False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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False: the Erdős-Ko-Rado bound holds without the hypothesis $n \geq 2 k$

Statement

For every $1 \leq k < n$ , every intersecting family of $k$ -subsets of an $n$ -element set has cardinality at most $(\binom{n - 1}{k - 1})$ .

Facts & Assumptions

Given: A three-element set $A$ and the family $F = [A]^{2}$ of all its two-element subsets.

[F1]

$(\binom{n}{k})$ is the cardinality of the family of $k$ -subsets of an $n$ -element set (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L1]

The Erdős-Ko-Rado theorem gives the star bound only under the hypothesis $1 \leq k$ and $n \geq 2 k$ (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).

Refutation

technique · direct

1.1

Any two members of $F$ intersect, since two disjoint two-element subsets would require at least four elements while $∣ A ∣ = 3$ . Thus $F$ is intersecting.

given

1.2

The family has cardinality $(\binom{3}{2}) = 3$ , while the claimed bound is $(\binom{2}{1}) = 2$ .

givenF1algebra

2.1

Hence the proposed bound fails at $k = 2$ , $n = 3$ , exactly where $k < n < 2 k$ and the hypothesis $n \geq 2 k$ of [L1] is absent.

step 1.1step 1.2L1∎

Depends on

Used by

When k<n<2k, the entire kth level is intersecting and exceeds the Erdős-Ko-Rado star bound Counterexample

Dependency tree · two levels

17 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

Erdős-Ko-Rado theorem (Wikipedia) (standard reference, not scraped)

False: the Erdős-Ko-Rado bound holds without the hypothesis n≥2kn≥2k