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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Iterated sparse restriction reaches the target sparsity threshold

Statement

Let c∈(0,1), b1>1, b2,b3>0, and assume b1b2≥b2+b3. Suppose that x∈(0,c) and that a graph G satisfies:

  1. G has a c-sparse induced subgraph with at least cb2∣V(G)∣ vertices; and
  2. for every λ∈[x,c] and every λ-sparse induced subgraph F of G with ∣V(F)∣≥λb2∣V(G)∣, there is a λb1-sparse induced subgraph of F with at least λb3∣V(F)∣ vertices.

Then G contains an x-sparse induced subgraph with at least xb1b2∣V(G)∣ vertices.

Facts & Assumptions

Given: The parameters and hypotheses in the statement.

[L1]

A λ-sparse vertex set is nonempty, and every vertex has degree at most λ times the size of that set inside the induced subgraph (c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · direct
1.1givenL1

Hypothesis 1 supplies a nonempty c-sparse vertex set, because c-sparse sets are nonempty by [L1]. Hence ∣V(G)∣>0.

1.2givenchooseL1algebra

For each nonempty induced subgraph E of G, let μ(E) be its maximum degree divided by ∣V(E)∣; by [L1], the graph E is λ-sparse exactly when μ(E)≤λ. Define λ(E):=max⁡(xb1,μ(E)). Hypothesis 1 gives a c-sparse induced subgraph E0 with at least cb2∣V(G)∣ vertices, so λ(E0)≤c and ∣V(E0)∣≥λ(E0)b2∣V(G)∣. Because G has only finitely many induced subgraphs, the set of values λ(E) with λ(E)≤c and ∣V(E)∣≥λ(E)b2∣V(G)∣ has a minimum. Choose an induced subgraph F for which that minimum is attained, and write λ:=λ(F).

2.1step 1.2givenalgebraassume-contradischarge-contradiction

Suppose that λ≥x. Then hypothesis 2 applies to F and produces a λb1-sparse induced subgraph F′⊆F with at least λb3∣V(F)∣≥λb2+b3∣V(G)∣≥λb1b2∣V(G)∣ vertices, where the last inequality uses b1b2≥b2+b3. Since F′ is λb1-sparse, we have μ(F′)≤λb1 and therefore λ(F′)=max⁡(xb1,μ(F′))≤λb1<λ. Also ∣V(F′)∣≥λ(F′)b2∣V(G)∣. This contradicts the minimal choice of λ. Therefore λ<x.

3.1step 1.2step 2.1algebra

Since λ=λ(F), step 1.2 gives xb1≤λ, and step 2.1 gives λ<x. Because F is λ-sparse, it is also x-sparse. Moreover ∣V(F)∣≥λb2∣V(G)∣≥(xb1)b2∣V(G)∣=xb1b2∣V(G)∣.

4.1step 3.1∎

The induced subgraph F from step 3.1 is the required x-sparse induced subgraph.

Depends on

Used by

Dependency tree · two levels

6 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