A simple proof for a forbidden subposet problem

The poset Yk,2 consists of k + 2 distinct elements x1, x2, . . . , xk, y1, y2, such that x1 6 x2 6 . . . 6 xk 6 y1, y2. The poset Y ′ k,2 is the dual poset of Yk,2. The sum of the k largest binomial coefficients of order n is denoted by Σ(n, k). Let La(n, {Yk,2, Y ′ k,2}) be the size of the largest family F ⊂ 2[n] that contains neither Yk,2 nor Y ′ k,2 as an induced subposet. Methuku and Tompkins proved that La(n, {Y2,2, Y ′ 2,2}) = Σ(n, 2) for n > 3 and conjectured the generalization that if k > 2 is an integer and n > k+1, then La(n, {Yk,2, Y ′ k,2}) = Σ(n, k). On the other hand, it is known that La(n, Yk,2) and La (n, Y ′ k,2) are both strictly greater than Σ(n, k). In this paper, we introduce a simple approach, motivated by discharging, to prove this conjecture. Mathematics Subject Classifications: 06A06 ∗Research supported in part by a Simons Foundation grant (# 353292) and by NSF-DMS Grants 1604458, 1604773, 1604697, and 1603823. †Research supported in part by the Hungarian National Research, Development and Innovation Office – NKFIH under the grant K116769 and by a generous grant from the Combinatorics Foundation. ‡Research supported in part by a generous grant from the Institute for Mathematics and its Applications and in part by NSF-DMS Grants 1604458, 1604773, 1604697, and 1603823. §Research supported in part by NSF-DMS Grants 1604458, 1604773, 1604697, and 1603823. the electronic journal of combinatorics 27(1) (2020), #P1.31 1


Introduction
The n-dimensional Boolean lattice, denoted B n , is the partially ordered set (poset) (2 [n] , ⊆), where [n] = {1, . . . , n}. For any 0 i n, let [n] i := {A ⊆ [n] : |A| = i} denote the ith level of the Boolean lattice. Let P be a finite poset and F be a family of subsets of [n]. We say that P is contained in F as a weak subposet if there is an injection α : P → F satisfying x 1 < p x 2 =⇒ α(x 1 ) ⊂ α(x 2 ) for all x 1 , x 2 ∈ P . F is called P -free if P is not contained in F as a weak subposet. We define the corresponding extremal function to be La(n, P ) := max{|F| : F is P -free}. Analogously, if P , Q are two posets, then La(n, {P, Q}) := max{|F| : F is P -free and Q-free}.
The linearly ordered poset on k elements, a 1 < a 2 < · · · < a k , is called a chain of length k, and is denoted by P k . Using this notation the well-known theorem of Sperner [15] can be stated as La(n, P 2 ) = n n/2 . Let us denote the sum of the k largest binomial coefficients of order n by Σ(n, k). Erdős [6] extended Sperner's theorem by showing that La(n, P k ) = Σ(n, k − 1), with equality if and only if the family is the union of k − 1 largest levels of the Boolean lattice. Notice that, since any poset P is a weak subposet of a chain of length |P |, Erdős's theorem implies that La(n, P ) (|P | − 1) n n/2 = O n n/2 . Later many authors, including Katona and Tarján [12], Griggs and Lu [9], and Griggs, Li, and Lu [8] studied La(n, P ) for various other posets P . (See the recent survey by Griggs and Li [7] for an excellent survey of all the posets that have been studied.) Let h(P ) denote the height (maximum length of a chain) of P . One of the first general results is due to Bukh [2], who showed that if T is a finite poset whose Hasse diagram is a tree of height h(T ) 2, then La(n, T ) = (h(T ) − 1 + O(1/n)) n n/2 . The most notorious poset for which the asymptotic value of the extremal function is still unknown is the diamond Q 2 , the poset on 4 elements with the relations a < b, c < d where b and c are incomparable. The best known bound is (2.20711 + o(1)) n n/2 , due to Grósz, Methuku, and Tompkins [10].
We say that P is contained in F as an induced subposet if and only if there is an injection α : P → F satisfying x 1 < p x 2 ⇐⇒ α(x 1 ) ⊂ α(x 2 ) for all x 1 , x 2 ∈ P . We say that F is induced-P -free if P is not contained in F as an induced subposet. We define the corresponding extremal function as La (n, P ) := max{|F| : F is induced P -free}. Analogously, if P , Q are two posets, then La (n, {P, Q}) := max{|F| : F is induced P -free and induced Q-free}.
Despite the considerable progress that has been made on the extremal function for forbidden weak subposets, little is known about forbidden induced subposets (except for P k , where the weak and induced containment are equivalent). The first results for forbidden induced subposets are due to Carroll and Katona [3] and to Katona [11], showing La (n, V r ) = (1 + o(1)) n n/2 where V r is the r-fork poset (x y i for all 1 i r). Boehnlein and Jiang [1] generalized the results of [3,11] by extending Bukh's result to induced containment of tree-shaped posets, T , proving La (n, T ) = (h(T )−1+o(1)) n n/2 . Only recently, Methuku and Pálvölgyi [14] showed that for every poset P , there is a constant c P depending only on P such that La (n, P ) c P n n/2 . the electronic journal of combinatorics 27(1) (2020), #P1.31 Even fewer exact results are known for forbidden induced subposets, which is the topic of this paper. Katona and Tarján [12] proved that La(n, {V, Λ}) = La (n, {V, Λ}) = 2 n−1 n− 1 2 , where V and Λ are the 2-fork and its dual, the 2-brush, respectively. Now we formally define the posets considered in this paper. Definition 1. Let k, r 2 be integers. The r-fork with a k-shaft poset consists of k + r elements x 1 , x 2 , . . . , x k , y 1 , y 2 , . . . , y r−1 , y r with x 1 x 2 · · · x k and x k y i for all 1 i r, and is denoted by Y k,r . Let Y k,r denote the reversed poset of Y k,r , also called the dual poset of Y k,r .
For simplicity, we will write Y k and Y k instead of Y k,2 and Y k,2 respectively.
The first result about Y k,r was due to Thanh [16] who showed that La(n, Y k,r ) = (k + o(1)) n n/2 . The lower order term in his upper bound was improved by De Bonis and Katona [4]. Thanh also gave a construction showing that La(n, Y k,r ) > Σ(n, k). Methuku and Tompkins [13] showed that if one forbids both Y k and Y k , then an exact result can be obtained: Using a cycle decomposition method, they also showed the following exact result for induced posets.
Theorem 2 strengthens the result of De Bonis, Katona, and Swanepoel [5] stating that La(n, B) = Σ(n, 2) where B is the butterfly poset which consists of 4 elements a, b, c, d with a, b c, d. Indeed if a family does not contain the butterfly as a subposet, then it contains neither Y 2 nor Y 2 as an induced subposet. However, a family might contain neither an induced Y 2 nor an induced Y 2 while still containing a butterfly.
In Section 3, we establish the following generalization of Theorem 2 by proving a conjecture from [13].
Note that forbidding only one of Y k and Y k is not enough to obtain an exact result. Indeed, again by Thanh's construction [16] we have La (n, Y k ) > Σ(n, k) and La (n, Y k ) > Σ(n, k).
In the course of proving Theorem 3, we establish the following LYM-type inequality in the case that ∅ and [n] are not in our family.
Theorem 4. Let k 2 be an integer and n k + 1.
In order to prove Theorem 4 we use a double-counting argument that is reminiscent of discharging-type arguments, as described in the next section. We then prove Theorem 3 by using Theorem 4 and induction on k.

Preliminaries and our approach
The following terminology will be used to prove Theorems 3 and 4. Let F be a family of subsets of [n] which is induced Y k -free and induced Y k -free. For sets U , V ⊆ [n], let the interval [U, V ] denote the Boolean lattice induced by the collection of all sets that contain U and are contained in Note that a spine may contain elements not from F.
Let C be the set of all full chains and let S be the set of all spines. We say that a full chain C ∈ C is associated with a spine S ∈ S or that C contains S as a spine if either 1. C has exactly k − 1 members of F, which we name F 1 , . . . , F k−1 . In this case, C is associated with the spine that is a subchain of C from F 1 to F k−1 ; or 2. C has exactly k + x elements of F (where x 1), which we name F 1 , . . . , F k+x . In this case, C is associated with x spines, namely S F i for 2 i x + 1, where S F i is the spine that is a subchain of C from F i to F i+k−2 . (Notice that a chain C with at least k + 1 elements of F is not associated with the spines that correspond to the first k − 1 elements of F ∩ C and to the last k − 1 elements of F ∩ C.) Let spine(C) denote the set of all spines that C contains. More precisely, spine(C) := {S : C contains S as a spine}.
We start by placing a weight (or charge) on a spine depending on the chains that contain it. More precisely, if S ∈ S is a spine and C ∈ C is a full chain, then we define a weight function w(S, C) as follows.
w(S, C) = 1, if S ∈ spine(C) and C contains at least k + 1 members of F, −1, if S ∈ spine(C) and C contains exactly k − 1 members of F.
The motivation behind the above weight function is that, when considering all the chains C associated with a fixed spine S, we would like to show that the total positive charge of these chains is outweighed by their total negative charge. This is confirmed by Lemma 6. Moreover, notice that for any C ∈ C with spine(C) = ∅, we have the following useful identity: Observe that any chain C having exactly k − 1 or at least k + 1 members of F must contain a spine S. It follows that if spine(C) = ∅ then C contains at most k members of F. In other words, the electronic journal of combinatorics 27(1) (2020), #P1.31 Before proving Lemma 6, we need the following straightforward counting lemma, the proof of which we provide for completeness. Proof. Let the set of full chains that contain at least one member of G be X and the set of full chains that contain no member of G be Y . To show that |X| |Y | we will construct an injection from X to Y . Consider any chain C ∈ X.
. , x n }. For simplicity, we will say the permutation corresponding to C is x 1 x 2 x 3 · · · x n .
If {x 1 , x 2 , . . . , x j } is the last set from G in C and x i = 1, then x 1 x 2 · · · x j is a permutation of {1, 2, . . . , j}. Hence, x j+1 j + 1 and 1 i j. Let us consider the chain C corresponding to the permutation obtained by swapping x j+1 with x i in the permutation corresponding to C. If C contains the set {1, 2, . . . , j + 1}, then it must be the case that x j+1 = j + 1. Thus, C contains the set {1, 2, . . . , j + 1}, which contradicts the maximality of j. Therefore, under this map, the full chain C does not contain any member of G. If we map C ∈ X to C ∈ Y in this way, the map is an injection, as desired. Proof. Let a spine S be the chain (Recall that, by definition of a spine, there are exactly k − 1 members of F in {A 1 , . . . , A } and A 1 , A ∈ F.) If all of the chains C ∈ C that contain S as a spine have at most k members of F then since w(S, C) = −1 for each of these chains, our lemma follows trivially. Therefore, we may assume that there is a chain C ∈ C that contains S as a spine such that w(S, C) = 1. By definition (of w(S, C) = 1), such a chain C has at least k + 1 members of F. Moreover, C must have sets P , Q ∈ F with P ⊂ A 1 and A ⊂ Q.
If two sets A, B ∈ F are unrelated to each other and A, B ⊂ A 1 then we have an induced copy of Y k consisting of A, B, the k − 1 members of F in S, and Q. Therefore, ] that contain a member of G 2 \ {A }. Now notice that the number of chains C ∈ C associated with the spine S that have exactly k −1 members of F is a 0 · b 0 and the number of chains C ∈ C associated with S that have at least k + 1 members of F is a 1 · b 1 (here we used that S is not associated with such a chain C unless C has a member of F both above and below S). Therefore, since a 1 a 0 and b 1 b 0 ,

Proofs of Theorem and Theorem 4
Recall that Σ(n, k) denotes the sum of the sizes of the largest k levels in the Boolean lattice B n . First we use a folklore lemma that establishes an inequality very similar to the LYM inequality.
The above lemma can be proved easily: if we fix |F|, the left-hand side of (3) is minimized when we choose the subsets with sizes as near to n/2 as possible. (See [17] for a more detailed proof.) Proof of Theorem 4. Observe that by Lemma 6, Now by double counting, Recall that for any C ∈ C with spine(C) = ∅, the identity (1) holds.
Proof of Theorem 3. The statement of Theorem 3 is true for k = 2 (base case) due to Theorem 2.
If neither ∅ nor [n] are in F, then we may apply Theorem 4 directly to obtain |F| Σ(n, k).
If both ∅ and [n] are in F, then F \{∅, [n]} is induced Y k−1 -free and induced Y k−1 -free. Therefore, it has size at most Σ(n, k − 1) by the induction hypothesis. Since 2 + Σ(n, k − 1) Σ(n, k) for n k + 1 and k 2, we are done. Now, without loss of generality, suppose that ∅ ∈ F and [n] ∈ F. Consider the family F := F \ {∅}. By Theorem 4, we have and, by Lemma 7, |F | Σ(n, k). Now suppose |F | = Σ(n, k). (Otherwise, we are done.) A consequence of the proof of Lemma 7 is that, in order for equality to hold in (3), the quantities n |F | (for F in F ) must be as large as possible-that is, the sets F ∈ F must have size as close to n/2 as possible. More precisely, in order for equality to hold in (3), the list of the quantities n |F | for F ∈ F in decreasing order (with multiplicities) must be the same as the list of the first Σ(n, k) quantities n |S| for S ⊆ 2 [n] in decreasing order (with multiplicities). First, if k and n have different parities, then |F | = Σ(n, k) can only occur if However, in that case, Y k−1 is an induced subposet of F . Hence, adding ∅ produces an induced copy of Y k in F, a contradiction. , then it is easy to see that Y k−1 is an induced subposet of F and adding ∅ produces an induced copy of Y k in F. Otherwise, F must contain all of the sets from [n] n−k 2 +k and n k + 2. But in this case, Y k−1 is again an induced subposet of F , giving an induced copy of Y k in F, again a contradiction.

Concluding Remarks
Our approach to proving Theorem 3 amounts to a weighted double-counting where the choice of the charge values were motivated by a simple discharging argument. However, we believe charge functions that are motivated by more sophisticated discharging arguments, especially with the identification and enumeration of "spines," can lead to progress on other forbidden subposet problems.
We believe that a more general result than Theorem 3 holds. Recall that Y k,r denotes the r-fork with a k-shaft poset and Y k,r denotes its dual.
Conjecture 8. For all k 2 and r 2, there is an n 0 = n 0 (k, r) such that if n n 0 , then La (n, {Y k,r , Y k,r }) = Σ(n, k).