TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Graph removal lemma for a fixed ordinary subgraph

Statement

For every finite simple graph $H$ and every $ϵ > 0$ there are $δ > 0$ and $n_{0}$ such that the following holds. If a graph $G$ of order $n \geq n_{0}$ has fewer than $δ n^{∣ V (H) ∣}$ injective edge-preserving maps from $H$ to $G$ , then one can delete fewer than $ϵ n^{2}$ edges from $G$ and obtain an $H$ -free graph.

Facts & Assumptions

Given: A fixed graph $H$ , a parameter $ϵ > 0$ , and a sufficiently large graph $G$ .

[L1]

For every $0 < γ < 1$ and every prescribed $m_{0} \geq 1$ there is $M = M (γ, m_{0})$ such that every graph of order $n \geq M$ has an equitable $γ$ -regular vertex partition into $k$ parts with $m_{0} \leq k \leq M$ (Szemerédi regularity lemma with an equitable partition and an explicit tower-type upper bound for graphs of order at least $m_{0}$ ).

[L2]

For fixed $H$ , regular pairs of density bounded below contain a positive product-proportion of injective part-respecting copies of $H$ (Counting lemma: regular dense pairs contain many part-respecting copies of every fixed graph).

[L3]

A finite simple graph has no loops or multiple edges, so deleting an edge changes one unordered vertex pair (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

Proof

technique · contraposition

1.1

We prove the contrapositive. Put $h = ∣ V (H) ∣$ , choose $0 < η < 1$ with $η ≪ ϵ$ , then choose a regularity parameter $0 < γ < η$ small enough for [L2], so $0 < γ < 1$ as [L1] requires. Set $m_{0} = ⌈ 8 / ϵ ⌉$ and require $n_{0} \geq \max (M (γ, m_{0}), 8 / ϵ)$ . Apply [L1] at $γ$ with minimum part count $m_{0}$ , obtaining an equitable $γ$ -regular partition into $k$ parts with $m_{0} \leq k \leq M$ .

contrapositive-reducegivenL1L2choose

2.1

From $G$ delete every edge inside a partition part, every edge across a $γ$ -irregular pair, and every edge across a regular pair of density below $η$ . Call the resulting graph $G^{'}$ .

step 1.1choose

3.1

Equitability bounds the within-part deletions by $\sum_{i} ∣ V_{i} ∣^{2} \leq n \max_{i} ∣ V_{i} ∣ \leq n (n / k + 1) = n^{2} / k + n$ , and $k \geq m_{0} \geq 8 / ϵ$ together with $n \geq n_{0} \geq 8 / ϵ$ make this at most $ϵ n^{2} / 4$ by [L1]; $γ$ -regularity bounds the irregular-pair deletions by $γ n^{2}$ ; and the low-density pairs contribute at most $η n^{2}$ . Choose $γ, η$ so that $γ + η < ϵ / 2$ . Hence fewer than $ϵ n^{2}$ edges were deleted, each changing one unordered vertex pair by [L3].

step 1.1step 2.1L1L3algebra

4.1

Assume that fewer than $ϵ n^{2}$ deletions cannot make $G$ $H$ -free. Then $G^{'}$ contains a copy of $H$ . Assign to each vertex of $H$ the partition part containing its image. Every edge of $H$ lies across a $γ$ -regular pair of density at least $η$ .

assume-hypstep 2.1step 3.1

5.1

Apply [L2] to those assigned parts. Since each has size at least $n / (2 M)$ for the fixed regularity bound $M$ , the original graph contains at least $c \prod_{i = 1}^{h} ∣ W_{i} ∣ \geq c (2 M)^{- h} n^{h}$ injective copies of $H$ . Put $δ = c (2 M)^{- h}$ .

step 4.1L2algebra

6.1

Thus every graph requiring at least $ϵ n^{2}$ edge deletions contains at least $δ n^{h}$ copies of $H$ , which is the contrapositive of the Statement.

step 1.1step 5.1discharge-contrapositive∎

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 21 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Y. Zhao, Graph Theory and Additive Combinatorics, Theorem 2.6.5 (standard reference, not scraped)
D. Conlon and J. Fox, Graph removal lemmas, Theorem 1.1 (standard reference, not scraped)