Improved bounds concerning the maximum degree of intersecting hypergraphs

For positive integers $n>k>t$ let $\binom{[n]}{k}$ denote the collection of all $k$-subsets of the standard $n$-element set $[n]=\{1,\ldots,n\}$. Subsets of $\binom{[n]}{k}$ are called $k$-graphs. A $k$-graph $\mathcal{F}$ is called $t$-intersecting if $|F\cap F'|\geq t$ for all $F,F'\in \mathcal{F}$. One of the central results of extremal set theory is the Erd\H{o}s-Ko-Rado Theorem which states that for $n\geq (k-t+1)(t+1)$ no $t$-intersecting $k$-graph has more than $\binom{n-t}{k-t}$ edges. For $n$ greater than this threshold the $t$-star (all $k$-sets containing a fixed $t$-set) is the only family attaining this bound. Define $\mathcal{F}(i)=\{F\setminus \{i\}\colon i\in F\in \mathcal{F}\}$. The quantity $\varrho(\mathcal{F})=\max\limits_{1\leq i\leq n}|\mathcal{F}(i)|/|\mathcal{F}|$ measures how close a $k$-graph is to a star. The main result (Theorem 1.5) shows that $\varrho(\mathcal{F})>1/d$ holds if $\mathcal{F}$ is 1-intersecting, $|\mathcal{F}|>2^dd^{2d+1}\binom{n-d-1}{k-d-1}$ and $n\geq 4(d-1)dk$. Such a statement can be deduced from the results of \cite{F78-2} and \cite{DF}, however only for much larger values of $n/k$ and/or $n$. The proof is purely combinatorial, it is based on a new method: shifting ad extremis. The same method is applied to obtain some nearly optimal bounds in the case of $t\geq 2$ (Theorem 1.11) along with a number of related results.


Introduction
For positive integers n ≥ k let [n] = {1, . . ., n} be the standard n-element set and [n]   k the collection of its k-subsets.A family F ⊂ [n]  k is called t-intersecting if |F ∩ F ′ | ≥ t for all F, F ′ ∈ F and t a positive integer.In the case t = 1 we usually omit t and speak of intersecting families.Let us recall one of the fundamental results of extremal set theory.
Let us recall some standard notation.Set ∩F = ∩{F : F ∈ F}.If | ∩ F| ≥ t then F is called a t-star, for t = 1 we usually omit the 1.
Let us define the shifting partial order ≺ where P ≺ Q for P = (x 1 , . . ., x k ) and Q = (y 1 , . . ., y k ) iff x i ≤ y i for all 1 ≤ i ≤ k.This partial order can be traced back to [3].
A family F ⊂ [n]   k is called initial if F ≺ G, G ∈ F always implies F ∈ F. Note that |F(1)| ≥ |F(2)| ≥ . . .≥ |F(n)| for an initial family.Initial families have some nice properties.Let us give here one simple example.Proposition 1.2.Suppose that F ⊂ 2 [n] is initial and t-intersecting.Then F([s]) is (t + s)-intersecting.
Proof.Suppose for contradiction that F, F ′ ∈ F([s]), Let us define the quantity Since ̺(F) = 1 if and only if F is a star, in a way ̺(F) measures how far a family is from a star.A set T is called a t-transversal of F if |T ∩ F | ≥ t for all F ∈ F. If F is t-intersecting then each F ∈ F is a t-transversal.Define τ t (F) = min{|T | : T is a t-transversal of F}.
For t = 1 we usually omit the 1.
Proof.Fix a t-transversal T of F with |T | = τ t (F).Then Obviously, τ t (F) = t iff F is a t-star.
The aim of the present paper is to prove some similar results concerning ̺(F) for t-intersecting families for n > ck with relatively small constants c.Let us state here our main result for the case t = 1.Let us stress once more that ̺(F) > 1 d follows from the results of [6] and [2] however only for much larger value of n.
On the other hand for initial families one can deduce ̺(F) > 1 d under much milder constraints (cf.[15]).The problem is that one cannot transform a general family into an initial family without increasing ̺(F).To circumvent this difficulty we are going to apply the recently developed method of shifting ad extremis that we will explain in Section 2.
Two families F, G are called cross-intersecting if F ∩ G = ∅ for all F ∈ F, G ∈ G. Our next result is for cross-intersecting families.
This shows that 4 n−3 k−3 in Theorem 1.6 cannot be replaced by c n−3 k−3 for c < 2.
Let T (n, k) denote the triangle family defined by For min{|F|, |G|} = Ω n−3 k−3 , we prove the following result.
Theorem 1.10.Suppose that For t-intersecting families, we obtain the following result.
Let us present some results that are needed in our proofs.Define the lexicographic order A < L B for A, B ∈ [n]  k by A < L B iff min{i : i ∈ A \ B} < min{i : i ∈ B \ A}.E.g., (1, 2, 9) < L (1,3,4).For n > k > 0 and n k ≥ m > 0 let L(n, k, m) denote the first m sets A ∈ [n]  k in the lexicographic order.For X ⊂ [n] with |X| > k > 0 and |X| k ≥ m > 0, we also use L(X, k, m) to denote the first m sets A ∈ X k in the lexicographic order.A powerful tool is the Kruskal-Katona Theorem ( [25,23]), especially its reformulation due to Hilton [20].Hilton's Lemma ( [20]).Let n, a, b be positive integers, n ≥ a + b.Suppose that A ⊂ [n]   a and B ⊂ [n]   b are cross-intersecting.Then L(n, a, |A|) and L(n, b, |B|) are cross-intersecting as well.
For F ⊂ [n]  k define the ℓth shadow of F, For ℓ = 1 we often omit the superscript.The Katona Intersecting Shadow Theorem gives an inequality concerning the sizes of a t-intersecting family and its shadow.
Katona Intersecting Shadow Theorem ( [22]).Suppose that n ≥ 2k − t, t ≥ ℓ ≥ 1.Let ∅ = A ⊂ [n]  k be a t-intersecting family.Then with equality holding if and only if A generalized version of the Katona Intersecting Shadow Theorem was proved in [4].
We need a notion called pseudo t-intersecting, which was introduced in [14].A family F ⊂ [n]  k is said to be pseudo t-intersecting if for every F ∈ F there exists 0 It is proved in [10] that (1.3) holds for pseudo t-intersecting families as well.
The following inequalities for cross t-intersecting families can be deduced from Proposition 1.14.
Proposition 1.19.Let n, k, i be positive integers.Then Proof.It is easy to check for all b > a > 0 that ba > (b + 1)(a − 1) holds.
(1.12) Note that Applying (1.12) repeatedly we see that the enumerator is greater than Thus (1.11) holds.
Corollary 1.20.Let n, k, t be positive integers.If n ≥ 2(t − 1)(k − t) and k > t ≥ 2, then By (1.11) we have Let us state one more result.
2 Shifting ad extremis and the proof of Theorem 1.5 Let us recall an important operation called shifting introduced by Erdős, Ko and Rado [3].For F ⊂ [n]  k and 1 ≤ i < j ≤ n, define where The following statement goes back to Katona [23].Let us include the very short proof.
Define the matching number ν(F) of F as the maximum number of pairwise disjoint edges in F. Note that ν(F) = 1 iff F is intersecting.
Proof.It is proved in [10] that if G has matching number s then its shadow has size at least |G|/s.By (2.1) we infer Let us define formally the notion of shifting ad extremis developed recently (cf.[18]).It can be applied to one, two or several families.For notational convenience we explain it for the case of two families in detail.
Let F ⊂ [n]  k , G ⊂ [n]   ℓ be two families and suppose that we are concerned, as usual in extremal set theory, to obtain upper bounds for |F| + |G|, |F||G| or some other function f of |F| and |G|.For this we suppose that F and G have certain properties (e.g., cross-intersecting and non-trivial).Since |S ij (H)| = |H| for all families H, it is convenient to apply S ij simultaneously to F and G. Certain properties, e.g., t-intersecting, cross-intersecting or ν(F) ≤ r are known to be maintained by S ij .However, some other properties may be destroyed, e.g., non-triviality, ̺(G) ≤ c, etc.Let P be the collection of the latter properties that we want to maintain.
For any family H, define the quantity Obviously w(S ij (H)) ≤ w(H) for 1 ≤ i < j ≤ n with strict inequality unless S ij (H) = H.
Definition 2.3.Suppose that F ⊂ [n]  k , G ⊂ [n]   ℓ are families having property P. We say that F and G have been shifted ad extremis with respect to P if S ij (F) = F and S ij (G) = G for every pair 1 ≤ i < j ≤ n whenever S ij (F) and S ij (G) also have property P.
Let us show that we can obtain shifted ad extremis families by the following shifting ad extremis process.Let F, G be cross-intersecting families with property P. Apply the shifting operation S ij , 1 ≤ i < j ≤ n, to F, G simultaneously and continue as long as the property P is maintained.By abuse of notation, we keep denoting the current families by F and G during the shifting process.If S ij (F) or S ij (G) does not have property P, then we do not apply S ij and choose a different pair (i ′ , j ′ ).However we keep returning to previously failed pairs (i, j), because it might happen that at a later stage in the process S ij does not destroy property P any longer.Note that the quantity w(F) + w(G) is a positive integer and it decreases strictly in each step.This guarantees that eventually we shall arrive at families that are shifted ad extremis with respect to P.
Let F, G be shifted ad extremis families.A pair (i, j) In the case of several families, It is essentially the same.One important property that is maintained by simultaneous shifting is overlapping, namely the non-existence of pairwise disjoint edges Let P 1 , . . ., P s be a maximal collection of pairwise disjoint shift-resistant pairs, For a pair of subsets E 0 ⊂ E, let us use the notation By Hilton's Lemma, we have It follows that Clearly |E| = 2(d 2 − d + 1).Then (as (2.4)) the following formula is evident.

By (2.5) we have
By (2.2) and (2.7), we obtain that It follows that implying T = ∅.By (2.5) we know that for every D ∈ Proof.Let B = F(S, X ∪ T ).Since P 1 , P 2 , . . ., P s is a maximal collection of pairwise disjoint shift-resistant pairs and P 1 ∪ P 2 ∪ . . .∪ P s ⊂ X, we infer that F is initial on [n] \ X.It follows that A, B are initial and cross-intersecting.For any . By a similar argument as in the proof of Proposition 1.2, we infer that by (2.9) and (1.5) we see that we infer that By (2.10), we have Thus We conclude But |T | = t ≤ d − 1 and we obtain that there exists some 3 Maximum degree results for cross-intersecting families In this section, we prove some maximum degree results for cross-intersecting families.
Let us first prove the s = 1 version of Proposition 1.2 for two families.
For initial cross-intersecting families we can obtain a stronger result.For the proof, we need a result of Tokushige concerning the product of cross t-intersecting families.
Let us mention that the proof in [27] is based on eigenvalues.In [16] for t ≥ 14 the best possible bound (3.2) is established via combinatorial arguments for the full range, that is, n ≥ (t + 1)(k − t + 1).Proposition 3.5.Let F, G ⊂ [n]  k be initial cross-intersecting families with Proof.Suppose to the contrary that ̺(F) ≤ 2 3 .Then It follows that Since F( 1), G(1) are cross-intersecting, by Hilton's Lemma we see that But F( 1), G( 1) are cross 2-intersecting, which contradicts (3.2) for Fact 3.6.For F ⊂ [n]  k and every {x, y} For the proofs of Lemma 3.3 and Theorem 1.10, we need the following two lemmas.Lemma 3.7.Suppose that F, G ⊂ [n]  k are cross-intersecting and both of F, G are initial on  Then for S = ∅, Thus As A 1 ( 1, 2) is not pseudo 3-intersecting and A 1 ( 1, 2), B 2 ( 1, 2) are cross 3-intersecting, Proposition 1.14 implies that B 2 ( 1, 2) is pseudo 4-intersecting.Then by (1.5), Now we are in a position to prove Lemma 3.3.
Proof of Lemma 3.3.Arguing indirectly assume that F, G ⊂ [n]  k are cross-intersecting, Without loss of generality, suppose that F, G are shifted ad extremis with respect to {̺(F) ≤ 1  2 , ̺(G) ≤ 1 2 } and let H 1 , H 2 be the graphs formed by the shift-resistant pairs for F and G, respectively.
Let H = F or G.For every {x, y} ∈ Claim 5. We may assume that for all P ∈ [n]   2 2 .For convenience assume that {x, y} = {n − 1, n}.Then by (1.9) we obtain It follows that Similarly, we have Since F(n − 1, n), G(n − 1, n) are cross-intersecting, by Hilton's Lemma Claim 6. H i has matching number at most 2 for i = 1, 2.
Set J r = F {ar,br} , r = 1, 2, 3. Then for r = 1, 2, 3, Note that (3.6) and n ≥ 36k imply for 1 ≤ r < r ′ ≤ 3 that Therefore, If both F and G are initial, then by (3.1) we are done.Thus we may assume that H 1 ∪ H 2 = ∅.Without loss of generality, assume that (n − 1, n) ∈ H 1 .By Claim 6, we may further assume that both F and G are initial on [n − 8].Let By (3.6) and (3.7), it follows that By Lemma 1.18, we also have By Lemma 3.7, we obtain that 2 |G|, by (3.9) it follows that Then for n ≥ 26k Then by (3.8) we infer Now by Lemma 3.8 we obtain that By following a similar approach as in the proof of Lemma 3.3, we prove Theorem 1.10.
Proof of Theorem 1.10.Arguing indirectly assume that F, G ⊂ [n]  k are cross-intersecting, Without loss of generality suppose that F, G are shifted ad extremis with respect to {̺(F) ≤ 1 2 − ε, ̺(G) ≤ 1 2 − ε} and let H 1 , H 2 be the graphs formed by the shift-resistant pairs in F and G, respectively.Claim 7. We may assume that for all P ∈ [n]   2 2 .Then by (1.8) we obtain By (3.4), it follows that contradicting our assumption (3.11).
Claim 8. H i has matching number at most 2 for i = 1, 2.
Proof.Suppose for contradiction that (a 1 , b 1 ), (a 2 , b 2 ), (a 3 , b 3 ) are pairwise disjoint and say Define J r = F {ar,br} , r = 1, 2, 3. Then for r = 1, 2, 3, Note that (3.12) implies for 1 ≤ r < r ′ ≤ 3 that By ε ≤ 1 58 < 1 54 , we have If both F and G are initial, then by Proposition 3.2 we are done.Thus we may assume that H 1 ∪ H 2 = ∅.Without loss of generality, assume that (n − 1, n) ∈ H 1 .By Claim 8, we may further assume that both F and G are initial on [n − 8].Let By (3.13) and (3.12), it follows that By Lemma 1.18 we also have By Lemma 3.7, we obtain that 2 |G|, by (3.15) it follows that Then for ε ≤ 1 58 < 1 42 we have Then by (3.14) we infer In the rest of this section, we give some examples and results for maximum degree problems concerning initial cross-intersecting families.
Example 3.11.Fix k, ℓ ≥ 1, n > k + ℓ and q < k + ℓ.Set Suppose that a + b = q + 1.Then E(n, k, q, a) and E(n, k, q, b) are initial, cross-intersecting and Let us mention an old result showing that the example is in some sense best possible.
Proposition 3.12 ( [9]).Suppose that F, G ⊂ [n]   k are initial and cross-intersecting.Then for any F ∈ F, G ∈ G, there exists q = q(F, G) such that For cross-intersecting families of relatively large size one can bound ̺ for each of them.Proposition 3.13.Suppose that F, G ⊂ [n]  k are initial, cross-intersecting and Proof.In view of (1.To prove (3.18) below we need a simple analytic inequality.
ℓ and F, G are non-empty, initial and cross-intersecting then Proof.Note that we do not assume n ≥ k+ℓ.Indeed, for n ≤ k+ℓ by averaging ̺(F) ≥ k n , ̺(G) ≥ ℓ n and these imply (3.18).From now on we may assume n > k+ℓ and use induction on n.
The cases k = 1, ℓ = 1 are trivial.Consider the two vs two families F(n), F(n) and G(n), G(n) which are cross-intersecting and initial.The cross-intersecting property holds trivially for three of the pairs without the assumption of initiality.As to the pair F(n), G(n) initiality implies the cross-intersecting property (cf.[9]).Set |F(n −1] .By the induction hypothesis, If Q 1 = 0, R 1 = 0, by the induction hypothesis we have The finite projective plane is the standard example showing that ̺(F) can be quite small even for intersecting families.
Example 3.16.Let q ≥ 2 be a prime power and consider F = G = {the lines of the projective plane of order q}.
Note that this is an intersecting (q + 1)-graph on q 2 + q + 1 vertices and regular of degree q + 1.Thus ℓ and F, G non-empty and cross-intersecting, one can only get and this is best possible as shown by the next example.

Maximum degree results for t-intersecting families
In this section we consider t-intersecting families.
Let us recall the t-covering number τ t (F): It should be clear that τ t (F) = t if and only if F is a t-star.Proposition 1.3 yields for any t-intersecting family F ⊂ [n]  k .In the case τ t (F) = t + 1 one can improve on (4.1).Proposition 4.1.Suppose that F ⊂ [n]  k is t-intersecting, n ≥ 2k, τ t (F) ≤ t + 1 and F is saturated.Then ̺(F) > t+1 t+2 .
Proof.Without loss of generality let [t + 1] be a t-transversal of F, i.e., |F ∩ [t + 1]| ≥ t for all F ∈ F. Define Remark.Considering all k-subsets of [2k − t] shows that without some conditions on |F| one cannot hope to prove better than ̺(F) ≥ k 2k−t .Recently, the first author and Katona [14] proved the following improved version of (1.3) In the case of cross t-intersecting families, t ≥ 2, we cannot apply Hilton's Lemma.To circumvent this difficulty we prove a similar albeit somewhat weaker inequality.Proposition 4.3.Let n, k, ℓ, t, s be integers, s > t ≥ 2, k, ℓ > s.Suppose that F ⊂ [n]   k and G ⊂ [n]   ℓ are cross t-intersecting.Assume that |G| > n ℓ−s .Then Thus We use it for R with |R| > t and R = t but s / ∈ R.

It follows that
Combining (4.7), (4.8) and (4.9), we conclude that Consider the obvious construction: Then (F, G) are cross t-intersecting and showing that (4.
Now F 1 ∈ F and F2 ∪ P ∈ F by definition and F 2 ≺ F2 ∪ P .Hence F 2 ∈ F contradicting the t-intersecting property.
Proof of Theorem 1.11.Suppose to the contrary that |F| > Shift F ad extremis for ̺(F) ≤ t t+1 and let H be the graph formed by the shift-resistant pairs.For every P ∈ [n]  2 , by (4.10) and n ≥ 2t(t + 2)k > 5t 2 +19t+24 6 k we infer Claim 11.H has matching number at most 1.

Maximum degree results for r-wise intersecting families
A family F ⊂ [n]  k is called r-wise t-intersecting if |F 1 ∩ . . .∩ F r | ≥ t for all F 1 , . . ., F r ∈ F. The investigation of r-wise t-intersecting families has a long history ([1], [5], [8], [11], [12], [17]).Many of those results concern general (non-uniform) families.One of the early gems of extremal set theory is the following.Theorem 5.1 (Brace-Daykin Theorem [1]).Suppose that F ⊂ 2 [n] is non-trivial rwise intersecting, r ≥ 3. Then For r-wise intersecting uniform families, F ⊂ [n]  k , n ≥ 2k, follows from the Erdős-Ko-Rado Theorem.In [5] it is proved for the best possible range n ≥ r r−1 k.In [9] it is shown that unlike the case r = 2, even if n = r r−1 k, the full star is the unique family attaining equality in (5.2).
Let us introduce the notation t j (F) = max {t : F is j-wise t-intersecting} .
Similarly, if t j (F) = 1 + t j+1 (F) then |F ∩ T | ≥ |T | − 1 holds for the above T and every F ∈ F. These considerations lead to Lemma 5.2.Suppose that F is non-trivial r-wise t-intersecting, r ≥ 3. Then Proof.Since F is non-trivial, t j (F) ≥ t j+1 (F) + 1 for 2 ≤ j < r.This implies (5.3).Suppose now that equality holds and choose T ∈ Remark.Since F is (r − 1)-intersecting, F x , F y are cross-intersecting.For n ≥ 2k we could use the product version of the Erdős-Ko-Rado Theorem (t = 1 case) due to Pyber [26] .However, to cover the range 2k > n ≥ r r−1 k as well we need [17], the product version of [5].