Generalized forbidden subposet problems
Dániel Gerbner, Balázs Keszegh, Balázs Patkós
Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences
P.O.B. 127, Budapest H-1364, Hungary.
Research supported by the János Bolyai Research Fellowship of the Hungarian Academy of Sciences and the National Research, Development and Innovation Office – NKFIH under the grant the grant K 116769.Research supported by the National Research, Development and Innovation Office – NKFIH under the grant the grant K 116769.Research supported by the János Bolyai Research Fellowship of the Hungarian Academy of Sciences and the National Research, Development and Innovation Office – NKFIH under the grants SNN 116095 and K 116769.
Abstract
A subfamily {F1,F2,…,F∣P∣}⊆F of sets is a copy of a poset P in F if there exists a bijection ϕ:P→{F1,F2,…,F∣P∣} such that whenever x≤Px′ holds, then so does ϕ(x)⊆ϕ(x′). For a family F of sets, let c(P,F) denote the number of copies of P in F, and we say that F is P-free if c(P,F)=0 holds. For any two posets P,Q let us denote by La(n,P,Q) the maximum number of copies of Q over all P-free families F⊆2[n], i.e. max{c(Q,F):F⊆2[n],c(P,F)=0}.
This generalizes the well-studied parameter La(n,P)=La(n,P,P1) where P1 is the one element poset, i.e. La(n,P) is the largest possible size of a P-free family. The quantity La(n,P) has been determined (precisely or asymptotically) for many posets P, and in all known cases an asymptotically best construction can be obtained
by taking as many middle levels as possible without creating a copy of P.
In this paper we consider the first instances of the problem of determining La(n,P,Q). We find its value when P and Q are small posets, like chains, forks, the N poset and diamonds. Already these special cases show that the extremal families are completely different from those in the original P-free cases: sometimes not middle or consecutive levels maximize La(n,P,Q) and sometimes no asymptotically extremal family is the union of levels.
Finally, we determine (up to a polynomial factor) the maximum number of copies of complete multi-level posets in k-Sperner families. The main tools for this are the profile polytope method and two extremal set system problems that are of independent interest: we maximize the number of r-tuples A1,A2,…,Ar∈A over all antichains A⊆2[n] such that (i) ∩i=1rAi=∅, (ii) ∩i=1rAi=∅ and ∪i=1rAi=[n].
1 Introduction
The very first theorem of extremal finite set theory is due to Sperner [22] and states that if F is a family of subsets of [n]={1,2,…,n} such that no two sets in F are in inclusion, then ∣F∣≤(⌈n/2⌉n) holds, and equality is achieved if and only if F consists of all the ⌈n/2⌉-element or all the ⌊n/2⌋-element subsets of [n]. Families consisting of all the k-element subsets of [n] are called (full) levels and we introduce the notation (k[n])={F⊆[n]:∣F∣=k} for them. Sperner’s theorem was generalized by Erdős [5] to the case when F is not allowed to contain k+1 mutually inclusive sets, i.e. a (k+1)-chain. He showed that among such families the ones consisting of k middle levels are the largest. In the early eighties, Katona and Tarján [18] introduced a generalization of the problem and started to consider determining the size of the largest family of subsets of [n] that does not contain a configuration defined by inclusions. Such problems are known as forbidden subposet problems and are widely studied (see the recent survey [12]).
In this paper, we propose even further generalizations: we are interested in the maximum number of copies of a given configuration Q in families that do not contain a forbidden subposet P. Before giving the precise definitions, let us mention that similar problems were studied by Alon and Shikhelman [1] in the context of graphs when they considered the problem of finding the most number of copies of a graph T that an H-free graph can contain.
Definition. Let P be an arbitrary poset and F a family of sets. We say that G⊆F is a copy of P in F if there exists a bijection ϕ:P→G such that whenever x≤Px′ holds, then so does ϕ(x)⊆ϕ(x′). Let c(P,F) denote the number of copies of P in F and for any pair of posets P,Q, let us define
[TABLE]
and for families of posets P,Q let us define
[TABLE]
We denote by Pk the chain of length k, i.e. the completely ordered poset on k elements. In particular, P1 is the poset with one element.
Let us state Erdős’s above mentioned result with our notation.
Theorem 1.1** (Sperner [22] for k=1, Erdős [5] for general k).**
For every positive integer k the following holds:
[TABLE]
The area of forbidden subposet problems deals with determining La(n,P)=La(n,P,P1) the maximum size of a P-free family. There are not many results in the literature where other posets are counted. Katona [16] determined the maximum number of 2-chains (copies of P2) in a 2-Sperner (P3-free) family F⊆2[n]. This was reproved in [21] and generalized by Gerbner and Patkós in [11].
Theorem 1.2** ([11]).**
For any l>k the quantity La(n,Pl,Pk) is attained for some family F that is the union of l−1 levels. Moreover, La(n,Pk+1,Pk)=(ikn)⋅(ik−1ik)⋅⋯⋅(i1i2), where i1<i2<⋯<ik<n are chosen arbitrarily such that the values i1,i2−i1,i3−i2,…,ik−ik−1,n−ik differ by at most one.
In this paper, we address the first non-chain instances of the general problem. We will consider the following posets (see Figure 1): let ⋁r denote the poset on r+1 elements 0,a1,a2,…,ar with 0≤ai for all i=1,2,…,r and we write ⋁ for ⋁2. Similarly, let ⋀r denote the poset on r+1 elements a1,a2,…,ar,1 with ai≤1 for all i=1,2,…,r and we write ⋀ for ⋀2. The poset N contains four elements a,b,c,d with a≤c and b≤c,d. The butterfly poset B consists of four elements a,b,c,d with a,b≤c,d.
Let the generalized diamond poset Dk be the poset on k+2 elements a,b1,b2,…,bk,c with a<b1,b2,…,bk<c.
A family that does not contain P2 is called an antichain. A family F that does not contain Pk can be easily partitioned into k−1 antichains F1,…,Fk−1 the following way: let Fi be the set of minimal elements of F∖∪j=1i−1Fj. We call this the canonical partition of F.
Our first theorem relies on some easy observations.
Theorem 1.3**.**
(a) La(n,⋁,P2)=La(n,⋀,P2)=(⌊n/2⌋n).
(b) La(n,{⋁,⋀},P2)=(⌊(n−1)/2⌋n−1).
(c) La(n,B,Dr)=(r(⌊n/2⌋n)).
(d) La(n,⋁,⋀r)=La(n,⋀,⋁r)=(r(⌊n/2⌋n))
The proof of our next theorem uses the notion of profile vectors (ordinary and l-chain profile vectors). Here and throughout the paper h(x) denotes the binary entropy function, i.e. h(x)=−xlog2x−(1−x)log2(1−x).
Theorem 1.4**.**
(a) La(n,P3,⋀r)=La(n,P3,⋁r)=(irn)(r(⌊ir/2⌋ir)) for some ir with ir=(1+o(1))2r+12rn.
(b) La(n,P4,Dr)=(jrn)(irjr)(r(⌊(jr−ir)/2⌋jr−ir)) for some ir=(1+o(1))2r+2n and either jr=n−ir or jr=n−ir−1.
(c) 2(c+o(1))n≤La(n,P3,N)≤o(23n),
*where c=h(c0)+3c0h(c0/(1−c0))=2.9502... with c0 being the real root of the equation 0=7x3−10x2+5x−1.
*
Let us return for a moment to the original forbidden subposet problems. The main conjecture of the area was first published by Griggs and Lu in [13].
Conjecture 1.5**.**
For a poset P let us denote by e(P) the largest integer m such that for any n, any family F⊆2[n] consisting of m consecutive levels is P-free. Then
[TABLE]
holds.
In words, Conjecture 1.5 states that to obtain an asymptotically largest P-free family F⊆2[n] one has to consider as many middle levels of 2[n] as possible without creating a copy of P. Again, we refer the interested Reader to the recent survey [12] to see for which families of posets Conjecture 1.5 has been verified.
However, already Theorem 1.2 shows that to make Conjecture 1.5 valid in the more general context one has to remove at least the word consecutive. All parts of both Theorem 1.3 and Theorem 1.4 suggest that a general conjecture stating that for any pair P,Q of posets La(n,P,Q) is asymptotically attained at a sequence of families consisting of full levels of 2[n]. But this is not the case at all! There are pairs of posets for which all families consisting of full levels are very far from being optimal. Let us consider La(n,D2,P3) the maximum number of 3-chains in diamond-free families. Every family that contains at least three full levels of 2[n] contains a copy of D2, while a family that is the union of at most two levels, does not contain any copy of P3. Therefore, if F is D2-free and is the union of full levels, then c(P3,F)=0, while there are D2-free families with lots of copies of P3. Note that D2 is the smallest poset for which Conjecture 1.5 has not been proved.
Theorem 1.6**.**
For the generalized diamond posets and integers k>l the following holds:
[TABLE]
Note that Theorem 1.6 implies La(n,Dk,Dl)=θk,l(La(n,P3,P2)) for any fixed k and l and the exact value of La(n,P3,P2) is given by Theorem 1.2. So it is a natural question whether the limit dk,l=lim∞La(n,P3,P2)La(n,Dk,Dl) exists and if so, what its value is. In the simplest case k=2,l=1 the above inequalities and Theorem 1.2 imply 1/3≤d2,1≤1.
So what can be saved from Conjecture 1.5 in the more general context? Let l(P) be the height of a poset P, i.e. the length of the longest chain in P. Clearly, if F is the union of any l(P)−1 full levels, it must be P-free. On the other hand if Fn=∪j=1l(Q)(ij[n]) is the union of l(Q) full levels with ij+1−ij≥cn for some constant c for all j=1,2,…l(Q)−1, then Fn contains many copies of Q.
Therefore we propose the following.
Conjecture 1.7**.**
For any pair P,Q of posets with l(P)>l(Q) there exist a sequence of P-free families Fn⊆2[n] all of which are unions of full levels such that
[TABLE]
holds.
As we have already seen, this conjecture often holds even if l(P)≤l(Q). We say that for a pair P,Q of posets Conjecture 1.7 strongly holds if for large enough n we have La(n,P,Q)=c(Q,Fn) and almost holds if La(n,P,Q)=O(nkc(Q,Fn)) for some k that depends only on P and Q. In both cases we also assume the family Fn is P-free and is the union of full levels, but we do not assume anything about l(P) and l(Q). Parts (a) (c), and (d) of Theorem 1.3 show that Conjecture 1.7 strongly holds for those pairs of posets.
In Theorem 1.3 and Theorem 1.4 we dealt with La(n,Pl(Q)+1,Q) for different posets Q. (In the case of La(n,B,Dr) it is implicit, as the B-free property implies P4-free property.) We knew the place of every element of every copy of Q in the canonical partition. In the following we deal with these kind of problems. We introduce the following binary operations of posets: for any pair Q1, Q2 of posets we define Q1⊗rQ2 by adding an antichain of size r between Q1 and Q2. More precisely, let us assume Q1 consists of q11,…,qa1 and Q2 consists of q12,…,qb2. Then R=Q1⊗rQ2 consists of q11,…,qa1,m1,m2,…,mr,q12,…qb2. We have qi1<Rqj1 if and only if qi1<Q1qj1 and similarly qi2<Rqj2 if and only if qi2<Q2qj2. Also we have qi1<Rmk<Rqj2 for every i, k, and j. Finally, the mk’s form an antichain. Note that l(Q1⊗rQ2)=l(Q1)+l(Q2)+1.
Let Q⊕r denote the poset Q⊗r0, where 0 is the empty poset, i.e. Q⊕r is obtained from Q by adding r elements that form an antichain and that are all larger than all elements of Q. Similar operations of posets were considered first in the area of forbidden subposet problems by Burcsi and Nagy [3].
We will obtain bounds on La(n,Pl(Q1⊗rQ2)+1,Q1⊗rQ2) involving bounds on La(n,Pl(Q1)+1,Q1) and La(n,Pl(Q2)+1,Q2). For this we will need the following auxiliary statement that can be of independent interest.
Theorem 1.8**.**
(a) For every r≥3 and antichain A⊆2[n] the number γ0,nr(A) of r-tuples A1,A2…,Ar with ∣⋂i=1rAi∣=0 and ⋃i=1rAi=[n] is at most n2rγ0,nr((⌊n/2⌋[n])). If r=2 and n is even, then γ0,n2(A)≤γ0,n2((n/2[n])), while if r=2 and n is odd, then γ0,n2(A)≤(⌊n/2⌋−1n−1).
(b) For every r there exists a sequence ln such that if A⊆2[n] is an antichain, then the number β0r(A) of r-tuples A1,A2…,Ar with ∣⋂i=1rAi∣=0 is at most n2r+1β0r((ln[n])).
(c) If A⊆2[n] is an antichain, then β02(A)≤21(⌊n/3⌋n)(⌊n/3⌋⌈2n/3⌉) and this is sharp as shown by (⌊n/3⌋[n]) if n≡0,1 mod 3 and by (⌈n/3⌉[n]) if n≡2 mod 3.
The reason for the strange indices is that we will prove a somewhat more general result in Section 4. The r=2 part of Theorem 1.8 (a) was proved by Bollobás [2].
Theorem 1.9**.**
Let Q1,Q2 be two non-empty posets.
(a) If r≥2, then we have
[TABLE]
Furthermore, if r≥3 and Conjecture 1.7 almost holds for the pairs Pl(Q1)+1,Q1 and Pl(Q2)+1, Q2, then so it does for the pair Pl(Q1⊗rQ2)+1,Q1⊗rQ2.
(b) If r=1, then we have
[TABLE]
Furthermore, if Conjecture 1.7 strongly/almost holds for the pairs Pl(Q1)+1,Q1 and Pl(Q2)+1, Q2, then so it does for the pair Pl(Q1⊗1Q2)+1,Q1⊗1Q2.
Theorem 1.10**.**
Let Q be a non-empty poset.
(a) If r≥2 and n∈N, then there exists an i=i(r,n) such that
[TABLE]
Furthermore, if Conjecture 1.7 almost holds for the pair Pl(Q)+1,Q, then so it does for the pair Pl(Q)+2,Q1⊕r.
(b) If r=1, then we have
[TABLE]
Furthermore, if Conjecture 1.7 strongly/almost holds for the pair Pl(Q)+1,Q, then so it does for the pair Pl(Q)+2,Q⊕1.
We can apply Theorem 1.9 and Theorem 1.10 to complete multi-level posets. Let Kr1,r2,…,rs denote the poset on ∑i=1sri elements a11,a21,…,ar11,a12,a22,…,ar22,…,a1s,a2s,…,arss with ahi<alj if and only if i<j. Observe that ⋁r=K1,r, B=K2,2 and Dr=K1,r,1.
Corollary 1.11**.**
For any complete multi-level poset Kr1,r2,…,rs Conjecture 1.7 almost holds for the pair Ps+1,Kr1,r2,…,rs.
Corollary 1.12**.**
Conjecture 1.7 strongly holds for the pair Ps+1,Kr1,r2,…,rs if for every i<s at least one of ri and ri+1 is equal to 1.
Corollary 1.12 does not tell us anything about the set sizes in the family containing the most number of copies of Kr1,r2,…,rs. The next theorem gives more insight for an even more special case.
Theorem 1.13**.**
The value of La(n,Pl+3,Kr,1,…,1,s) is attained for a family F=∪j=1l+2(ij[n]), where i1=⌊i2/2⌋, il+2=⌊(n+il+1)/2⌋ and i3−i2, i4−i3,…,il+1−il differ by at most 1.
The rest of the paper is organized as follows: we prove Theorem 1.6 and Theorem 1.3 in Section 2. We explain the profile polytope method and prove Theorem 1.4 in Section 3. Theorem 1.8, Theorem 1.9, Theorem 1.10, and their corollaries are proved in Section 4, while Section 5 contains some concluding remarks.
2 Proofs of Theorem 1.6 and Theorem 1.3
Proof of Theorem 1.6.
We start by proving the upper bound. Let F⊆2[n] be a Dk-free family. As for any poset P the canonical partition of a P-free family can consist of at most ∣P∣−1 subfamilies, we can assume that the canonical partition of F is ∪i=1k+1Fi. In any copy of Dl in F, the sets corresponding to the top and bottom element of Dl come from Fi and Fj with i−j≥2. The number of such pairs of indices is (2k+1)−k. Let us bound the number of copies of Dl with top element from Fi and bottom element from Fj. As Fi∪Fj is P3-free, there are at most La(n,P3,P2) many ways to choose the top and the bottom elements FB⊂FT. As F is Dk-free there can be at most k−1 sets in F lying between FB and FT, so the number of copies of Dl with FB,FT being top and bottom is at most (lk−1). The upper bound on La(n,Dk,Dl) follows.
For the lower bound we need a construction. Let F1∪F2⊆2[n−k+1] be the canonical partition of the P3-free family F with c(P2,F)=La(n−k+1,P3,P2). For j=3,4,…,k+1 let Fj={F∪[n−k,n−k+j−1]:F∈F2}. We claim that G=∪i=1k+1Fi is Dk-free with c(Dl,F)≥(lk−1)La(n−k+1,P3,P2). Indeed, every set G∈G is contained in a set Fk+1∈Fk+1 and contains a set F1∈F1, therefore if there was a copy of Dk, we could assume that its bottom element is from F1 and its top element is from Fk+1. But any Fk+1∈Fk+1 contains exactly one element from each Fi where i=2,3,…,k, so there is no space for a copy of Dk. On the other hand, for every pair F1⊂F2 in F1∪F2 we can add l sets from {F2∪[n−k,n−k+j−1]:j=3,4,…,k+1} to form a copy of Dl. For each such pair we will obtain (lk−1) such copies.
∎
Proof of Theorem 1.3.
To prove (a), by symmetry, it is enough to show La(n,⋁,P2)=(⌊n/2⌋n). Consider any ⋁-free family F⊆2[n] and its canonical partition F1∪F2. By the ⋁-free property of F, elements of F1 are contained in at most one copy of P2. Also, as the ⋁-free property implies the P3-free property, every copy of P2 in F must contain a set from F1. Sperner’s theorem yields c(P2,F)≤∣F1∣≤(⌊n/2⌋n). On the other hand F:={F⊆[n]:∣F∣=⌊n/2⌋}∪{[n]} is ⋁-free and every ⌊n/2⌋-element set forms a copy of P2 with [n].
We continue with proving (b). We will need the following definition. For any family F, the comparability graph of F has vertex set F and two sets F,F′∈F are joined by an edge if F⊆F′ or F′⊆F holds. The connected components of the comparability graph of F are said to be the components of F. If a family F is both ⋁-free and ⋀-free, then its components are either isolated vertices or isolated edges in the comparability graph. Therefore c(P2,F) is the number of components that are isolated edges. It follows that La(n,{⋁,⋀},P2)≤21La(n,{⋁,⋀},P1)=(⌊(n−1)/2⌋n−1) where the result in the last equation was proved by Katona and Tarján [18]. The construction (given also in [18]) F:=(⌊(n−1)/2⌋[n−1])∪{{n}∪F:F∈(⌊(n−1)/2⌋[n−1])} shows that the above upper bound can be attained.
To prove (c), let us consider a B-free family F⊆2[n] and let M={M∈F:∃F′,F′′∈Fsuch thatF′⊊M⊊F′′}. As B-free implies P4-free, we obtain that M is an antichain, thus ∣M∣≤(⌊n/2⌋n), by Theorem 1.1. Moreover, if M∈M, then there do not exist two elements F1,F2∈F with M⊊F1,F2. Indeed, by the definition of M there exists F′∈F with F′⊊M, and F′,M,F1,F2 would form a copy of B. Similarly, for every M∈M there exists exactly one element F∈F with F⊊M. Therefore a copy of Dr contains r elements of M, and they determine the remaining two elements, which implies c(Dr,F)≤(r∣M∣)≤(r(⌊n/2⌋n)).
The construction F:={∅,[n]}∪(⌊n/2⌋[n]) shows that this upper bound can be attained.
To prove (d), by symmetry, it is enough to show La(n,⋁,⋀r)=(r(⌊n/2⌋n)). If F is ⋁-free, then it is in particular P3-free. Consider its canonical partition. Then a copy of ⋀r contains r elements from F1 and one from F2. Moreover, an r-tuple from F1 may form a copy of ⋀r with at most one element from F2, otherwise there is a copy (actually r copies) of ⋁ in F. As F1 is an antichain, by Theorem 1.1, the upper bound La(n,⋁,⋀r)≤(r(⌊n/2⌋n)) follows and F:={[n]}∪(⌊n/2⌋[n]) shows that this can be attained.
∎
3 The profile polytope method
In this section we prove Theorem 1.4 after introducing the notions of profile vectors and profile polytopes. For a family F⊆2[n] of sets, let α(F)=(α0,α1,…,αn) denote the profile vector of F, where αi=∣{F∈F:∣F∣=i}∣. Many problems in extremal finite set theory ask for the largest size of a family in a class A⊆22[n]. This question is equivalent to determining maxF∈Aα(F)⋅1, where 1 is the vector of length n+1 with all entries being 1, and ⋅ denotes the scalar product.
More generally, consider a weight function w:{0,1,…,n}→R, and assume we want to maximize w(F):=∑F∈Fw(∣F∣). Then this is equivalent to maximizing α(F)⋅w, where w=(w(0),w(1),…,w(n)).
As A⊆22[n] holds, we have {α(F):F∈A}⊆Rn+1 and therefore we can consider its convex hull μ(A) that we call the profile polytope of A.
It is well known that any weight function with the above property is maximized by an extreme point of μ(A) (a point that is not a convex combination of other points of μ(A)), moreover if such a weight function is non-negative, then it is maximized by an essential extreme point, i.e. an extreme point which is maximal with respect to the coordinate-wise ordering. First results concerning profile polytopes were obtained in [17, 7, 8, 9, 10] and the not too recent monograph of Engel [6] contains a chapter devoted to this topic.
Using this we can determine La(n,P3,P2), and using induction with this as the base case one can determine La(n,Pk+1,Pk), but in other cases we will need a more powerful tool than ordinary profile vectors. The notion of l-chain profile vector αl(F) of a family F⊆2[n] was introduced by Gerbner and Patkós [11] and denotes a vector of length (ln+1). The coordinates are indexed by l-tuples of [0,n] and αl(F)(i1,i2,…,il) is the number of chains F1⊊F2⊊⋯⊊Fl such that Fj∈F and ∣Fj∣=ij for all 1≤j≤l.
For a set A⊆22[n] one can define the l-chain profile polytope μl(A), its extreme points and essential extreme points analogously to the above. Note that for l=1 we get back the definition of the original profile polytope.
Let Sn,k be the class of all k-Sperner families on [n].
Lemma 3.1** (Gerbner, Patkós, [11]).**
The essential extreme points of μl(Sn,k) are the l-chain vectors of k-Sperner families that consist of the union of k full levels.
Let us state the immediate consequence of the above lemma that we will use in our proofs in the remainder of this section.
Corollary 3.2**.**
Let l≤k and w:(l2[n])→R+ be a weight function such that w({F1,F2,…Fl}) depends only on ∣F1∣,∣F2∣,…,∣Fl∣. Then the maximum of
[TABLE]
over all families F∈Sn,k is attained at some family that consists of k full levels.
Proof of Theorem 1.4..
To prove (a) we show La(n,P3,⋀r)=(irn)(r(⌊ir/2⌋ir)) as the other statement follows by symmetry. Let us consider the canonical partition of a P3-free family F. Note that a copy of ⋀r contains exactly one element F from F2 and r elements F1,F2,…,Fr∈F1 with Fi⊊F for all 1,2,…,r. Let us consider a set F∈F2. The sets of F1 contained in F form an antichain, thus by Theorem 1.1, their number is at most (⌊∣F∣/2⌋∣F∣). Therefore, the number of copies of ⋀r that contain F is at most (r(⌊∣F∣/2⌋∣F∣)) and we obtain
[TABLE]
Therefore if we set w(i):=(r(⌊i/2⌋i)), then we can apply Corollary 3.2 with l=k=1 to obtain c(⋀r,F)≤max0≤i≤n(in)w(i). On the other hand, the families F(i)=(i[n])∪(⌊i/2⌋[n]) are P3-free and c(⋀r,F(i))=(in)w(i) showing La(n,P3,⋀r)=max0≤i≤n(in)w(i).
To obtain the value of ir we need to maximize f(i):=(in)w(i). Considering
[TABLE]
when i tends to infinity with n. For constant values of i, the ratio f(i)/f(i+1) is easily seen to be smaller than 1 (if n is big enough), therefore the maximum of f(i) is attained at ir=(1+o(1))2r+12rn as stated.
To prove (b) we consider the canonical partition of a P4-free family F⊆2[n]. Any copy of Dr in F must contain one set from F1,F3 each and r sets from F2. For any F1∈F1,F3∈F3 with F1⊂F3, the number of copies of Dr containing F1 and F3 is (rm), where m=∣MF1,F3∣ with MF1,F3={F∈F2:F1⊂F⊂F3}. As MF1,F3′={M∖F1:M∈MF1,F3} is on antichain in F3∖F1, we have m≤(⌊(∣F3∣−∣F1∣)/2⌋∣F3∣−∣F1∣). Therefore, we obtain
[TABLE]
where to obtain the last inequality we applied Corollary 3.2 with l=k=2 and w(i,j)=(r(⌊(j−i)/2⌋j−i)). Observe that if ir and jr are the values for which this maximum is taken, then for the family F=(ir[n])∪(jr[n])∪(⌊(ir+jr)/2⌋[n]) we have c(Dr,F)=(jrn)(irjr)(r(⌊(jr−ir)/2⌋jr−ir)).
To obtain the value of ir and jr let us fix x=j−i first. Note that (jn)(ij)=(xn)(in−x), so we have
[TABLE]
which implies that ir+jr=n or ir+jr=n−1 holds.
Let g(i)=(in)(in−i)(r(⌊(n−i)/2⌋n−i)), then
[TABLE]
This implies the maximum of g(i) is attained at i=(1+o(1))⋅2r+21 .
To prove (c) we again consider the canonical partition of a P3-free family F⊆2[n]. A copy of N in F must contain two sets from F1 and two from F2. Let a,b,c,d be the four elements of N with a≤c and b≤c,d. For every copy of N in F there is a bijection ϕ from N to that copy. Then we count the copies of N in F according to the images ϕ(b),ϕ(c). Clearly, they form a 2-chain in F, the possible images of d form an antichain among those sets of F2 that contain ϕ(b) and the possible images of a form an antichain among those sets of F1 that are contained by ϕ(c). Therefore, we obtain
[TABLE]
where to obtain the last inequality we applied Corollary 3.2 with l=k=2 and w(i,j)=(⌊2n−i⌋n−i)(⌊2j⌋j). We have (jn)(ij)=(j−in)(in−j+i), thus we get
[TABLE]
Note that j=⌈3n/4⌉ and i=⌈n/4⌉ show the exponent cannot be improved with this method.
To obtain the lower bound consider the P3-free families Fi,j=(i[n])∪(j[n]) with 0≤i<j≤n. Observe that we have c(N,Fi,j)≥41(jn)(ij)2(j−in−i)=:g(i,j) (the 1/4-factor is due to the fact that copies of B are counted 4 times as copies of N). To maximize g(i,j) we first fix j−i and consider
[TABLE]
It is easy to see that this fraction becomes smaller than 1 when i is roughly n−j+1. Thus the maximum of g(i,j) is asymptotically achieved when i=n−j.
Similarly, we have
[TABLE]
Therefore, if we write j=(c+o(1))n, we obtain that g is maximized when (2c−1)6c2(1−c)4=1. After taking the square root of the expression on the left hand side, this is equivalent to that 0=7c3−10c2+5c−1 holds. The solution of this equation is c0=0.69922... As g(n−j,j)=41(jn)(n−jj)3=Ω(2h(j/n)+3njh((n−j)/j)/n2), the lower bound follows by plugging in j=0.69922n.
∎
4 The ⊗r operation and copies of complete multi-level posets
In this section we prove results concerning the binary operation Q1⊗rQ2. We introduce two types of profile vectors.
For a family F⊆2[n] of sets, let βr(F)=(β0r,β1r,…,βn−1r) denote the r-intersection profile vector of F, where \beta^{r}_{i}=\beta^{r}_{i}({\mathcal{F}})=|\{\{F_{1},F_{2},\dots,F_{r}\}:F_{j}\in{\mathcal{F}},\textrm{these are r different sets and}\ ∣⋂j=1rFj∣=i}∣.
For a family F⊆2[n] of sets, let γr(F)=(γ0,1r,γ0,2r,…,γ0,nr,γ1,2r,…,γn−1,nr) denote the r-intersection-union profile vector of F, where \gamma_{i,j}^{r}=\gamma_{i,j}^{r}({\mathcal{F}})=|\{\{F_{1},\dots,F_{r}\}:F_{1},\dots,F_{r}\in{\mathcal{F}},\textrm{these are r different sets},|F_{1}\cap\dots\cap F_{r}|=i,|F_{1},\cup\dots\cup F_{r}|=j\}|. Note that if A⊆2[n] is an antichain, then γi,jr(A)>0 implies j−i≥2, therefore the number of non-zero coordinates in γr(A) is at most (2n+1)−n=(2n)≤n2.
Let us illustrate with two examples why these profile vectors can be useful in counting copies of different posets. Let F be a P3-free and G be a P4-free family. We will estimate c(Kp,r,F) and c(Kp,r,s,G). If we consider the canonical partitions of F=F1∪F2 and G=G1∪G2∪G3, then a copy of Kp,r in F contains p sets from F1 and r sets from F2. If we fix F1,…,Fr∈F2, then the p ”bottom” sets of the copies of Kp,r in F containing F1,…,Fr form an antichain in {F∈F1:F⊆⋂j=1rFj}. Therefore, by Theorem 1.1, the number of these copies is at most (p(⌊∣⋂j=1rFj∣/2⌋∣⋂j=1rFj∣)), so summing up for all possible r-tuples of F2 we obtain c(Kp,r,F)≤βr(F)⋅wp and consequently
[TABLE]
where the jth entry of the vector wp is (p(⌊j/2⌋j)).
Similarly, if we consider the canonical partition of G=G1∪G2∪G3, then a copy of Kp,r,s in G contains p sets from G1, r sets from G2 and s sets from G3. If we fix G1,…,Gr∈G2, then the bottom p and top s sets of copies of Kp,r,s containing G1,…,Gr form antichains in {G∈G1:G⊆⋂j=1rGj} and {G∈G3:G⊇⋃j=1rGj}. Therefore, using again Theorem 1.1, we obtain c(Kp,r,s,G)≤γr(G2)⋅wp,s and consequently
[TABLE]
where the (i,j)th entry of the vector wp,s is (p(⌊i/2⌋i))(s(⌊(n−j)/2⌋n−j)).
Therefore determining the convex hull (and more importantly its extreme points) of the r-intersection profile vectors and r-intersection-union profile vectors of antichains would yield upper bounds on La-functions of complete multi-level posets.
We are not able to determine these convex hulls, we will only obtain upper bounds on the coordinates of these profile vectors. Note that Theorem 1.8 is about special coordinates, so the next two theorems imply that result.
Theorem 4.1**.**
(a) If F⊆2[n] is an antichain and j−i is even, then γi,j2(F)≤γi,j2(((i+j)/2[n]))=21(jn)(ij)((j−i)/2j−i). If j−i is odd, then γi,j2(F)≤(jn)(ij)((⌊j−i)/2⌋−1j−i−1).
(b) If F is an antichain and r≥3, then γi,jr(F)≤n2rγi,jr((⌊(i+j)/2⌋[n])).
During the proof we will use several times that the number of pairs A⊂B⊂[n] with ∣A∣=a,∣B∣=b is (bn)(ab)=(b−an)(an−(b−a)). The first calculation is obvious, for the second pick first B∖A from [n] and then A from [n]∖(B∖A).
Proof.
To see (a), we first consider the special case i=0,j=n. Observe that γ0,n2(F) is the number of complement pairs in F. In an antichain, by Theorem 1.1, this is at most ∣F∣/2≤(⌊n/2⌋n)/2. If n is even, then this is achieved when F=(n/2[n]), while the case of odd n was solved by Bollobás [2], who showed that the number of such pairs is at most (⌊n/2⌋−1n−1) and this is sharp as shown by {F∈(⌊n/2⌋[n]):1∈F}∪{F∈(⌈n/2⌉[n]):1∈/F}.
To see the general statement observe that for pair I⊂J writing FI,J={F∈F:I⊆F⊆J} we have γi,j2(F)=∑I∈(i[n]),J∈(j[n])γ0,j−i2(FI,J). Therefore if j−i is even we obtain γi,j2(F)≤(jn)(ij)γ0,j−i2(((j−i)/2[j−i]))=γi,j2(((j+i)/2[n])), while if j−i is odd, we obtain γi,j2(F)≤(jn)(ij)((⌊j−i)/2⌋−1j−i−1).
To show (b) it is enough to prove the statement fo i=0,j=n. Indeed, γi,jr(F)=∑I,Jγ0,j−ir(FI,J)≤(jn)(ij)γ0,j−ir((⌊(j−i)/2⌋[j−i])=γi,jr(⌊(j+i)/2⌋[n]).
We proceed by induction on r. We postpone the proof of the base case r=3, and assume the statement holds for r−1 and any i<j. Let us consider r−1 sets F1,…,Fr−1 of F and examine which sets can be added to them as Fr to get empty intersection and [n] as the union. Let F′ be the family of those sets. Let A=∩l=1r−1Fl and B=∪l=1r−1Fl with a=∣A∣ and b=∣B∣. Then members of F′ contain the complement of B and do not intersect A, and F′ is Sperner. If we remove B from them, the resulting family is Sperner on an underlying set of size b−a, thus have cardinality at most (⌊(b−a)/2⌋b−a)=:w(a,b). Note that we count every r-tuple F1,…,Fr exactly r times. It implies
[TABLE]
.
By induction this is at most n2n2r−2maxa<bγa,br−1((⌊(a+b)/2⌋[n]))w(a,b). Let
[TABLE]
If we fix b−a and consider f(a+1,b+1)f(a,b)=n−(b−a)−aa+1, we can see that the maximum is taken when b+a=n or b+a=n−1 depending on the parity of b−a and n.
Let a∗,b∗ be the values for which the above maximum is taken. Note that for any a∗<p<b∗ we have rγ0,nr((p[n]))≥(b∗n)(a∗b∗)γ0,b∗−a∗r−1((p−a∗[b∗−a∗]))(p−n+b∗b∗−a∗), by counting only those r-tuples where the first r−1 sets have intersection of size a∗ and union of size b∗. (This way we count those r-tuples at most r times). This is exactly f(a∗,b∗) if p=⌊n/2⌋=⌊(a∗+b∗)/2⌋, so we obtained rγ0,nr(F)≤n2rrγ0,nr((⌊n/2⌋[n])) as required.
For r=3 we similarly consider two members of F and examine which sets can be added to them to get empty intersection and [n] as the union. This leads to
[TABLE]
Note that if the maximum is taken at a′ and b′ with b′−a′ even, then
part (a) of the theorem gives γa′,b′2(F)≤γa′,b′2(((b′+a′)/2[n])) so 3γ0,n3(F)≤n2γa′,b′2(((b′+a′)/2[n]))w(a′,b′). This essentially lets us use r=2 as the base case of induction, and finish the proof of this case similarly to the induction step above.
Let us choose a∗,b∗ that maximizes this upper bound with b∗−a∗=b′−a′. Similarly to the computation about f(a,b) , we have a∗+b∗=n or n−1 depending on the parity of n. Then we obtain
3γ0,n3(F)≤n2γa∗,b∗2(((b∗+a∗)/2[n]))w(a∗,b∗)≤n2(b∗n)(a∗b∗)γ0,b∗−a∗2(((b∗+a∗)/2[b∗−a∗]))w(a∗,b∗). The lower bound on 3γ0,n3((⌊n/2⌋[n])) is (b∗n)(a∗b∗)γ0,b∗−a∗2((p−a∗[b∗−a∗]))(p−a∗b∗−a∗)
as in the inductive step.
However, if b′−a′ is odd, then γa′,b′2((⌊(a′+b′)/2⌋[n]))=0. But we know by part (a)
[TABLE]
Similarly to the previous cases, if b′−a′ is fixed, then the maximum of the right hand side is taken for some a∗,b∗ with b∗−a∗=b′−a′ and a′+b′=n if n is odd, and a′+b′=n−1 or a′+b′=n+1 if n is even. Thus we can assume ⌊n/2⌋=(a∗+b∗−1)/2.
On the other hand, since b∗−a∗ is odd, we have
[TABLE]
by counting only those triples where two of the sets have intersection of size a∗ and union of size b∗−1. We can pick first the (b∗−1)-set B and the a∗-set A in (b∗−1n)(a∗b∗−1) ways, then among {G∈(⌊(a∗+b∗)/2⌋[n]):A⊂G⊂B} we can pick a pair G1,G2 with G1∩G2=A, G1∪G2=B in ((b∗−a∗−1)/2b∗−a∗−1)/2 ways and then the third set contains the complement of B and does not intersect A. Using that (2a∗+b∗−1−n+b∗−1b∗−a∗−1)≥((b∗−a∗−1)/2−1b∗−a∗−1)
this implies
[TABLE]
as b∗≥n/2.
∎
Theorem 4.2**.**
(a) For any antichain F⊆2[n] we have βi2(F)≤βi2((j(i)[n])), where j(i)=i+⌊(n−i)/3⌋ if n−i≡0,1 mod 3 and j(i)=i+⌈(n−i)/3⌉ if n−i≡2 mod 3.
(b) For every r≥3 and i≤n there exists j(r,i,n) such that βir(F)≤n2r+1βir((j(r,i,n)[n])) holds for any antichain F⊆2[n].
Proof.
First we prove (a) for the special case i=0. Let F⊆2[n] be an antichain, and let F={F:F∈F}, where F=[n]∖F. As F is an antichain, so is F, and thus F∪F is P3-free. Note that for every pair F1,F2∈F with ∣F1∩F2∣=0 and F1∪F2=[n], we have two 2-chains F1⊊F2 and F2⊊F1. Also, every 2-chain in F∪F comes from a pair F1,F2∈F with ∣F1∩F2∣=0 and F1∪F2=[n].
Therefore, if we take the canonical partition of F∪F into F1∪F2 and introduce the weight function
w(F)=21(⌈∣F∣/2⌉∣F∣) if F∈F2,F∈/F2 and w(F)=1/2 if F∈F2,F∈F2, then the number of disjoint pairs in F equals ∑F∈F2w(F). This weight function does not depend only on the size of F, but w′(f)=21(⌈∣F∣/2⌉∣F∣) does and obviously w(F)≤w′(F) holds for all F’s. As proved by Katona in Theorem 3 of [16] this weight function is maximized when F2=(⌈2n/3⌉[n]). As F2 does not contain complement pairs, it also maximizes w.
To see the general statement of (a), we can apply the special case to any I⊆[n] and FI={F∖I:I⊆F∈F}. We obtain
[TABLE]
To see (b), let F⊆2[n] be antichain. Observe
[TABLE]
where j(r,i,n)=⌊(i+j∗)/2⌋ with j∗ being the value of j that maximizes γi,jr((⌊(i+j)/2⌋[n])). The penultimate inequality follows from Theorem 4.1.
∎
Proof of Theorem 1.9.
Let Q1,Q2 be non-empty posets and let us consider the canonical partition of a Pl(Q1⊗rQ2)+1-free family F⊆2[n]. Then in any copy of Q1⊗rQ2 in F, if F1,…,Fr correspond to the r middle elements forming an antichain, we must have F1,…,Fr∈Fl(Q1)+1. Also, if a copy of Q1⊗rQ2 contains F1,…,Fr, then the sets corresponding to the Q1 part of Q1⊗rQ2 must be contained in ⋂l=1rFl, while the sets corresponding to the Q2 part of Q1⊗rQ2 must contain ⋃l=1rFl. Therefore the number of copies of Q1⊗rQ2 in F that contain F1,…,Fr is at most La(∣⋂l=1rFl∣,Pl(Q1)+1,Q1)⋅La(n−∣⋃i=1rFl∣,Pl(Q2)+1,Q2). We obtained that the total number of copies of Q1⊗rQ2 in F is at most
[TABLE]
If r≥2, then grouping the summands in (1) according to the pair (∣⋂l=1rFl∣,∣⋃l=1rFl∣) we obtain
[TABLE]
where the (i,j)th coordinate of w is La(i,Pl(Q1),Q1)⋅La(n−j,Pl(Q2)+1,Q2). Clearly, we have
[TABLE]
where the last inequality follows from Theorem 4.1. We can calculate γi,jr((⌊(i+j)/2⌋[n])) by picking the union of size j and the intersection of size i first, which finishes the proof of the upper bound of part (a).
To see the furthermore part suppose that the above maximum is obtained when i takes the value i∗ and j takes the value j∗. We know that there exist two families F1,i∗⊆2[i∗] and F2,n−j∗⊆2[n−j∗], both unions of full levels, integers k1,k2 and constants C1,C2 such that C1(i∗)k1c(Q1,F1,i∗)≥La(i∗,Pl(Q1)Q1) and C2(j∗)k2c(Q2,F2,n−j∗)≥La(n−j∗,Pl(Q2),Q2) hold. Therefore by the upper bound already proven, we know that La(n,Pl(Q1⊗rQ2)+1,Q1⊗rQ2) is at most n2r+2C1(i∗)k1C2(j∗)k2γi∗j∗r((⌊(i∗+j∗)/2⌋n))La(i∗,Pl(Q1),Q1)La(n−i∗,Pl(Q2),Q2).
If F1,i∗ consists of levels of set sizes h1,…,hl(Q1) and F2,n−i∗ consists of levels of set sizes h1′,…,hl(Q2′)′, then for the family
[TABLE]
we have c(Q1⊗Q2,F)≥γi∗j∗r((⌊(i∗+j∗)/2⌋n))La(i∗,Pl(Q1),Q1)La(n−i∗,Pl(Q2),Q2). Therefore with k=2r+2+k1+k2 the family F shows that Conjecture 1.7 almost holds for the pair Pl(Q1⊗rQ2)+1,Q1⊗rQ2.
If r=1, then ∪F1=∩F1=F1, so (1) becomes
[TABLE]
We can apply Corollary 3.2 with l=k=1 and w(i)=La(i,Pl(Q1),Q1′)⋅La(i,Pl(Q2),Q2′) to obtain
[TABLE]
as required.
As the proofs are almost identical we only show the ’strongly holds’ case of the furthermore part of (b). Suppose that the above maximum is obtained when i takes the value i∗. We know that there exist two families F1,i∗⊆2[i∗] and F2,n−i∗⊆2[n−i∗], both unions of full levels, such that c(Q1,F1,i∗)=La(i∗,Pl(Q1)Q1) and c(Q2,F2,n−i∗)=La(n−i∗,Pl(Q2),Q2) hold. If F1,i∗ consists of levels of set sizes j1,…,jl(Q1) and F2,n−i∗ consists of levels of set sizes j1′,…,jl(Q2′)′, then for the family
[TABLE]
we have c(Q1⊗Q2,F)=(i∗n)La(i∗,Pl(Q1),Q1)La(n−i∗,Pl(Q2),Q2).
∎
Proof of Theorem 1.10.
The proof goes very similarly to the proof of Theorem 1.9. Let us consider the canonical partition of a Pl(Q⊗r)+1-free family F⊆2[n]. Then in any copy of Q⊗r in F, if F1,…,Fr correspond to the r top elements forming an antichain, we must have F1,…,Fr∈Fl(Q)+1. Also, if a copy of Q⊗r contains F1,…,Fr, then the sets corresponding to the other elements of the poset must be contained in ⋂l=1rFl. Let j=∣⋂l=1rFl∣. Then the number of copies of Q⊗r in F that contain F1,…,Fr is at most La(j,Pl(Q)+1,Q). If r≥2, we obtain
[TABLE]
where the jth coordinate of w is La(j,Pl(Q)+1,Q). Clearly we have βr(Fl(Q)+1)⋅w≤nmaxiβir(Fl(Q)+1)w(i)≤n2r+2maxiβir((j(r,i,n)[n]))w(i), where the last inequality follows from Theorem 4.2. We have βir((j(r,i,n)[n]))=(in)β0r((j(r,i,n)−i[n−i])) by picking the intersection of size i first.
To see the furthermore part of (a), let i∗ be the value of i for which the above maximum is attained. Then if Fi∗=(h1[i∗])∪⋯∪(hl(Q)[i∗]) is a family with Cnkc(Q,Fi∗)≥La(i∗,Pl(Q)+1,Q), then for the family F∗=(h1[n])∪⋯∪(hl(Q)[n])∪(j(r,i∗,n)[n]), we have
[TABLE]
therefore F∗ with C′=C and k′=2r+2+k shows that Conjecture 1.7 almost holds for the pair Pl(Q)+2,Q⊕r.
If r=1, then ∣∩F1∣=∣F1∣, so applying Corollary 3.2 with l=k=1 we obtain
[TABLE]
The proof of the furthermore part of (b) is analogous to the previous ones and is left to the reader.
∎
Proof of Corollary 1.11.
We proceed by induction on the number of levels. The base case is guaranteed by Sperner’s Theorem 1.1. The inductive step follows by applying Theorem 1.10 as Kr1,…,rl=Kr1,…,rl−1⊕rl.
∎
Proof of Corollary 1.12.
We proceed by induction on the number of levels. The base case is guaranteed by Sperner’s Theorem 1.1. Suppose the statement has been proved for all complete multipartite posets satisfying the condition with height smaller than l and consider Kr1,r2,…,rl. We know that there exists an i with 1≤i≤l such that ri=1. If 1<i<l, then the inductive step will follow by applying the furthermore part of Theorem 1.9 to Q1=Kr1,…,ri−1 and Q2=Kri+1,…,rl. If i=l, then the inductive step will follow by applying the furthermore part of Theorem 1.10 to Q=Kr1,…,rl−1 and r=1. The case i=1 follows from c(Kr1,…,rl,F)=c(Krl,…,r1,F), where F={[n]∖F:F∈F}.
∎
Proof of Theorem 1.13.
We only give the sketch of the proof as it is very similar to previous ones. Consider a Pl+3-free family F⊆2[n] and its canonical partition. If l=1, then we count the number of copies of Kr,1,s according to the set F∈F2 that plays the role of the middle element of Kr,1,s. The number of copies that contain F is not more than (r(⌊∣F∣/2⌋∣F∣))(s(⌊(n−∣F∣)/2⌋n−∣F∣)). Applying Corollary 3.2 with l=k=1 and w(i)=(r(⌊i/2⌋i))(s(⌊(n−i)/2⌋n−i)) yields c(Kr,1,s,F)≤maxi(in)(r(⌊i/2⌋i))(s(⌊(n−i)/2⌋n−i)). Let i∗ be the value of i for which this maximum is attained. Then the family (⌊i∗/2⌋[n])∪(i∗[n])∪(⌊(n+i∗)/2⌋[n]) contains exactly that many copies of Kr,1,s.
If l≥2, then we count the number of copies of Kr,1,1,…,1,s according to the sets F2∈F2 and Fl+1∈Fl+1 playing the role of the elements on the second and (l+1)st level of Kr,1,1,…,1,s. For a fixed pair F2∈F2 and Fl+1∈Fl+1 with F2⊂Fl+1 the number of copies of Kr,1,1,…,1,s containing F2 and Fl+1 is at most (r(⌊∣F2∣/2⌋∣F2∣))(s(⌊(n−∣Fl+1∣)/2⌋n−∣Fl+1∣))La(∣Fl+1∣−∣F2∣,Pl−1,Pl−2). The value of La(∣Fl+1∣−∣F2∣,Pl−1,Pl−2) is given by Theorem 1.2. So we can apply Corollary 3.2 with l=k=2 and w(i,j)=(r(⌊i/2⌋i))(s(⌊(n−j)/2⌋n−j))La(j−i,Pl−1,Pl−2) to obtain c(Kr,1,1…,1,s,F)≤maxi,j(jn)(ij)(r(⌊i/2⌋i))(s(⌊(n−j)/2⌋n−j))La(j−i,Pl−1,Pl−2). Let i∗ and j∗ be the values of i and j for which this maximum is attained. Then the family consisting of (⌊i∗/2⌋[n]),(i∗[n]),(j∗[n]),(⌊(n+j∗)/2⌋[n]) and the l−2 full levels determined by Theorem 1.2 contains exactly that many copies of Kr,1,1,…,1,s.
∎
There are several other complete multi-partite posets for which one can determine the levels that form an almost optimal family. For example, using Theorem 4.2 (a) one can prove that La(n,P3,Kp,2)≤nc(Kp,2,F) where F=(i[n])∪(j[n]) with i=(3+2p2p−1+o(1))n and j=(3+2p1+2p+o(1))n. In particular La(n,P3,Kp,2)=2(cp+o(1))n, where cp=3+2p2+p⋅2p+h(3+2p2p)+3+2p3h(2/3).
Let us finish this section by some remarks about K2,2=B as there exist several extremal results concerning B. Let us consider the following two posets that contain B: B+ and B++ have five elements a,b,c,d,e such that a<B+c,e and b<B+c,d and also d<B+e, while a,b<B++c,d and d<B++e. By results of DeBonis, Katona, and Swanepoel [4] and Methuku and Tompkins [20] we know that La(n,B,P1)=La(n,B+,P1)=(⌊n/2⌋n)+(⌊n/2⌋+1n), and as B++ contains a copy of B+ we have La(n,B+,P1)≤La(n,B++,P1). It is natural to ask how many copies of B can a B+-free or B++-free family in 2[n] contain, especially that the largest B+-free family does not contain any. Obviously, a P3-free poset is both B+-free and B++-free, therefore we obtain the inequality La(n,P3,B)≤La(n,B+,B)≤La(n,B++,B). The next proposition shows that these functions are asymptotically equal.
Proposition 4.3**.**
[TABLE]
Proof.
The first two inequalities are true by definition. Note that La(n,P3,B)=2(c2+o(1))n, where c2=10/7+h(4/7)+3h(2/3)/7 by the remarks made after the proof of Theorem 1.13.
Let F⊆2[n] be B++-free, and consider its canonical partition (note that P5 contains a copy of B++, thus F is P5-free).
Consider first the family S⊆F2 of sets that appear in 4-chains in F (they must be the second smallest in those chains). Note that if S∈S with F1⊊S⊊F3⊊F4, then S is not comparable to any other set F of F as F,F1,S,F3,F4 would form a copy of B++ both if S⊊F or F⊊S. Therefore every set S∈S is contained in at most one copy of B in F. As S⊆F2 is an antichain, we obtain c(B,F)−c(B,F∖S)≤(n/2n).
Clearly, F′=F∖S is P4-free. Let us consider its canonical partition and denote the resulting antichains by F1′,F2′,F3′. Let S′⊆F2′ be the family of middle sets of all 3-chains in F′. We know that for any S∈S′ there exist F1′,F3′∈F′ with F1′⊊S⊊F3′. Also, there cannot exist F1′′,F3′′ with F1′⊊S⊊F3′ as then F1′,F1′′,S,F3′,F3′′ would form a copy of B++. So either there is a unique set F′ that contains S and potentially several sets that are contained in S or there exists a unique F′ contained in S and several sets containing S. In the former case, if S is contained in a copy of B, it can only be one of the top sets. Furthermore, if a copy of B contains S, then it contains F′ as otherwise this copy could be extended by F′ to form a B++. As the sets contained in S form an antichain (they are a subfamily of F1′), we obtain that the number of copies of B containing S is at most (2(⌊2∣S∣⌋∣S∣)). Similarly, if S contains exactly one other set of F′, then the number of copies of B containing S is at most (2(⌊2n−∣S∣⌋n−∣S∣)). So introducing w(i)=max{(2(⌊i/2⌋i)),(2(⌊(n−i)/2⌋n−i)})} we obtain that the total number of copies containing at least one element of S′ is at most ∑S∈S′w(∣S∣). By the special case k=l=1 of Corollary 3.2 we obtain that this expression is maximized over all antichains when S is a full level of 2[n].
The weight function w is symmetric, i.e. w(i)=w(n−i) holds for any i, therefore it is enough to maximize (in)w(i) over n/2≤i≤n. It is a routine exercise to see that (in)(2(⌊i/2⌋i)) is maximized when i=(4/5+o(1))n. Therefore the number of copies of B that contain an element of S′ is at most 2h(4/5)+8/5+o(1). We obtained that
[TABLE]
as h(4/5)+8/5=2.3219...<c2.
∎
5 Remarks
One can define an even more general parameter LaR(P,Q). For three posets, R,P and Q we are interested in the maximum number of copies of Q in subposets R′ of R that do not contain P. Analogously to what we had for set families, we say that R′⊆R is a copy of Q in R if there exists a bijection ϕ:Q→R′ such that whenever x≤Qx′ holds, then so does ϕ(x)≤R′ϕ(x′). Let c(Q,R) denote the number of copies of Q in R and for any three posets R,P and Q we define
[TABLE]
and for a poset R and families of posets P,Q let us define
[TABLE]
Note that La(n,P,Q)=LaBn(P,Q), where Bn is the poset with elements of 2[n] ordered by inclusion.
Very recently Guo, Chang, Chen, and Li [14] introduced LaR(Q,P1), as a general approach to forbidden subposet problems. That is to solve the analogous question in a less complicated structure like the cycle, chain or double chain, and then to apply an averaging argument.
In many parts of Theorem 1.3, the construction yielding the lower bound that matches the upper bound contained the empty set and/or the set [n]. One might wonder whether the La-function remains the same if we do not allow these elements to be included. In other words, if Bn− denotes the subposet of Bn with ∅ and [n] removed, then how LaBn−(∨,P2) relates to La(n,∨,P2), LaBn−(B,P3) to La(n,B,P3) and so on. The {⋁,⋀}-free construction (⌊(n−1)/2⌋[n−1])∪{F∪{n}:F∈(⌊(n−1)/2⌋[n−1])} and the B-free construction (⌊(n−2)/2⌋[n−2])∪{F∪{n−1}:F∈(⌊(n−2)/2⌋[n−2])}∪{F∪{n−1,n}:F∈(⌊(n−2)/2⌋[n−2])} show that
(⌊(n−1)/2⌋n−1)≤LaBn−(⋁,P2)=LaBn−(⋀,P2)≤(⌊n/2⌋n),
(⌊(n−2)/2⌋n−2)≤LaBn−(B,P3)≤(⌊n/2⌋n).
There is a longstanding (folklore) conjecture which would imply the existence of constructions in both cases that asymptotically match the upper bounds. Let Mk+1⊆(k+1[n]) be a family of sets with the property that for every K∈(k[n]) there exists at most one set M∈Mk+1 with K⊊M. Obviously, for any such set we have ∣Mk+1∣≤(kn)/(k+1) and Rk+1:=Mk+1∪(k[n]) is ⋁-free with c(P2,Rk+1)=(k+1)∣Mk+1∣. It is conjectured that there exists a family M⌊n/2⌋+1 with the above property such that ∣M⌊n/2⌋+1∣=(1−o(1))(⌊n/2⌋[n])/(⌊n/2⌋+1) holds. Similarly, writing Mn−k+1 for {[n]∖M:M∈Mn−k+1} the construction Tk:=Mk+1∪(k[n])∪Mn−k+1 is B-free. The above conjecture would yield c(P3,T⌊n/2⌋)=(1−o(1))(⌊n/2⌋n).
In Section 4, we proved that apart from a polynomial factor Conjecture 1.7 holds for complete multi-level posets, i.e. there exists a sequence Fn of families that consists of full levels such that La(n,Pl+1,Kr1,r2,…,rl)≤nkc(Fn,Kr1,r2,…,rl) for some constant k=k(Kr1,r2,…,rl). To improve this result or to completely get rid of the polynomial factor one would need to improve Theorem 1.8 or rather to determine the intersection profile polytope of antichains.
