX has degG[X](x)2cX.

step 1.1algebrachoose
3.1

For xX one has degG[X](x)degG[X](x)2cX4cX, because XX/2. Thus [L3] makes X 4c-sparse.

step 2.1L3algebra

Depends on

Used by

Dependency tree · two levels

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Sources