TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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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.
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Fisher's inequality, nonuniform form: distinct nonempty $A_{1}, \dots, A_{m} \subseteq [n]$ with $∣ A_{i} \cap A_{j} ∣ = t$ for all $i \neq j$ satisfy $m \leq n$

Statement

Let $A_{1}, \dots, A_{m}$ be distinct nonempty subsets of $[n]$ . If there is a natural number $t$ such that

$∣ A_{i} \cap A_{j} ∣ = t for every i \neq j,$

then $m \leq n$ .

Facts & Assumptions

Given: distinct nonempty subsets $A_{1}, \dots, A_{m} \subseteq [n]$ and a natural number $t$ with $∣ A_{i} \cap A_{j} ∣ = t$ for every $i \neq j$ .

[L1]

If real vectors have a common pairwise inner product $t \geq 0$ and larger diagonal entries, then they are linearly independent (If $v_{1}, \dots, v_{m} \in R^{n}$ satisfy $⟨ v_{i}, v_{j} ⟩ = t \geq 0$ for $i \neq j$ and $⟨ v_{i}, v_{i} ⟩ > t$ , they are linearly independent).

[L2]

For incidence vectors over $R$ , the pairing is the intersection size ( $⟨ v_{A}, v_{B} ⟩$ is the image of $∣ A \cap B ∣$ in $F$ ; over $F_{2}$ it is $0$ or $1$ according to the parity of $∣ A \cap B ∣$ ).

[L3]

A family with independent incidence vectors has at most $n$ members (If the incidence vectors of $A_{1}, \dots, A_{m} \subseteq [n]$ are linearly independent over $F$ then $m \leq n$ ).

Proof

technique · direct

1.1

First suppose that some set, say $A_{1}$ , has size exactly $t$ . Then $t \geq 1$ because the sets are nonempty. For every $j > 1$ , the equality $∣ A_{1} \cap A_{j} ∣ = t = ∣ A_{1} ∣$ forces $A_{1} \subseteq A_{j}$ , so the differences $A_{j} ∖ A_{1}$ are nonempty. If $(A_{j} ∖ A_{1}) \cap (A_{k} ∖ A_{1})$ were nonempty for $j \neq k$ , then $A_{j} \cap A_{k}$ would properly contain $A_{1}$ , contradicting $∣ A_{j} \cap A_{k} ∣ = t$ . Thus these differences are pairwise disjoint, so there are at most $n - ∣ A_{1} ∣$ of them and therefore at most $n - 1$ indices $j > 1$ . Hence $m \leq n$ .

givenalgebra

1.2

Now suppose every $∣ A_{i} ∣ > t$ . By [L2], the incidence vectors in $R^{n}$ satisfy $⟨ v_{A_{i}}, v_{A_{j}} ⟩ = t$ for $i \neq j$ and $⟨ v_{A_{i}}, v_{A_{i}} ⟩ = ∣ A_{i} ∣ > t$ . So [L1] makes them linearly independent.

L1L2given

2.1

The master lemma [L3] then gives $m \leq n$ . Together with step 1.1, this proves the theorem in every case.

L3step 1.1step 1.2∎

Remarks

The two-case split is essential. If some set has size $t$ , the linear-algebra argument does not apply because the diagonal entry is not larger than the off-diagonal one.

Depends on

Used by

Dependency tree · two levels

39 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

L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, §4.1 (standard reference, not scraped)
J. Matousek, Thirty-three Miniatures, Miniature 4 (standard reference, not scraped)

Fisher's inequality, nonuniform form: distinct nonempty A1,…,Am⊆[n]A1​,…,Am​⊆[n] with ∣Ai∩Aj∣=t∣Ai​∩Aj​∣=t for all i≠ji=j satisfy m≤nm≤n