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An Eventown family that no further set can be added to has exactly $2^{⌊ n / 2 ⌋}$ members

Statement

Let $F$ be an Eventown family on $[n]$ that is maximal under inclusion among Eventown families on $[n]$ . Then

$∣ F ∣ = 2^{⌊ n / 2 ⌋} .$

Facts & Assumptions

Given: a maximal Eventown family $F$ on $[n]$ .

[L1]

Every Eventown family has at most $2^{⌊ n / 2 ⌋}$ members (Eventown: distinct $A_{1}, \dots, A_{m} \subseteq [n]$ with every $∣ A_{i} ∣$ and every $∣ A_{i} \cap A_{j} ∣$ even satisfy $m \leq 2^{⌊ n / 2 ⌋}$ ).

[L2]

For a subspace $U \leq F_{2}^{n}$ , one has $\dim U^{⊥} = n - \dim U$ (For a subspace $U \leq F^{n}$ , $\dim_{F} U^{⊥} = n - \dim_{F} U$ , where $U^{⊥} = {x : ⟨ x, u ⟩ = 0 for all u \in U}$ ).

[L3]

A $d$ -dimensional vector space over $F_{2}$ has $2^{d}$ elements (A $d$ -dimensional vector space over a field with $q$ elements has exactly $q^{d}$ elements).

[L4]

If $Φ : V \to F_{2}$ is a nonzero linear map on a finite-dimensional $F_{2}$ -vector space, then $\dim \ker Φ = \dim V - 1$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · direct

1.1

The family $F$ contains $\emptyset$ and is closed under symmetric difference: if $A, B \in F$ , then $∣ A △ B ∣ = ∣ A ∣ + ∣ B ∣ - 2 ∣ A \cap B ∣$ is even, and for every $C \in F$ the intersection $(A △ B) \cap C$ has even size as well; maximality therefore forces $A △ B \in F$ .

given

2.1

Hence the set $U : = {v_{A} : A \in F}$ is a subspace of $F_{2}^{n}$ . Every $A \in F$ has even size, so the all-ones vector $1$ is orthogonal to every $v_{A}$ and therefore lies in $U^{⊥}$ . The Eventown hypotheses give $⟨ v_{A}, v_{B} ⟩ = 0$ for every $A, B \in F$ ; bilinearity then gives $⟨ u, u^{'} ⟩ = 0$ for all $u, u^{'} \in U$ . Consequently $U \subseteq H : = {x : ⟨ x, x ⟩ = 0}$ and $U \subseteq U^{⊥}$ .

step 1.1given

3.1

If $x \in U^{⊥} \cap H$ , then the subset $X \subseteq [n]$ with incidence vector $x$ has even size and even intersection with every member of $F$ , so maximality forces $X \in F$ and thus $x \in U$ . Therefore $U = U^{⊥} \cap H$ .

step 2.1

4.1

Let $d = \dim U$ . If $n$ is odd, then $1 \notin H$ while step 2.1 gives $1 \in U^{⊥}$ , so the linear map $Φ : U^{⊥} \to F_{2}$ given by $Φ (x) = ⟨ x, x ⟩ = ⟨ x, 1 ⟩$ is nonzero and has kernel $U^{⊥} \cap H$ . Hence step 3.1, [L2] and [L4] give $d = \dim (U^{⊥} \cap H) = \dim U^{⊥} - 1 = n - d - 1$ , so $d = (n - 1) / 2$ . If $n$ is even, then the whole set $[n]$ has even size and even intersection with every member of $F$ , so maximality forces $[n] \in F$ and therefore $1 \in U$ . Every $x \in U^{⊥}$ then satisfies $⟨ x, 1 ⟩ = 0$ , hence $U^{⊥} \subseteq H$ ; step 3.1 gives $U = U^{⊥} \cap H = U^{⊥}$ , and [L2] yields $d = n - d$ , so $d = n / 2$ . In both cases $d = ⌊ n / 2 ⌋$ .

L2L4step 2.1step 3.1algebra

5.1

The subspace $U$ has $2^{d} = 2^{⌊ n / 2 ⌋}$ elements by [L3], and step 1.1 identified those elements with the members of $F$ . The upper bound [L1] is therefore attained by every maximal Eventown family.

L1L3step 4.1∎

Remarks

Maximality is used only once, in step 3.1, to turn the orthogonality conditions back into actual set membership.

Depends on

Used by

The pairing construction gives an Eventown family of size 2^⌊ n/2⌋ Example

Dependency tree · two levels

40 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, §2.3.2 (standard reference, not scraped)

An Eventown family that no further set can be added to has exactly 2⌊n/2⌋2⌊n/2⌋ members