False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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

Statement

For every $1\le k<n$ , 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 $\mathcal F=[A]^2$ of all its two-element subsets.

[F1]

$\binom nk$ 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 $\binom{n}{k} := \lvert [n]^{k}\rvert$ ).

[L1]

The Erdős-Ko-Rado theorem gives the star bound only under the hypothesis $1\le k$ and $n\ge2k$ (Erdős-Ko-Rado theorem: for $1\le k$ and $n\ge 2k$ , 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 $\mathcal F$ intersect, since two disjoint two-element subsets would require at least four elements while $|A|=3$ . Thus $\mathcal F$ is intersecting.

given

1.2

The family has cardinality $\binom32=3$ , while the claimed bound is $\binom21=2$ .

givenF1algebra

2.1

Hence the proposed bound fails at $k=2$ , $n=3$ , exactly where $k<n<2k$ and the hypothesis $n\ge2k$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 23 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

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

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