Erdős–Ko–Rado Theorem for a Restricted Universe

A family F of k-element subsets of the n-element set [n] is called intersecting if F ∩F ′ 6= ∅ for all F, F ′ ∈ F . In 1961 Erdős, Ko and Rado showed that |F| 6 ( n−1 k−1 ) if n > 2k. Since then a large number of results providing best possible upper bounds on |F| under further restraints were proved. The paper of Li et al. is one of them. We consider the restricted universeW = { F ∈ ([n] k ) : |F ∩ [m]| > ` } , n > 2k, m > 2` and determine max |F| for intersecting families F ⊂ W. Then we use this result to solve completely the problem considered by Li et al. Mathematics Subject Classifications: 05D05


Introduction
Let n, k be positive integers, n 2k. Let [n] = {1, 2, . . . , n} be the standard n-element set and [n] k the collection of all its k-element subsets. Subsets F of [n] k are called k-uniform families.
A family F is called intersecting if F ∩ F = ∅ holds for all F, F ∈ F. The simplest example of a large intersecting family is the star: Obviously, |S| = n−1 k−1 . The classical Erdős-Ko-Rado Theorem [EKR] states that no k-uniform intersecting family can surpass |S|.
Hilton and Milner [HM] showed that for n > 2k up to isomorphism the star is the unique maximal family. The Erdős-Ko-Rado Theorem was the origin of a lot of research, there are many papers strengthening, generalizing or extending it. We refer the interested reader to the recent monograph by Gerbner and Patkos [GP].
Motivated by a paper of Li, Chen, Huang and Li [LCHL] we consider the following problem.
Let (m, ) be a pair of integers, m 2 > 0. Let us consider In order to guarantee that W is not empty, we suppose that k . Since increasing m beyond n does not change W, we suppose n m as well. Whenever we need to stress the parameters, we write W(n, k, m, ) instead of the short form W. We call W the restricted universe.
Definition 1. Let g(n, k, m, ) denote the maximum of |F| where F ⊂ W(n, k, m, ) is intersecting. Let us define the restricted star R = R(n, k, m, ) by R = S ∩ W, that is, For the case m k + we are going to show that (1) g(n, k, m, ) = |R|.
In the case m k + − 1 one can add further sets to R. Define t = m + 1 − and note < t k (the first part follows from m 2 ). Define P = P ∈ W : |P ∩ [m]| t , P 0 = P ∈ P : 1 / ∈ P .
Then |R ∪ P| = |R| + |P 0 | and R ∪ P is still intersecting. Indeed, if R ∈ R, P ∈ P, then + t > m implies R ∩ P ∩ [m] = ∅. For P, P ∈ P the same follows from t + t > + t > m.
We prove Theorem 2 in Section 2. In Section 3 we use it for the complete solution of the following problem that was raised in [LCHL].
k , F is intersecting and m-complete . For m = k and k + 1 the Erdős-Ko-Rado Theorem and the Hilton-Milner Theorem, respectively, give the answer: Theorem 4. If n 2k > m k, then We should mention that Li, Chen, Huang and Li [LCHL] determined h(n, m, k) for n > n 0 (k) and also for m = 2k − 1, 2k − 2, 2k − 3. Their construction corresponding to (3) is 2 The proof of Theorem 2 We are going to apply induction on n. The base of the induction is the case n = 2k.
In the non-restricted universe, Since the upper bound 2k−1 k−1 is still valid in the restricted universe, we just have to show that our constructions match this bound. That is, for every complementary pair (U, V ) one of the sets is in our family. By symmetry assume 1 ∈ U . If m k + , then i.e., V ∈ P. This concludes the proof of (1) and (2) for the case n = 2k.
In the induction step we are going to prove (1) and (2) for the pair (n + 1, k) assuming its validity for the pairs (n, k) and (n, k − 1). Formally there could be a problem in the case k = , because k − 1 is no longer greater or equal to . However, if k = , then our restricted universe is simply [m] k and (1) directly follows from the Erdős-Ko-Rado Theorem. Thus we assume k > in the sequel.
The inductive step is fairly simple, however it relies on shifting, an operation on families that was introduced by Erdős, Ko and Rado [EKR]. To keep this paper short let us refer to [F87] for the details concerning shifting. The point is that this operation does not destroy the intersecting property and it maintains the property |F ∩ [m]| as well. Applying it repeatedly eventually produces a family F with the following property.
then the set (F \ {j}) ∪ {i} is also in F.
Families satisfying (4) are called shifted. In view of the above discussion, in proving Theorem 2 we may assume that the intersecting family F ⊂ W is shifted. Shifted families have many useful properties but we only need one, namely Claim 5 below.
Recall the standard definitions The equality |F| = |F(i)| + |F(i)| should be obvious. Since F(i) ⊂ F, F(i) is intersecting.
k is shifted and intersecting, n + 1 2k, then F(n + 1) is intersecting as well.
What about the case n + 1 = m? Fortunately, in this case k implies W(n + 1, k, n + 1, ) = [n+1] k and the statement of Theorem 2 follows directly from the Erdős-Ko-Rado Theorem. Thus we may assume both n > m and k > . By Lemma 6, to conclude the proof in these cases the only thing that we have to show is that for the families F = R(n + 1, k, m, ) or F = R(n + 1, k, m, ) ∪ P 0 (n + 1, k, m, ) |F(n + 1)| = g(n, k, m, ) and |F(n + 1)| = g(n, k − 1, m, ).
However, both these facts are immediate from the definitions of R and P 0 . The only case that needs a slight verification is P 0 = P 0 (n + 1, k, m, ). Recall that t = m + 1 − is independent of k and P 0 = P ∈ [n + 1] k : 1 / ∈ P, |P ∩ [m]| t .