ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For $0 \leq ϵ < 1 / 2$ , every sufficiently large $ϵ$ -restricted set lies in one side

Example

Let $G$ be the disjoint union of two cliques of the same order. For $0 \leq ϵ < 1 / 2$ , every $ϵ$ -restricted set of unbounded size is concentrated in one of the two cliques.

Facts & Assumptions

Given: A graph $G$ that is the disjoint union of two cliques $A$ and $B$ of the same order, a real $0 \leq ϵ < 1 / 2$ , and a nonempty set $X \subseteq V (G)$ with $a = ∣ X \cap A ∣$ and $b = ∣ X \cap B ∣$ .

[L1]

A nonempty set contained in one clique is $0$ -dense (The $0$ -sparse sets are exactly the stable sets and the $0$ -dense sets exactly the cliques).

[L2]

A nonempty set $Y$ is $ϵ$ -restricted when either every vertex of $Y$ has at most $ϵ ∣ Y ∣$ neighbours in $Y$ , or every vertex of $Y$ has at most $ϵ ∣ Y ∣$ non-neighbours in $Y$ other than itself ( $c$ -sparse, $c$ -dense and $c$ -restricted vertex sets).

Verification

technique · direct

1.1L1

If $a = 0$ or $b = 0$ , then $X$ lies in one clique, so [L1] makes it $0$ -dense and hence $ϵ$ -restricted.

1.2given

Suppose $a, b > 0$ . The largest internal degree in $X$ is $\max {a, b} - 1$ , while the largest number of non-neighbours in $X$ is $\max {a, b}$ , attained by a vertex in the smaller trace.

2.1step 1.2L2algebra

If $X$ is $ϵ$ -restricted, then [L2] and step 1.2 force either $\max {a, b} - 1 \leq ϵ (a + b)$ in the sparse case or $\max {a, b} \leq ϵ (a + b)$ in the dense case. Either implies $\frac{1}{2} (a + b) - 1 \leq ϵ (a + b)$ , so $∣ X ∣ = a + b \leq 2 / (1 - 2 ϵ)$ . Thus a restricted set meeting both sides has size bounded solely in terms of $ϵ$ .

3.1step 1.1step 2.1∎

Therefore every sufficiently large $ϵ$ -restricted set is concentrated on one side.

Depends on

Used by

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For 0≤ϵ<1/20≤ϵ<1/2, every sufficiently large ϵϵ-restricted set lies in one side