TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Erdős-Ko-Rado theorem: for $1 \leq k$ and $n \geq 2 k$ , an intersecting family of $k$ -subsets of an $n$ -set has size at most $(\binom{n - 1}{k - 1})$ , and a star attains the bound

Statement

Let $A$ be an $n$ -element set, where $1 \leq k$ and $n \geq 2 k$ . If $F \subseteq [A]^{k}$ is intersecting, then

$∣ F ∣ \leq (\binom{n - 1}{k - 1}) .$

For every fixed $a \in A$ , the star ${S \in [A]^{k} : a \in S}$ is intersecting and has cardinality $(\binom{n - 1}{k - 1})$ , so the bound is attained. No uniqueness of extremal families is asserted.

Facts & Assumptions

Given: An $n$ -element set $A$ , natural numbers $1 \leq k$ and $n \geq 2 k$ , and an intersecting family $F \subseteq [A]^{k}$ .

[L1]

In any cyclic order of $A$ , at most $k$ length- $k$ cyclic intervals can belong to a pairwise intersecting family (At most $k$ cyclic intervals of length $k$ in a cyclic order are pairwise intersecting when the ground-set size is at least $2 k$ ).

[F1]

A $k$ -uniform family is intersecting when every two members meet, and a star consists of the $k$ -sets through one fixed point (Intersecting uniform families of finite sets, The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L2]

An $m$ -element set has $m!$ orderings, and independent finite choices multiply (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality, The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[L3]

The binomial closed formula gives $(n - 1)! = (k - 1)! (n - k)! (\binom{n - 1}{k - 1})$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

Proof

technique · direct

1.1

A cyclic order of $A$ is a linear ordering modulo cyclic rotation. There are $(n - 1)!$ cyclic orders: fix one element in the first position and order the remaining $n - 1$ elements.

givenL2

1.2

Fix $S \in [A]^{k}$ . Exactly $k! (n - k)!$ cyclic orders make $S$ a cyclic interval: arrange the elements of $S$ within one consecutive block and arrange the elements of $A ∖ S$ in the complementary block.

givenL2

1.3

For a fixed $a \in A$ , deleting $a$ is a bijection from the star centred at $a$ to the $(k - 1)$ -subsets of $A ∖ {a}$ . The star is intersecting because all its members contain $a$ , and its size is $(\binom{n - 1}{k - 1})$ .

2.1

Count pairs $(ω, S)$ where $ω$ is a cyclic order and $S \in F$ is a length- $k$ interval in $ω$ . By step 1.2 there are $∣ F ∣ k! (n - k)!$ pairs.

step 1.2L2

3.1

By [L1], each of the $(n - 1)!$ cyclic orders occurs in at most $k$ pairs. Hence $∣ F ∣ k! (n - k)! \leq k (n - 1)!$ .

step 1.1step 2.1L1

4.1

Cancelling the positive factor $k! (n - k)! = k (k - 1)! (n - k)!$ in step 3.1 and using [L3] gives $∣ F ∣ \leq (\binom{n - 1}{k - 1})$ .

step 3.1L3algebra

5.1

Step 4.1 proves the upper bound and step 1.3 exhibits an intersecting family attaining it.

step 4.1step 1.3∎

Remarks

The hypothesis $n \geq 2 k$ is essential. At the boundary $n = 2 k$ , choosing exactly one set from each complementary pair already gives many extremal families, so the theorem deliberately does not claim that stars are the only extremizers.

Depends on

Used by

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34 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

J. Matoušek and J. Vondrák, The Probabilistic Method, pp. 14-15 (standard reference, not scraped)

Erdős-Ko-Rado theorem: for 1≤k1≤k and n≥2kn≥2k, an intersecting family of kk-subsets of an nn-set has size at most (n−1k−1)(k−1n−1​), and a star attains the bound