LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Successive small integral geometric layers contradict a large X-part

Statement

Let $(A_{1}, \dots, A_{t})$ be a decreasing partition of $X$ with $t \geq ℓ \geq 4$ and every $∣ A_{j} ∣ \leq w / (2 ℓ)$ . Form its integral geometric layers $C_{1}, \dots, C_{q}$ . If, for every $r < q$ , the layer $C_{r}$ contains a block of size less than $w / ℓ^{5 r / 2}$ , then $∣ X ∣ < w / 2$ .

Facts & Assumptions

Given: The decreasing partition, its layers, and one stated small block in every preterminal layer.

[F1]

The first layer has at most $ℓ^{1 / 2}$ blocks, and layer $C_{r + 1}$ has at most $ℓ^{(r + 1) / 2}$ blocks (Integral geometric layers exist, cover the partition, and retain the required cutoff bounds).

[F2]

The layers partition the blocks of $X$ in their original nonincreasing order (Integral geometric layers of a decreasing block partition).

[F3]

For $∣ z ∣ < 1$ , the infinite geometric series sums to $1 / (1 - z)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Proof

technique · direct

1.1

The first-layer contribution is at most $ℓ^{1 / 2} w / (2 ℓ) = w / (2 \sqrt{ℓ}) \leq w / 4$ .

F1givenalgebra

1.2

A small block in $C_{r}$ has size less than $w / ℓ^{5 r / 2}$ ; by the nonincreasing order and [F2], every block in $C_{r + 1}$ is no larger. Hence the contribution of $C_{r + 1}$ is less than $w ℓ^{(r + 1) / 2 - 5 r / 2} = w ℓ^{1 / 2 - 2 r}$ .

F1F2givenalgebra

2.1

Since $ℓ \geq 4$ , the sum of these latter bounds is at most $w \sum_{r \geq 1} 4^{1 / 2 - 2 r} = \frac{w}{8} \sum_{s \geq 0} 16^{- s} = \frac{2 w}{15}$ by [F3].

F3step 1.2algebra

3.1

Adding steps 1.1 and 2.1 gives $∣ X ∣ < w (1 / 4 + 2 / 15) = 23 w / 60 < w / 2$ , as required.

step 1.1step 2.1F2algebra∎

Depends on

Used by

The structural comb-partition criterion implies property (*) Theorem

Dependency tree · two levels

35 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

Huang, Ju, and Zhou, Erdős–Hajnal beyond the five-vertex path, final sum in Lemma 5.1 (standard reference, not scraped)