Almost intersecting families

Let $n>k>1$ be integers, $[n] = \{1, \ldots, n\}$. Let $\mathcal F$ be a family of $k$-subsets of~$[n]$. The family $\mathcal F$ is called intersecting if $F \cap F' \neq \emptyset$ for all $F, F' \in \mathcal F$. It is called almost intersecting if it is not intersecting but to every $F \in \mathcal F$ there is at most one $F'\in \mathcal F$ satisfying $F \cap F' = \emptyset$. Gerbner et al. proved that if $n \geq 2k + 2$ then $|\mathcal F| \leq {n - 1\choose k - 1}$ holds for almost intersecting families. The main result implies the considerably stronger and best possible bound $|\mathcal F| \leq {n - 1\choose k - 1} - {n - k - 1\choose k - 1} + 2$ for $n>(2 + o(1))k$.


Introduction
Let [n] = {1, . . . , n} be the standard n-element set, 2 [n] its power set and [n] k the collection of all its k-subsets. Subsets of 2 [n] are called families. A family F is called intersecting if F ∩G = ∅ for all F, G ∈ F . One of the fundamental results in extremal set theory is the Erdős-Ko-Rado Theorem: Theorem 1.1 ( [EKR]). Suppose that F ⊂ [n] k is intersecting, n ≥ 2k > 0. Then Gerbner et al. [GLPPS] proved an interesting generalisation of (1.1). To state it we need a definition.
Definition 1.2. A family F ⊂ 2 [n] is called almost intersecting if it is not intersecting, but to every F ∈ F there is at most one G ∈ F satisfying F ∩ G = ∅. Theorem 1.3 ( [GLPPS]). Suppose that n ≥ 2k + 2, k ≥ 1, F ⊂ [n] k . If F is intersecting or almost intersecting then (1.1) holds.
A natural example of almost intersecting families is [2k] k . For n = 2k and 2k + 1 the best possible bound |F | ≤ 2k k is proven in [GLPPS]. To present another example let us first define some k-uniform intersecting families. For integers 1 ≤ a ≤ b ≤ n set [a, b] = {a, a + 1, . . . , b}. For a fixed x ∈ [n] let S = S(n, k, x) be the full star with center in x, i.e., S = S ∈ [n] k : x ∈ S . Every non-empty family F ⊂ S for some x is called a star.
The family B k+1 is called the Hilton-Milner family. It has a single set, namely [2, k + 1], which does not contain 1. For x, y ∈ [n] let us recall the standard notation: The maximum degree ∆(F ) of a family F ⊂ 2 [n] is max{|F (x)| : Hilton and Milner [HM] proved the following stability result for intersecting families.
Theorem 1.4 ( [HM] The case k = 2 is easy. Suppose that G ⊂ [n] 2 is almost intersecting and let F, G ∈ G be pairwise disjoint. Set X = F ∪ G and note |X| = 4. Claim 1.7. G ⊂ X 2 . Proof. If G = {F, G} then we have nothing to prove. On the other hand, for any further edge H ∈ G, both F ∩ H and G ∩ H must be non-empty. Since |H| = 2, H ⊂ X follows.
Let us make two simple but important observations.
k be almost intersecting. Then there is a unique The above partition of F is called the canonical partition. The function ℓ(F ) = ℓ is an important parameter of F . Definition 1.9. A family T = {T 1 , . . . , T ℓ } satisfying T i ∈ P i , is called a full tail (of F ).
Proposition 10. There are 2 ℓ full tails T and for each of them F 0 ∪ T is intersecting.
Let us close this section by a short proof of (1.3) for the special case ℓ(F ) = 1.
There are two cases to consider according whether the families F 0 ∪ {P 1 }, F 0 ∪ {Q 1 } are stars or not. Suppose first that one of them, say F 0 ∪ {P 1 } is not a star. By Theorem 1.4, F 0 ∪ {P 1 } = |F | − 1 ≤ B k+1 , implying (1.3). For k ≥ 4 uniqueness in the Hilton-Milner Theorem implies uniqueness in Theorem 1.6 as well. In the case k = 3, one has the extra possibility F 0 ∪ {P 1 } = B 3 . However, it is easy to check that adding a new 3-set to B 3 will never produce an almost intersecting family.
The second case is even easier. If both F 0 ∪ {P 1 } and F 0 ∪ {Q 1 } are stars then P 1 ∩ Q 1 = ∅ implies that there are two distinct elements (the centres of the stars) x, y such that {x, y} ⊂ F for all F ∈ F 0 . Consequently,

Preliminaries
Let us first prove an inequality on the size ℓ = ℓ(F ) of full tails.
The proof of (2.1) depends on a classical result of Bollobás [B].

Proof of Proposition 11. Define
Another ingredient of the proof of Theorem 1.6 is the following Let us note that if A is not a star then for all (1)|. This shows that Theorem 2.3 extends the Hilton-Milner Theorem.
The last ingredient of the proof is the Kruskal-Katona Theorem ([Kr], [Ka2]). We use it in a form proposed by Hilton [H].
For fixed n and k let us define the lexicographic order For an integer 1 ≤ m ≤ n k let L(m) = L(m, n, k) denote the family of the first m subsets A ∈ [n] k in the lexicographic order. Let a, b be positive integers, Theorem 2.4 ( [Kr], [Ka2], [H]). Let X ⊂ [n] and |X| ≥ a + b. If A ⊂ X a and B ⊂ X b are cross-intersecting then L(|A|, X, a) and L(|B|, X, b) are cross-intersecting as well.
k are cross-intersecting. Usually we apply Theorem 2.4 to these families (with X = [2, n]).
In our situation with F ⊂ [n] k being almost intersecting and F 0 ⊂ F defined by Proposition 8, F 0 (1) and F (1) are cross-intersecting.
Using Theorem 2.4 one easily deduces the following.
Proof. Let us first show that for n, k fixed the function f (r) = n−r+1 n−k−r+1 < 1 as both factors are less than 1 for n > 2k + 1.
Consequently it is sufficient to check (3.5) in the case r = t + 1 where t = √ k + 4. Fixing k and thereby r, t, define Claim 3.4. For n ≥ 2k, g(n) is a monotone decreasing function of n. Indeed, where we used t ≥ 3 and ab > (a − 2)(b + 2) for a > b + 2 > 0.
In view of the claim it is sufficient to prove (3.5) for the case n = 2k + 2 √ k + 4. (3.6) To estimate the RHS, note that the first part is at most 2 × 2 = 4. As to the product part, we can use the inequality (a−i)(a+i) To prove (3.5) we need to show that this quantity is at most 1/4. We show the stronger upper bound e − 3 2 . Using the inequality 1 − x < e −x , it is sufficient to show n + 1 − 2k .
Lemma 3.5. Suppose that n ≥ 3k + 3, k ≥ 4 then Proof. Let us first prove (3.7) in the case n = 3k + 3, The cases k = 4, 5, 6 can be checked directly. Let k ≥ 7. Note that Thus it is sufficient to show In view of k ≥ 7, 2k−1 k−1 2k k−2 is less than 1. Thus (3.9) will follow from 2k+i is a decreasing function of k, it is sufficient to check (3.10) for k = 7. Plugging in k = 7 we obtain 143 612 < 1 4 , as desired. To prove (3.7) for n > 3k + 3, we show that passing from n to n + 1 the RHS increases more than the LHS. More exactly we show: Using n > 3k, n−4 n−k < 2 and k−3 n−k−1 < 1 2 , we get (3.11).
Proof. Suppose by symmetry |D(a 1 , b 1 )| ≥ 2 and let x, y ∈ D(a 1 , b 1 ). The almost intersecting property implies (a i , b i , z) / ∈ F for i = 2, 3 and z / ∈ {x, y}. This already proves (ii). To continue with the proof of (i) choose x 2 , x 3 ∈ {x, y}, not necessarily distinct elements so that (a i , b i , x i ) ∈ F for i = 2, 3.
There are two simple cases to consider. Either x 2 = x 3 or x 2 = x 3 . By symmetry assume x 3 = y. In the first case (a 1 , b 1 , x) is disjoint to both (a 2 , b 2 , y) and (a 3 , b 3 , y). While in the latter case (a 3 , b 3 , y) is disjoint to both (a 1 , b 1 , x) and (a 2 , b 2 , x). These contradict the almost intersecting property. Now (iii) follows in the same way.
How many choices of (a, b), 1 ≤ a ≤ 3, 4 ≤ b ≤ 6 can be that satisfy |D(a, b)| ≥ 3 ? In view of Lemma 4.1 (ii), {a, b} ∩ {a ′ , b ′ } = ∅ must hold for distinct choices. Recall the easy fact that every bipartite graph without two disjoint edges is a star. Consequently, by symmetry, we may assume that |D(a, b)| ≥ 3 implies a = 1. Let us distinguish four cases.
For n = 9 as well we obtain the inequality |F | ≤ 3n − 7. However, to obtain uniqueness would require some extra case analysis.
In case of equality, [6] 3 ⊂ F . However, that would immediately imply F = [6] 3 . Thus the proof of the case k = 3, n ≥ 13 is complete.
5 The proof of (1.3) for k ≥ 4 We are going to distinguish three cases according to ∆(F 0 ).
Let us suppose n ≥ 2k + 5. In view of (3.2), Consequently, for any choice of a full tail T , Thus we may apply (2.4) with r = 4: From (5.1) and ℓ ≤ 2k−1 k−1 we infer , it is sufficient to show that the RHS is not larger than |B 5 |. Equivalently Since (5.3) is the same as (3.7), for n ≥ 3k + 3 we are done.
To deal with the case (iii), we cannot be so generous. We assume that n ≤ 3k + 2. Note that Using (5.2) and the inequality above, it is sufficient for us to show that Thus, it is sufficient for us to show that Let us define 2p = n − 2k − 4 and note p > √ k and thus p ≥ 4. In view of (3.3) and n ≤ 3k + 2 we have Thus, putting t := √ k, we are done if t ≤ (4/3) 2t−1 for any t ≥ 4. The latter is verified via an easy calculation. This concludes the proof of (1.3) in this case.
Let 1 be the vertex of highest degree in F 0 .
k be any intersecting family containing F 0 . Then 1 is the unique vertex of highest degree in G.
Proof. By assumption |G(1)| ≥ |F 0 (1)| > n−2 k−2 + n−3 k−2 . Let 2 ≤ x ≤ n be an arbitrary vertex. In view of Corollary 2.5, The inequality Define the parameter r, 4 ≤ r ≤ k by Let us choose the full tail T so that 1 / ∈ T for all T ∈ T . Applying Claim 5.1 to G = F 0 ∪ T yields ∆(F 0 ∪ T ) = ∆(F 0 ). Thus Theorem 2.3 implies Let us first prove (1.3) in the case n ≥ 3k + 3. Using |B r | ≤ |B k | and For n ≥ 3k + 3 the RHS is an increasing function of n. Thus it is sufficient to check the case n = 3k + 3: This inequality is true by (3.2) and k − 3 ≥ 1. Now let us turn to the case k ≥ 10, 3k + 2 ≥ n ≥ 2 k + √ k + 2 . Recall the definition of r from (5.5).
Using (2.1) and Corollary 2.5 we have Let us first consider the case We are going to prove (1.3) in the form We want to apply (3.4) to the RHS. Note that n − s ≥ 2k − 4 is satisfied if is a lower bound for the RHS. As to 2k−1 k−1 , in view of (3.1) and (3.3) it is very small, e.g., As to the main term, n−r k−r+1 , using r ≥ 4 we have Both factors in the coefficient of n−r−1 k−2 are decreasing functions of n. Thus the maximum is attained for n = 2k + 2 √ k + 4 and its value is To prove (5.7) it is sufficient to show we are done.
Using (5.6) and (5.7) one sees that the following inequality is sufficient: This inequality is the sum of (3.5) applied once for r and once for r + 1. The final subcase is r = k. Using (5.6) and (5.7) we obtain To show |F | < |B + | it is sufficient to show (5.9) The second half of (5.9) is evident from k ≥ 10 and n > 2k + 4. To show the first half note that n − k + 1 1 + n − k + 1 2 = n − k + 2 2 < 2 n − k − 1 2 , where the last inequality is true for n − k − 1 ≥ 8. On the other hand, for n − k − 1 ≥ 8 one has also 2 n−k−1 2 ≤ n−k−1 3 , concluding the proof of (5.9).
This concludes the entire proof.