TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Counting lemma: regular dense pairs contain many part-respecting copies of every fixed graph

Statement

Let $H$ be a finite simple graph on labelled vertices $[h]$ , and let $0 < η < 1$ . There are constants $γ = γ (H, η) > 0$ , $c = c (H, η) > 0$ , and $N = N (H, η)$ such that the following holds for every finite simple graph $G$ . Let $W_{1}, \dots, W_{h} \subseteq V (G)$ be nonempty, not necessarily distinct, with $∣ W_{i} ∣ \geq N$ . Whenever $i j \in E (H)$ , require $W_{i} \neq W_{j}$ and require $(W_{i}, W_{j})$ to be $γ$ -regular in $G$ of density at least $η$ . Then at least $c \prod_{i = 1}^{h} ∣ W_{i} ∣$ maps $ϕ : [h] \to V (G)$ with $ϕ (i) \in W_{i}$ carry every edge of $H$ to an edge of $G$ . At least half this many such maps are injective, after increasing $N$ if necessary.

Facts & Assumptions

Given: $H, η$ , a host graph $G$ , and host sets $W_{i} \subseteq V (G)$ satisfying the Statement.

[L1]

If $(X, Y)$ is $ϵ$ -regular of density $d$ and $Y^{'} \subseteq Y$ satisfies $∣ Y^{'} ∣ \geq ϵ ∣ Y ∣$ , then fewer than $ϵ ∣ X ∣$ vertices $x \in X$ have $∣ N (x) \cap Y^{'} ∣ < (d - ϵ) ∣ Y^{'} ∣$ , and separately fewer than $ϵ ∣ X ∣$ have $∣ N (x) \cap Y^{'} ∣ > (d + ϵ) ∣ Y^{'} ∣$ (In a regular pair, fewer than $ϵ ∣ X ∣$ vertices have too small a degree into a large subset, and fewer than $ϵ ∣ X ∣$ have too large a degree).

[L2]

Restricting a regular pair to sufficiently large subsets preserves regularity, with a controlled new parameter and density loss (Slicing lemma: large subpairs remain regular and their density shifts by at most $ϵ$ ).

[L3]

A finite simple graph has finitely many vertices, no loops, and unordered edges between distinct vertices (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

Proof

technique · direct

1.1givenL3choose

If $h = 0$ there is exactly one map $ϕ : \emptyset \to V (G)$ , it is injective and carries every edge of $H$ to an edge of $G$ vacuously, and $\prod_{i = 1}^{0} ∣ W_{i} ∣ = 1$ , so $γ = c = 1 / 2$ and $N = 1$ give both assertions; assume $h \geq 1$ from here. Put $e = ∣ E (H) ∣$ . Choose $γ > 0$ with $γ \leq (η / 2)^{h} / (2 h)$ ; since $h \geq 1$ and $0 < η < 1$ this also gives $γ \leq (η / 2)^{h}$ and $γ \leq η / 2$ . Set $c = 2^{- h} (η / 2)^{e} > 0$ .

2.1step 1.1induction

Embed the vertices in the order $1, \dots, h$ . After choosing images for $1, \dots, r$ , maintain for each $j > r$ the candidate set obtained by intersecting $W_{j}$ with the neighbourhoods of the already embedded neighbours of $j$ . The invariant is $∣ C_{j} ∣ \geq (η / 2)^{d_{r} (j)} ∣ W_{j} ∣$ , where $d_{r} (j)$ is the number of neighbours of $j$ among $[r]$ .

3.1step 1.1step 2.1L1L2

Suppose the invariant holds before embedding $r + 1$ , and let $j > r + 1$ be a neighbour of $r + 1$ . The invariant and step 1.1 give $∣ C_{j} ∣ \geq (η / 2)^{h} ∣ W_{j} ∣ \geq γ ∣ W_{j} ∣$ , so $C_{j}$ meets the size threshold [L1] asks of a subset of $W_{j}$ inside the original $γ$ -regular pair $(W_{r + 1}, W_{j})$ ; no reslicing of that pair is needed for this bound, and [L2] confirms that the restricted pairs carried along remain regular. Write $d$ for the density of $(W_{r + 1}, W_{j})$ , so $d \geq η$ . Applying [L1] to this one subset $C_{j}$ , fewer than $γ ∣ W_{r + 1} ∣$ vertices of $W_{r + 1}$ have fewer than $(d - γ) ∣ C_{j} ∣$ neighbours in $C_{j}$ , and $(d - γ) ∣ C_{j} ∣ \geq (η - γ) ∣ C_{j} ∣ \geq (η / 2) ∣ C_{j} ∣$ because $γ \leq η / 2$ . This is a bound for the single subset $C_{j}$ , giving one exceptional set for each neighbour of $r + 1$ .

4.1step 1.1step 2.1step 3.1inductionalgebra

The union of the at most $h$ exceptional sets of step 3.1 has size at most $h γ ∣ W_{r + 1} ∣ \leq \frac{1}{2} (η / 2)^{h} ∣ W_{r + 1} ∣ \leq \frac{1}{2} ∣ C_{r + 1} ∣$ , so at least half of $C_{r + 1}$ survives; note that the exceptional sets are measured against $W_{r + 1}$ , not against $C_{r + 1}$ . Every remaining choice preserves the invariant after replacing each future-neighbour candidate set by its neighbourhood. Vertex $v$ is therefore chosen from a set of size at least $\frac{1}{2} (η / 2)^{d_{v - 1} (v)} ∣ W_{v} ∣$ , and $\sum_{v} d_{v - 1} (v) = e$ because each edge is counted once, at its later endpoint. Thus the greedy induction supplies at least $2^{- h} (η / 2)^{e} \prod_{i} ∣ W_{i} ∣ = c \prod_{i} ∣ W_{i} ∣$ edge-preserving maps.

5.1step 4.1L3algebrachoose

A noninjective map identifies at least one pair $i < j$ and hence there are at most $(\binom{h}{2}) N^{- 1} \prod_{i} ∣ W_{i} ∣$ collision choices. Choose $N$ so that this is at most $(c / 2) \prod_{i} ∣ W_{i} ∣$ whenever all $∣ W_{i} ∣ \geq N$ .

6.1step 4.1step 5.1∎

Removing these collision maps leaves at least $(c / 2) \prod_{i} ∣ W_{i} ∣$ injective part-respecting copies of $H$ .

Depends on

Used by

Dependency tree · two levels

5 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

Y. Zhao, Graph Theory and Additive Combinatorics, Theorem 2.6.2 (standard reference, not scraped)
D. Conlon and J. Fox, Graph removal lemmas, sec. 2.1 (standard reference, not scraped)