A note on saturation for $k$-wise intersecting families

A family $\mathcal{F}$ of subsets of $\{1,\dots,n\}$ is called $k$-wise intersecting if any $k$ members of $\mathcal{F}$ have non-empty intersection, and it is called maximal $k$-wise intersecting if no family strictly containing $\mathcal{F}$ satisfies this condition. We show that for each $k\geq 2$ there is a maximal $k$-wise intersecting family of size $O(2^{n/(k-1)})$. Up to a constant factor, this matches the best known lower bound, and answers an old question of Erd\H{o}s and Kleitman, recently studied by Hendrey, Lund, Tompkins, and Tran.


Introduction
Given positive integers k ≥ 2 and n, we say that a family F of subsets of [n] = {1, 2, . . . , n} is k-wise intersecting if whenever X 1 , . . . , X k ∈ F then X 1 ∩ · · · ∩ X k = ∅. It is well known (and easy to see) that any (2-wise) intersecting family over [n] has size at most 2 n−1 , so the largest possible size of a k-wise intersecting family is clearly 2 n−1 for all k. This is achieved, for example, by taking F = {A ∈ P([n]) : 1 ∈ A}, where P(X) denotes the set of subsets of X.
However, the corresponding saturation problem of finding the smallest possible size of a maximal k-wise intersecting family is more interesting for k ≥ 3. (A family F of subsets of [n] is maximal k-wise intersecting if it is k-wise intersecting but no family F ′ over [n] strictly containing F is k-wise intersecting.) This problem was originally raised by Erdős and Kleitman [4] in 1974. Very recently, Hendrey, Lund, Tompkins, and Tran [6] studied this problem for k = 3. They determined the smallest possible size of a maximal 3-wise intersecting family exactly when n is sufficiently large and even. For general k, they showed that the smallest possible size f k (n) of a maximal k-wise intersecting family satisfies c k · 2 n/(k−1) ≤ f k (n) ≤ d k · 2 n/⌈k/2⌉ (for some constants c k , d k > 0). They asked about closing the exponential gap between the lower and upper bounds.
In this note we prove the following result, which shows that the lower bound gives the right order of magnitude. Theorem 1.1. For each k ≥ 3 there exists some constant C k such that for all n there is a maximal k-wise intersecting family over [n] of size at most C k · 2 n/(k−1) .
In the case when n ≥ 2(k − 1) is a multiple of k − 1, the exact value of our upper bound will be 2 n/(k−1)+k−3 (k − 1) − (2 k−1 − 1)(k − 2). In the special case k = 3 this upper bound is tight for n sufficiently large (as shown by Hendrey, Lund, Tompkins, and Tran [6]), but for k ≥ 4 the construction has more complicated structure than for k = 3. In fact, in [6] it was shown that for k = 3 (and n large and even), the unique maximal 3-wise intersecting families of smallest possible size are given by {A c : A ∈ (P(X) \ {X}) ∪ (P(Y ) \ {Y })} for some partition X ∪ Y of [n] into two equal parts. This was proved by first obtaining a stability result stating that for any 'small' maximal 3-wise intersecting family F, the familyF = {A c : A ∈ F} must be 'close' to the union of two cubes P(X) ∪ P(Y ) (with X, Y as above).
However, the following result shows that, for k ≥ 4, directly generalising this approach cannot work, and it is necessary to have more complicated structure.
We mention that many other saturation problems have already been studied in the context of set systems and intersection properties. For example, several authors gave bounds for the smallest possible size m(r) of a set system which is maximal among (2-wise) intersecting families F ⊆ N r consisting of sets of size r -see, for example, [3,1,2]. A linear lower bound follows from a result of Erdős and Lovász [5], and Dow, Drake, Füredi, and Larson [3] showed that in fact m(r) ≥ 3r for r ≥ 4. Blokhuis [1] proved a polynomial upper bound of m(r) ≤ r 5 , and for certain values of r quadratic upper bounds are also known -see, e.g., [1,2]. Finding the order of magnitude of m(r) is still an open problem. See the introduction and the references in [6] for other related saturation problems.

The construction
We now prove Theorem 1.1 by describing a family of size O(2 n/(k−1) ) and showing that it is maximal k-wise intersecting over [n]. For simplicity, we will work with complements, using the observation thatḠ = {X c : X ∈ G} is k-wise intersecting if and only if no k elements of G have union [n].
Fix some k ≥ 3 and n ≥ 2(k − 1) integers. Partition [n] into k − 1 sets A 1 , . . . , A k−1 which are as close as possible in size, and pick 'special' elements a i ∈ A i for each i. Consider the following families. (All indices will be understood mod k − 1, so, for example, a 0 = a k−1 .) We will show that {A c : A ∈ F} is maximal k-wise intersecting. Note that if k = 3 then This was shown to be (up to isomorphism) the unique minimal-sized construction when k = 3 and n is even and sufficiently large by Hendrey, Lund, Tompkins, and Tran [6]. Furthermore, note that if k − 1 divides n. Indeed, the number of subsets of {a 1 , . . . , a k−1 } appearing in F is 2 k−1 − 1, the number of sets which are not subsets of {a 1 , . . . , a k−1 } but appear in some F 2 (i) is (k − 1) · (2 n/(k−1) − 4) · 2 k−3 , and finally, the only elements of F we have not yet counted are the (k − 1) sets A i \ {a i } for i = 1, . . . , k − 1. Summing these contributions gives the formula above. Proof. Suppose we have sets Y 1 , . . . , Y k ∈ F satisfying Y 1 ∪ · · · ∪ Y k = [n]. Then clearly at least one of them must come from some F 2 (i), we may assume Y 1 ∈ F 2 (1). If there is some j = 1 and i = 1 such that Y j ∈ F 2 (i), then for each t we can pick b t ∈ A t \ {a t } such that b t ∈ Y 1 ∪ Y j . Then no element of F contains more than one b t , but {b 1 , . . . , b k−1 } ⊆ ℓ =1,j Y ℓ , giving a contradiction. On the other hand, if there is no such pair (i, j), then {Y ℓ : ℓ = 1} must contain at least one non-empty set from F 1 (t) (to cover A t \ {a t }) for t = 2, . . . , k − 2, at least two different sets from F 1 (k − 1) (to cover A t−1 ), and at least one set having an element in A 1 , again giving a contradiction since these k sets must all be different.
Proof. We first consider the following five cases, then check that each choice of X belongs to at least one of these cases.
• Case 1: X ∩ A i = ∅ for all i. Then let X i = A i \ X, so X i ∈ F 1 (i) and the X i satisfy the conditions.
• Case 2: There is some i such that X ∩A i contains an element b i with b i = a i and X ∩A i−1 = ∅. We may assume that i = 1. Then let ). These clearly satisfy the conditions.
• Case 3: Then X i ∈ F 2 (i), X j ∈ F 2 (j), and X ℓ ∈ F 1 (ℓ) for ℓ = i, j, so it is easy to see that the conditions are satisfied.
• Case 4: X ⊇ (A i \ {a i }) ∪ {a j } for some i, j with j = i. Then let X i = {a t : t = j} (so X i ∈ F 2 (j + 1)), and let X ℓ = A ℓ \ {a ℓ } for ℓ = i (so X ℓ ∈ F 1 (ℓ)). It is easy to see that the conditions are satisfied.
• Case 5: X = A i for some i. Then let X i = {a t : t = i} (so X i ∈ F 2 (i+1)) and X ℓ = A ℓ \{a ℓ } for ℓ = i (so X ℓ ∈ F 1 (ℓ)). Then the conditions are again satisfied.
We now check that any X ∈ P([n]) \ F belongs to at least one of these cases. If X ⊆ {a 1 , . . . , a k−1 } then in fact X = {a 1 , . . . , a k−1 } and we are in Case 1. Otherwise there is some i and some b i ∈ A i such that b i ∈ X and b i = a i . If X ∩ X i−1 = ∅ then we are in Case 2. Otherwise, if X is not a subset of A i ∪ {a ℓ : ℓ = i, i − 1} then we are in Case 3. Finally, if X is a subset of A i ∪ {a ℓ : ℓ = i, i − 1}, then we are in Case 4 or Case 5 since X ∈ F 1 (i) ∪ F 2 (i).

Non-existence of certain types of maximal families
Finally, we prove Lemma 1.2 stating that there can be no construction for k ≥ 4 which is close to the union of k − 1 cubes.
Proof of Lemma 1.2. Suppose that F is as in the statement of the lemma, andF is maximal k-wise intersecting. Then at least one X i does not appear in F, we may assume X k−1 ∈ F. Let G be the family of sets S over [n] satisfying the following conditions.
• The set S cannot be written as S 1 ∪ · · · ∪ S k−1 such that S i ∈ F for all i and S 1 ∈ F \ Q.
• For all S ′ ∈ F \ Q and i ≤ k − 2, we have S ′ ∩ X i = S ∩ X i .
• For all i, S ∩ X i is non-empty.