Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-09-26 (gpt-6-sol)
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For large n, the log-log choice of x still leaves a dense-or-sparse set of order at least n

Example

Fix a nonempty finite graph H, let CH>0 be the constant from Loglog quantitative induced density bound, and let G be an H-free graph on n≥3 vertices. Write L:=log⁡2n, and set β:=1/(4CH) and x:=2−βLlog⁡2L.

Facts & Assumptions

Given: The data in the Example.

[L1]

For nonempty H,G and 0<x<1/2, the proved quantitative induced-density theorem gives every H-free n-vertex graph a nonempty set S of order at least 2−CH(log⁡2(1/x))2/log⁡2log⁡2(1/x)n whose induced graph or complement has at most x(∣S∣2) edges (Loglog quantitative induced density bound).

Verification

technique · direct
1.1givenalgebra

For the chosen x, one has log⁡2(1/x)=βLlog⁡2L.

2.1step 1.1algebra

For all sufficiently large L, the inequality βLlog⁡2L≥L>1 holds. Hence 0<x<1/2 and log⁡2log⁡2(1/x)≥12log⁡2L.

3.1step 2.1L1algebra

Hence, for all sufficiently large n, CH(log⁡2(1/x))2/log⁡2log⁡2(1/x)≤2CHβ2L=L/8, and therefore 2−CH(log⁡2(1/x))2/log⁡2log⁡2(1/x)n≥2−L/8n=27L/8≥n.

4.1step 3.1L1∎

So for all sufficiently large n, this choice of x still leaves a dense-or-sparse set of order at least n.

Depends on

Used by

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Dependency tree · two levels

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Sources