Induced and non-induced forbidden subposet problems

The problem of determining the maximum size $La(n,P)$ that a $P$-free subposet of the Boolean lattice $B_n$ can have, attracted the attention of many researchers, but little is known about the induced version of these problems. In this paper we determine the asymptotic behavior of $La^*(n,P)$, the maximum size that an induced $P$-free subposet of the Boolean lattice $B_n$ can have for the case when $P$ is the complete two-level poset $K_{r,t}$ or the complete multi-level poset $K_{r,s_1,\dots,s_j,t}$ when all $s_i$'s either equal 4 or are large enough and satisfy an extra condition. We also show lower and upper bounds for the non-induced problem in the case when $P$ is the complete three-level poset $K_{r,s,t}$. These bounds determine the asymptotics of $La(n,K_{r,s,t})$ for some values of $s$ independently of the values of $r$ and $t$.


Introduction
We use standard notation: 2 X denotes the power set of X, X k denotes the set of k-element subsets of X and for two sets A ⊂ B the interval {G : A ⊆ G ⊆ B} is denoted by [A, B].
The very first theorem in extremal finite set theory is due to Sperner [14] and it states that if F ⊆ 2 [n] is a family of sets that does not contain two sets F 1 , F 2 with F 1 F 2 , then |F | ≤ n ⌊ n 2 ⌋ holds. Such families are called antichains or Sperner families. A first generalization is due to Erdős [6], who proved that if F does not contain any (k+1)-chains, i.e. k + 1 sets F 1 , F 2 , . . . , F k+1 with F 1 F 2 · · · F k+1 , then |F | ≤ Σ(n, k) := k i=1 n ⌊ n−k 2 ⌋+i holds. Such families are called k-Sperner families. These two theorems have many applications and generalizations. One such generalization is the topic of forbidden subposet problems first introduced by Katona and Tarján [11]. We say that a poset Q contains another poset P if there is an injection i : P → Q such that for every p 1 , p 2 ∈ P the fact p 1 ≤ p 2 implies i(p 1 ) ≤ i(p 2 ). If Q does not contain P , then it is said to be P -free. If P is a set of posets, then Q is P-free if it is P -free for all P ∈ P. The parameter introduced by Katona and Tarján is the quantity La(n, P ) that denotes the maximum size of a P -free subposet of B n , the Boolean poset of all subsets of [n] ordered by inclusion. With this notation Erdős's theorem states that La(n, P k+1 ) = Σ(n, k), where P k+1 denotes the path on k + 1 elements, i.e. a total ordering on k + 1 elements.
In the same paper, Katona and Tarján introduced the induced version of the problem. We say that Q contains an induced copy of P if there is an injection i : P → Q such that for any p 1 , p 2 ∈ P we have p 1 ≤ p 2 if and only if i(p 1 ) ≤ i(p 2 ). If Q does not contain an induced copy of P , then Q is said to be induced P -free. The analogous extremal number is denoted by La * (n, P ) and obviously the inequality La(n, P ) ≤ La * (n, P ) holds for any poset P . The notation for multiple forbidden subposets is La(n, P) and La * (n, P).
As any poset P is contained by P |P | , we clearly have La(n, P ) ≤ La(n, P |P | ) = Σ(n, |P | − 1). Strengthenings of this general bound were obtained by Burcsi and Nagy [2], Chen and Li [4] and recently by Grósz, Methuku and Tompkins [10]. Therefore it is natural to compare La(n, P ) to n ⌊ n 2 ⌋ . Unfortunately, it is not known whether π(P ) = lim n→∞ La(n,P ) exists.
The following conjecture was first stated in [9]. Conjecture 1.1. For any poset P let e(P ) denote the largest integer k such that for any j and n the family j+i is P -free. Then π(P ) exists and is equal to e(P ). This conjecture has been verified for many classes of posets. The most remarkable result is due to Bukh.
Much less is known about the induced version of the problem. It has only been proved recently by Methuku and Pálvölgyi [13] that for every poset P there exists a constant c P such that La * (n, P ) ≤ c P n ⌊ n 2 ⌋ holds. (For a special class of posets this has already been established by Lu and Milans [12].) As the list of known results on forbidden induced subposet problems is very short here we enumerate all such theorems.
Finally, the induced version of Theorem 1.2 has been proved, but only with an o(1) error term instead of O( 1 n ).
holds. Before we state our results, let us formulate the induced analogue of Conjecture 1.1. Conjecture 1.6. Let P be a poset and let e * (P ) denote the largest integer k such that for any j and n the family j+i is induced P -free. Then π * (P ) = lim n→∞ exists and is equal to e * (P ).
In the present paper, we address both the induced and the non-induced problem for complete multi-level posets. Let K r 1 ,r 2 ,...,rs denote the poset on s i=1 r i elements a 1 1 , a 1 2 , . . . , a 1 r 1 , a 2 1 , a 2 2 , . . . , a 2 r 2 , . . . , a s 1 , a s 2 , . . . , a s rs with a i α < a j β if and only if i < j. Our first result gives not only the asymptotics of La * (n, K r,s ), but also the order of magnitude of the second order term of the extremal value. Theorem 1.7. For any positive integers 2 ≤ r, s we have Σ(n, 2)+( r+s−2 n −O r,t ( 1 n 2 )) n ⌊n/2⌋ ≤ La * (n, K r,s ) ≤ (2 + 2(r+s−2) n + O r,s ( 1 n 2 )) n ⌊n/2⌋ .
Note that the same upper bound for La(n, K r,s ) follows from Theorem 1.2 as K r,s is an (induced) subposet of K r,1,s and K r,1,s is a tree poset. By the same argument, Theorem 1.5 implies the asymptotics of La * (n, K r,s ) but its error term is worse than that of Theorem 1.7. Let us remark that La(n, K 2,2 ) = Σ(n, 2) was shown by De Bonis, Katona, Swanepoel, [5]. As they also showed the uniqueness of the extremal family, it was known that the strict inequality La(n, K 2,2 ) < La * (n, K 2,2 ) holds. Theorem 1.7 tells us the order of magnitude of the gap between these two parameters.
Then we turn our attention to the three level case of K r,s,t . To do so we need to introduce the following notation: for positive integers r, t let Also, for any integer s ≥ 2 let us define m = m s = ⌈log 2 (s − f (r, t) + 2)⌉ and m ′ = m ′ s = min{m : s ≤ m ⌈m/2⌉ } and for any real number z, let z + denote max{0, z}.
It is easy to verify that three consecutive levels in B n form an unextendable family of K 1,2,2 -free and K 2,2,2 -free family of sets, but from our proofs it does not follow that they are of largest possible size. However we formulate the following conjecture. Conjecture 1.10. If n is large enough, then La(n, K 1,2,2 ) = La(n, K 2,2,1 ) = La(n, K 2,2,2 ) = Σ(n, 3) holds.
Then we turn our attention to the general case of K r,s 1 ,s 2 ,...,s j ,t . As there are more technical details in calculating e(K r,s 1 ,s 2 ,...,s j ,t ) than in calculating e * (K r,s 1 ,s 2 ,...,s j ,t ) we will only consider the induced problem in its full generality.
Proof. To see (i) let F consist of f (r, t) + j i=1 m ′ s i consecutive levels of 2 [n] and suppose we find an induced copy of K r,s 1 ,s 2 ,...,s j ,t . If F 1 , . . . , F r and F ′ 1 , . . . , F ′ t play the role of the bottom r and the top t sets, then . . , F j ′ s j ′ play the role of the sets of the j ′ th middle level of K r,s 1 ,s 2 ,...,s j ,t , then their union has size at least s j ′ more than the union of the sets on the (j ′ − 1)st level. Thus one would need j i=1 m ′ s i more levels for the j middle levels of K r,s 1 ,s 2 ,...,s j ,t . It is easy to see that f (r, t) and also U ⊂ I. Thus even in the 'most economic' U = I case U can play the role of the set on the j ′ th level. If s i−1 = s i = 1, then the set representing level i of K r,s 1 ,s 2 ,...,s j ,t requires a new level.
There exists an integer s 0 such that for any r, s, t with s ≥ s 0 we have Σ(n, n +O r,t ( 1 n 2 )) n ⌊n/2⌋ . (iv) For any constant c with 1/2 < c < 1 there exists an integer s c such that if all s i 's satisfy that either s i = 4 or s i ≥ s c and s ≤ c m ′ ⌈m ′ /2⌉ , then we have La * (n, K r,s 1 ,s 2 ,...,s j ,t ) = (e * (K r,s 1 ,s 2 ,...,s j ,t ) + O r,t ( 1 n )) n ⌊n/2⌋ .
Our main technique to prove all four theorems is the chain partition method [8,7]. The remainder of the paper is organized as follows: in Section 2 we prove some preliminary lemmas that will be used in the proofs of Theorem 1.7, Theorem 1.8, Theorem 1.9, and Theorem 1.12. Then in Section 3 we prove our results.

Preliminary lemmas
Let C n denote the set of maximal chains in [n]. For a family F ⊆ 2 [n] of sets and A ⊆ [n] we define s − F (A) to be the maximum size of an antichain in F ∩ 2 A and s + F (A) to be the maximum size of an antichain in {F ∈ F : A ⊆ F }.
Proof. The property possessed by A and F ensures that F A := {F ∈ F : F ⊂ A} contains at most k − 1 sets of each possible size. Thus the number of pairs (F, C) in question is at most Let us fix i with 1 ≤ i ≤ j and consider pairs (F, C) with F ∈ F ∩ C and C ∈ C A i ,A . As s − F (A i ) < k, we can apply Lemma 2.1 to F and A i , and obtain that the number of such pairs with F A i is at most 2 Even if all A i 's belong to F , then every chain C ∈ C k,A can contain one more set from F , namely one of the A i 's. This completes the proof.
be a family of sets such that any antichain A ⊂ G has size at most 3. Then the number of pairs (G, C) with G ∈ G ∩ C and C ∈ C k is at most 4k!.
(ii) For any constant c with 1/2 < c < 1 there exists an integer s c such that if s ≥ s c and s ≤ c m ′ ⌈m ′ /2⌉ , then the following holds: if G ⊆ 2 [k] is a family of sets such that any antichain A ⊂ G has size less than s, then the number of pairs (G, C) with G ∈ G ∩ C and C ∈ C k is at most m ′ k!.
(iii) There exists an integer s 0 such that if s ≥ s 0 and G ⊆ 2 [k] is a family of sets such that any antichain A ⊂ G has size at most s, then the number of pairs (G, C) with G ∈ G ∩ C and C ∈ C k is at most (m ′ + 1)k!.
Proof. First we prove (i). We may assume that ∅, [k] ∈ G holds as adding them will not result in violating the condition of the lemma and the number of pairs to be counted can only increase. These two sets are in k! maximal chains each, thus giving 2k! pairs. All other sets belong to |G|!(k − |G|)! = k! ( k |G| ) chains in C k . Sets of same size form an antichain, therefore for every 1 ≤ i ≤ k − 1 there exist at most 3 sets of size i in G and thus the total number of pairs (G, C) is at most For k = 3, 4, 5 the sum S(k) equals 4k!, 4k!, 3.8m!, respectively (for k = 1, 2 the number of pairs counted is 2k! and 3k!, respectively). Furthermore, if k is at least 5, then 1 ( k i ) ≥ 1 ( k+1 i ) holds for all i and also the inequality is valid. Thus, S(k) k! is monotone decreasing for k ≥ 5 and therefore S(k) k! ≤ 4 holds for all positive integer k. This completes the proof of (i). Now we prove (ii). Clearly, as long as k < m ′ we can have G = 2 [k] and then the number of pairs is (k + 1)k! ≤ m ′ k!. When k ≥ m ′ we again use the observation that for any 0 ≤ j ≤ k we have |{G ∈ G ∩ [k] j | < s and thus the number of pairs (G, C) is at most holds for all k ≥ m ′ . Consider the case k = m ′ . If s is large enough (and thus m ′ and k), Therefore, by the assumption on s and c we have at least 2 √ m ′ / log m ′ summands in R(m ′ ) that are not more than 1+c 2 , a constant smaller than 1. Thus, if m ′ is large enough, their subsum is less than ⌈m ′ /2⌉ + 2 √ m ′ / log m ′ − 1 and since all other summands are not more than 1, we obtain R(m ′ ) < m ′ .
Finally, we prove (iii). Clearly, as long as k ≤ m ′ for any family G ⊆ 2 [k] the number of pairs is (k + 1)k! ≤ (m ′ + 1)k!. We need to show that R(k) ≤ m ′ + 1 holds for all k > m ′ . As in (ii), the proof of R(k + 1) ≤ R(k) for k ≥ m ′ did not require the assumption on s and c, we obtain that R(k) ≤ m ′ + 1 holds for all k.
Our last auxiliary lemma was proved by Griggs, Li and Lu [8]. ( a K 1,s,1 -free family of sets, then the number of pairs (G, C) with G ∈ G ∩ C and C ∈ C k is at most m * k!. ( a K 1,s,1 -free family of sets, then the number of pairs (G, C) with G ∈ G ∩ C and C ∈ C k is at most )k!.

Proofs
In this section we prove our main theorems. Let us start with constructions to see the lower bounds. Let us partition [n] k into n classes: k r,mod denote the union of the r largest classes. Clearly, | [n] k r,mod | ≥ r n n k . Furthermore, it has the property that for any distinct r + 1 sets • For Theorem 1.7 consider the family F := We claim that F is K r,s,t -free. Assume not and let A 1 , A 2 , . . . , A r , B 1 , B 2 , . . . , B s , C 1 , C 2 , . . . , C t ∈ F form a copy of K r,s,t . If r ≥ 2, then | ∪ r i=1 A i | ≥ k + 2 and if r = 1, then |A 1 | ≥ k + 1 (note that if r = 1, 2, then (r − 2) + = 0 and thus the smallest set size in F is k + 1). Similarly, if t ≥ 2, then | ∩ t j=1 C j | ≤ l − 2 and if t = 1, then and thus there is no place for B 1 , B 2 , . . . , B s -a contradiction.
• For Theorem 1.9 let k be the index of the level below the three middle levels, i.e. k = ⌈ n−3 2 ⌉ − 1. Write l = k + 4 and let us consider the family If f (r, t) = 2, then for any A 1 , A 2 , . . . , A r ∈ F and C 1 , C 2 , . . . , C t ∈ F we have | ∩ t i=1 C i | − | ∪ r j=1 A j | ≤ 0, thus we cannot have two sets in between. While if f (r, t) = 1, say t = 1, then for any A 1 , A 2 , . . . , A r ∈ F and C ∈ F we have |C| − | ∪ r j=1 A j | ≤ 1, thus we cannot have two sets in between them and below C.
• For Theorem 1.12 (i), (ii) and (iii), let k be the index of the level below the m ′ + f (r, t) middle levels, i.e. k = ⌈ n−m ′ −f (r,t) 2 ⌉ − 1. Write l = k + m ′ + f (r, t) + 1 and let us consider the family One can see that for any antichains A 1 , A 2 , . . . , A r ∈ F and C 1 , As we mentioned in the Introduction, for all proofs we will use the chain partition method. This works in the following way: for a family F ⊆ 2 [n] suppose we can partition C n into C n,1 , C n,2 , . . . C n,l such that for all 1 ≤ i ≤ l the number of pairs (F, C) with F ∈ F ∩ C and C ∈ C n,i is at most b|C n,i |. Then clearly the number of pairs (F, C) with F ∈ F ∩ C and C ∈ C n is at most b|C n |. Since the number of such pairs is exactly F ∈F |F |!(n − |F |)! we obtain the LYM-type inequality and thus |F | ≤ b n ⌈n/2⌉ holds. Therefore, in the proofs below we will end our reasoning whenever we reach a bound on the appropriate partition as mentioned above.
Proof of the upper bound in Theorem 1.7. Let F be an induced K r,s -free family. We can assume that F contains an antichain of size at least r as otherwise Lemma 2.1 can be applied to F and [n] to obtain that |F | ≤ o( n ⌈n/2⌉ ). Now we define the min * r -partition of C n and F . For a set A with s − F (A) ≥ r we define C F ,A,r = {C ∈ C n : A ∈ C, ∀C ⊂ A, C ∈ C : s − F (C) < r}. Note that every C ∈ C n belongs to exactly one set C F ,A,r as by our assumption s − F ([n]) ≥ r holds. We claim that that the number of pairs (F, C) with F ∈ F ∩ C and C ∈ C F ,A,r is at most (2 + 2(r+s−2) n + O( 1 n 2 )|C F ,A,r |. We distinguish three types of pairs: 1. if A ∈ F , then there are exactly |C F ,A,r | pairs with F = A (otherwise there is none), 2. any chain in C r,A can be extended to (n − |A|)! chains in C F ,A,r , thus by Corollary 2.2 there are (1 + 2(r−1) n + O r ( 1 n 2 ))|C F ,A,r | pairs with F A, 3. finally, any maximal chain from A to [n] can be extended to |C r,A | chains in C F ,A,r , thus Lemma 2.1 implies that there are ( 2(s−1) This gives us a total of at most (2 + 2(r+s−2) n + O r,s ( 1 n 2 ))|C F ,A,r | pairs, which completes the proof. Now we turn our attention to complete three level posets.
Proof of the upper bound in Theorem 1.8. Let F be a K r,s,t -free family. We can assume that F contains an antichain of size at least max{r, t} as otherwise Lemma 2.1 can be applied to F and [n] to obtain that |F | ≤ o( n ⌈n/2⌉ ). Now we define the min * r − max * t partition of C n . Let S = {S ∈ 2 [n] : s − F (S) ≥ r}, S − = {S ∈ S : s + F (S) < t} and finally S + = S \ S − . For any set S ∈ S − let C S denote the set of those maximal chains C in C n in which For any set A ∈ S + and B with A ⊆ B let C A,B = C A,r,B,t denote the set of those maximal chains C in C n in which • if t = 1, then B is the largest set in F ∩ C, Consider a maximal chain C ∈ C n . By the assumption s − F ([n]) ≥ max{r, t}, there is a smallest set H of C with s − F (H) ≥ r. If H ∈ S − , then C belongs to C H . If not, then H ∈ S + and thus for the largest set H ′ of C with s + F ≥ t we have H ⊆ H ′ and therefore C ∈ C H,H ′ holds. We obtained that the min * r − max * t partition of C n is indeed a partition. We claim that that the number of pairs (F, C) with F ∈ F ∩ C and C ∈ C S , C ∈ C A,B is at most b|C S |, b|C A,B |, respectively, where b is the bound stated in Theorem 1.8.
First consider the "degenerate" case of C S with S ∈ S − . A chain C ∈ C S goes from ∅ until one of the subsets S 1 , S 2 , . . . , S k of B with size |B| − 1 for which s − F (S i ) < r. Then C must go through S, and finally C must contain a maximal chain from S to [n]. Thus |C S | = k(|S| − 1)!(n − |S|)!. We distinguish two types of pairs to count.
1. If r ≥ 2, then applying Corollary 2.2 we obtain that there are at most (1 + 2(r−1) n + O r ( 1 n 2 ))|C S | pairs (F, C) with F S. Together with {(S, C) : C ∈ C S } we have (2 + 2(r−1) n + O r ( 1 n 2 ))|C S | pairs. If r = 1, then by definition the number of pairs (F, C) with F ⊆ S is at most |C S | as for all such pairs we must have F = S.
2. Applying Lemma 2.1 we obtain that there are at most ( 2(t−1) This gives a total of at most (2 + 2(r+t−2) n + O r,t ( 1 n 2 ))|C S | pairs. We now consider the "more natural" A ∈ S + , A ⊆ B case. As there are sets in the interval [A, B], this time we distinguish three types of pairs: 1. If r = 1, then there is no pair (F, C) with F A. If r ≥ 2, then applying Corollary 2.2 we obtain that there are at most (1 + 2(r−1) n )|C A,B | pairs (F, C) with F A. 2. If t = 1, then there is no pair (F, C) with B F . If t ≥ 2, then applying Corollary 2. 2 we obtain that there are at most (1 + 2(t−1) n )|C A,B | pairs (F, C) with B F .
3. If F is a K r,s,t -free family, then {F ∈ F : A ⊆ F ⊆ B} is a K 1,s−f (r,t),1 -free family.
the min * r − max * t partition of C n and count the number of pairs (F, C) with F ∈ F ∩ C and C ∈ C n .
The degenerate case is identical to what we had in the proof of Theorem 1.8, thus we only consider the case when A ∈ S + , A ⊆ B. The three types of pairs: 1. If r = 1, then there is no pair (F, C) with F A. If r ≥ 2, then applying Corollary 2. 2 we obtain that there are at most (1 + 2(r−1) n )|C A,B | pairs (F, C) with F A. 2. If t = 1, then there is no pair (F, C) with B F . If t ≥ 2, then applying Corollary 2. 2 we obtain that there are at most (1 + 2(t−1) n )|C A,B | pairs (F, C) with B F . 3. Note that {F ∈ F : A ⊆ F ⊆ B} cannot contain an antichain of size s as otherwise F would contain an induced copy of K r,s,t .
(a) If F is an induced K r,4,t -free family, then by Lemma 2.3 (i) the number of pairs  Altogether these bounds yield that the total number of pairs is at most 3. (m ′ + 1 + f (r, t) + 2(r+t−2) n + O( 1 n 2 ))|C n | if F is induced K r,s,t -free and s large enough. Now we prove (iv). Let F be an induced K r,s 1 ,s 2 ,...,s j ,t -free family. We can assume that F contains an antichain of size at least max{r, t} as otherwise Lemma 2.1 can be applied to F and [n] to obtain that |F | ≤ o( n ⌈n/2⌉ ). We consider the following partition of C n : for any (2) if i < j − 1 and s + F (A i ) < s i+1 but for all k < i we have s + F (A k ) < s k+1 or if (3) i = j − 1 and s + F (A i ) < t. In case (1) the set C A 0 ,A 1 ,...,A j consists of those maximal chains C in C n that satisfy • if r = 1, then A 0 is the smallest set in F ∩ C, • if r ≥ 2, then A 0 is the smallest set in C with s − F (A 0 ) ≥ r, • for any k with 1 ≤ k ≤ j − 1 A k is the smallest set in F ∩ C such that there exists an antichain of size s k in F ∩ [A k−1 , A k ], • if t = 1, then A j is the largest set in F ∩ C, • if t ≥ 2, then A j is the largest set in C with s + F (A j ) ≥ t.
In case (2) and (3) the definition of C A 0 ,A 1 ,...,A i is modified to the set of maximal chains C in C n that satisfy • if r = 1, then A 0 is the smallest set in F ∩ C, • if r ≥ 2, then A 0 is the smallest set in C with s − F (A 0 ) ≥ r, • for any k with 1 ≤ k ≤ i A k is the smallest set in F ∩ C such that there exists an antichain of size s k in F ∩ [A k−1 , A k ], • in case (2) we have s + F (A i ) < s i+1 , while in case (3) we have s + F (A j−1 ) < t.
One can verify along the lines of the proof of Theorem 1.8 that the above definition results a partition of C n . Let us note that a chain C in C A 0 ,...,A i contains all A i 's and for every 0 ≤ k ≤ j − 1 it goes through one of the (|A k | − 1)-subsets A k 1 , . . . A k l k of A k for which [A k−1 , A k l ] does not contain an antichain of size s k where A −1 = ∅ and s −1 = r. We now count the pairs (F, C) with F ∈ F ∩ C and C ∈ C A 0 ,...,A i . First we consider the cases (2) and (3) • Applying Lemma 2.1 and/or Corollary 2.2 we obtain that below A 0 and above A i there are at most (f (r, t) + O r,t,s i+1 ( 1 n ))|C A 0 ,...,A i | such pairs, • applying Lemma 2.3 to A k and all A k 1 , . . . , A k l k we obtain that the number of such pairs above A k (including A k ) and below A k+1 there are at most m ′ s k |C A 0 ,...,A i | such pairs.
Altogether we obtained that the number of pairs is at most (f (r, t) + 1 + i k=1 m ′ s k + O( 1 n ))|C A 0 ,...,A i | ≤ (f (r, t) + j k=1 m ′ s k + O( 1 n ))|C A 0 ,...,A i |. Finally, in case (1) everything below A j−1 and above A j is as before. As F is induced K r,s 1 ,...,s j ,t -free the interval [A j−1 , A j ] cannot contain an antichain of size s j . Thus the number of pairs with F ∈ [A j−1 , A j ] is at most m ′ s j |C A 0 ,...,A i |. Thus in this case the total number of pairs is at most (f (r, t) + j k=1 m ′ s k + O( 1 n ))|C A 0 ,...,A j |.