A Generalization of Sperner's Theorem on Compressed Ideals

Let [n] = {1, 2,. .. , n} and B n = {A : A ⊆ [n]}. A family A ⊆ B n is a Sperner family if A B and B A for distinct A, B ∈ A. Sperner's theorem states that the density of the largest Sperner family in B n is n n/2 /2 n. The objective of this note is to show that the same holds if B n is replaced by compressed ideals over [n].


Introduction
Let B n be the poset of subsets of [n] = {1, 2, . . ., n} ordered by inclusion.A family A ⊆ B n is a Sperner family if A B and B A for distinct A, B ∈ A .A famous result due to Sperner [5] states that the density of the largest Sperner family in B n is n n/2 /2 n .Sperner's theorem is one of the central results in extremal finite set theory and it has many generalizations and extensions (see [1,2] for instance).
For P ⊆ B n , we say that P is a convex family if A, B ∈ P and A ⊆ C ⊆ B imply that C ∈ P. A family I ⊆ B n is an ideal if A ∈ I and B ⊆ A imply B ∈ I .Clearly, an ideal is a convex family.In [3,  The conjecture seems difficult to prove and no progress was made in more than 30 years.Since no progress for the general case was made, it is quite natural to consider the special case of ideals.Here, we will restrict our research to the compressed ideals of B n .On B n we consider the reverse lexicographic order , which is defined by A with respect to .Here, we use the compress notation from [1, Ch. 7.5] and [2, p. 41].An ideal n for all 0 k n.Clearly, B n is a compressed ideal.In this paper, we will prove the following result.
Theorem 2. Let I be a compressed ideal in B n and A the largest Sperner family in I .Then 2 Proof of Theorem 2 As usual, we let To prove Theorem 2, we need two lemmas.
Proof of Theorem 2. To simplify the notation, let us write the electronic journal of combinatorics 23(3) (2016), #P3.24 It can be verified that T (2n − 1) = T (2n) and T (2n)/T (2n + 1) = (2n + 2)/(2n + 1).Hence we have We use induction on n.The case n = 1 is trivial.So we proceed to the induction step.Let I be a compressed ideal in B n .Then I = I 1 ∪ I 2 , where I 1 = {A ∈ I : n / ∈ A} and I 2 = {A ∈ I : n ∈ I }.Denote by I 2 (n) the collection of all sets A \ {n}, with A ∈ I 2 .Clearly, I 1 and I 2 (n) are compressed ideals in B n−1 .We therefore use the induction hypothesis for B n−1 , assuming that there exists the largest Sperner families Let Then A 2 is the largest Sperner family in I 2 and Denote by the collection of all sets I i which occur in n , and the collection of all sets A i which occur in n−1 by the definition of compressed ideal.Hence I 1 can be written as We now prove that r n/2 .Note that n and then , we obtain a larger Sperner family than A 2 (n) in I 2 (n).Thus r − 1 (n − 1)/2 , i.e., r n/2 .In the following, we show that there is the largest Sperner family A in I such that (1) holds.We distinguish two cases.
Case 1: we consider the case that n is even.Let n = 2m.Then r m.We show that s r.Assume that s < r.
the electronic journal of combinatorics 23(3) (2016), #P3.24 By (4), ∇ (r) (A 1 ∩ I (s) 1 ) ⊂ I 1 and Ā1 is still a Sperner family in I 1 .By Lemma 3, which contradicts the maximality of A 1 in I 1 .Hence we have s r, which means that A 1 ∪ A 2 is still a Sperner family in I .Hence by ( 2) and ( 3), we have and thus A = A 1 ∪ A 2 is the family as desired.
Case 2: we consider the case that n is odd.Let n = 2m + 1.Then r m + 1.If r < m + 1, by ( 4) we similarly have s r, and thus A = A 1 ∪ A 2 is the family as desired.If r = m + 1, then Then Ā2 is still a Sperner family in I 2 .Moreover, B (m) 2m ∪ Ā2 is also a Sperner family in I .We then show that We first claim that Note that , and .
So, to show that ( 6) is correct, it suffices to show that the following inequality is correct.

Remarks
Let I be an ideal in

, 3 ,
5} and {3, 5} {1, 3, 5}.Let C(m, B n ) be the family of the first m minimal elements of B n with respect to .The family C(m, B n ) is called compressed and the operation of exchanging an m-element family of B n by C(m, B n ) is called compression.Denote by B (k) n the collection of all k-subsets of B n .Similarly, we define C(F ), where F ⊆ B (k) n , to be the first |F | elements of B (k) n B n .The sequence f (I ) = (f 0 (I ), f 1 (I ), . . ., f t (I )), with f k (I ) = |I B (k) n |, is called the profile of the ideal I .It is known that there exists a compressed ideal I sharing the same profile with the ideal I in B n (see [1, Theorem 8.2.1] for details).By Theorem 2, there exists the largest Sperner family A in compressed ideal I such that |A |/|I | n n/2 /2 n .So, a key step to show that the ideal I satisfies Conjecture 1 should be to find the relationship between the largest Sperner family A ∈ I and A .
Conjecture 1.3], Frankl conjectured that the density of the largest Sperner family in any convex subfamily of B n is at least n n/2 /2 n .