On some extremal and probabilistic questions for tree posets

Given two posets $P,Q$ we say that $Q$ is $P$-free if $Q$ does not contain a copy of $P$. The size of the largest $P$-free family in $2^{[n]}$, denoted by $La(n,P)$, has been extensively studied since the 1980s. We consider several related problems. Indeed, for posets $P$ whose Hasse diagrams are trees and have radius at most $2$, we prove that there are $2^{(1+o(1))La(n,P)}$ $P$-free families in $2^{[n]}$, thereby confirming a conjecture of Gerbner, Nagy, Patk\'os and Vizer [Electronic Journal of Combinatorics, 2021] in these cases. For such $P$ we also resolve the random version of the $P$-free problem, thus generalising the random version of Sperner's theorem due to Balogh, Mycroft and Treglown [Journal of Combinatorial Theory Series A, 2014], and Collares Neto and Morris [Random Structures and Algorithms, 2016]. Additionally, we make a general conjecture that, roughly speaking, asserts that subfamilies of $2^{[n]}$ of size sufficiently above $La(n,P)$ robustly contain $P$, for any poset $P$ whose Hasse diagram is a tree.


Introduction
In this paper we will consider several related extremal and probabilistic questions for posets.All posets we consider will be viewed as finite collections of finite subsets of N equipped with the containment relation.Indeed, we will usually be working inside the power set 2 [n] for some n ∈ N.
Let P, Q be posets.A poset homomorphism from P to Q is a function φ : P → Q such that for every A, B ∈ P , if A ⊆ B then φ(A) ⊆ φ(B).We say that P is a subposet of Q if there is an injective poset homomorphism from P to Q; otherwise, Q is said to be P -free.The size of the largest P -free family in 2 [n] is denoted by La(n, P ).Further we say P is an induced subposet of Q if there is an injective poset homomorphism φ from P to Q such that for every A, B ∈ P , φ(A) ⊆ φ(B) if and only if A ⊆ B; otherwise, Q is said to be induced P -free.
The systematic study of La(n, P ) was initiated by Katona and Tarján in 1983 [14], though Sperner's classical theorem [20] was the first result in the area.The latter asserts that the largest C 2 -free family 1 in 2 [n] has size n ⌊n/2⌋ .Whilst the asymptotic value of La(n, P ) has been determined for a range of posets P , the general problem remains open; see [10] or Chapter 7 of [8] for a survey on the topic and [3,12] for a conjecture on the asymptotic value of La(n, P ) for all posets P .
The Hasse diagram of a poset P is the directed graph with vertex set P and for A, B ∈ P , there is a directed edge from A to B in the Hasse diagram if A ⊂ B and there does not exist a C ∈ P such that A ⊂ C ⊂ B. The undirected Hasse diagram is the (undirected) graph obtained from the Hasse diagram by removing all orientations on the edges.A tree poset is a poset whose undirected Hasse diagram is a tree.A tree poset P is upward (resp.downward) monotone if there exists an element B ∈ P with B ⊆ A (resp.A ⊆ B) for any A ∈ P .A tree poset is called monotone if it is either upward or downward monotone.
The following result of Bukh [3] determines La(n, P ) asymptotically for all tree posets P .Given a poset P , we write h(P ) for the height of P , i.e., the length t of the longest chain C t in P .
BP: Alfréd Rényi Institute of Mathematics, Budapest.Email: patkos@renyi.hu.Research supported by NKFIH grant FK 132060.AT: University of Birmingham, United Kingdom.Email: a.c.treglown@bham.ac.uk.Research supported by EPSRC grant EP/V002279/1. 1 We write Ct to denote the chain on t elements.Thus, a C 2 -free family is simply an antichain.
A phenomenon often exhibited in combinatorial problems is the property that once one is sufficiently above a 'threshold' for guaranteeing the existence of a structure, one can actually 'robustly' find this structure.This behaviour is often articulated in terms of supersaturation.For example, in recent years there has been significant attention on an old conjecture of Kleitman on the minimum number of chains C t of length t in a subfamily of 2 [n] of a given size.Indeed, after earlier progress [2,5,6], this conjecture was proven by Samotij [19].
The first question we consider takes a different viewpoint on robustness.For this we need the following definition.
Definition 1.2.Let P be a tree poset and x ∈ P , and let d ≥ 2 be a positive integer.The d-blow-up P (x, d) rooted at x is the tree poset whose Hasse diagram is defined as follows: for each u ∈ P , if u is at distance ρ from x in the undirected Hasse diagram of P , then u is replaced with d ρ elements u 1 , u 2 , . . ., u d ρ ; furthermore, if uv is an edge of the Hasse diagram of P with v being at distance ρ − 1 to x in P , then the u i s are partitioned into d ρ−1 pairwise disjoint sets U 1 , U 2 , . . ., U d ρ−1 each of size d, and for every j ∈ [d ρ−1 ], v j is joined to all members of U j .The orientation of all such edges is the same as that of uv.
x x Figure 1.A poset rooted at x and its 2-blow-up.
The following conjecture states that once sufficiently above the threshold in Theorem 1.1, one can robustly guarantee a copy of a given tree poset P .Conjecture 1.3.Let P be any tree poset of height h.Given any ε > 0 there exists δ > 0 such that the following holds for any sufficiently large n ∈ N and any x ∈ P .Consider the δn-blow-up P (x, δn) of P rooted at x.If F ⊆ 2 [n] has size at least (h − 1 + ε) n ⌊n/2⌋ , then F contains a copy of P (x, δn).Remark 1.First observe that, by considering the complement family F := {[n] \ F : F ∈ F }, it is clear that if the conjecture holds for some poset P , then it also holds for the dual poset P d obtained from P by reversing all relations.
Note that at the cost of decreasing δ by a constant factor, if Conjecture 1.3 holds for a 'simple' tree poset P , then it holds for some more complicated tree posets P ′ .More precisely, suppose u, v are leaves of the Hasse diagram of P ′ , w is an element of P ′ lying on the paths from x to both u and v, and for the unique . Then if we obtain P from P ′ by removing v 1 , v 2 , . . ., v k , then P ′ (x, δn/2) ⊂ P (x, δn) for sufficiently large n; so it is enough to verify Conjecture 1.3 for P .
Similarly, suppose P 1 , P 2 are connected components of P ′ \ {x} such that y 1 ∈ P 1 , y 2 ∈ P 2 are the only elements of P 1 ∪ P 2 adjacent to x in the Hasse diagram of P ′ , and also x ⊆ y 1 if and only if x ⊆ y 2 .Suppose further P 2 contains a copy of P 1 with y 2 playing the role of y 1 .If we obtain P from P ′ by removing P 1 , then again P ′ (x, δn/2) ⊂ P (x, δn) for sufficiently large n; so it is enough to verify Conjecture 1.3 for P .
Perhaps the most interesting case of Conjecture 1.3 is when P is a monotone tree.For example, it implies that if one is sufficiently above the threshold for guaranteeing a binary monotone tree of height h (or even a chain of length h), then actually one can find a δn-ary monotone tree of height h.
As well as being interesting in its own right, Conjecture 1.3 has implications to a couple of other wellstudied questions.The first of these concerns counting P -free families in 2 [n] .Kleitman [15] proved that the number of C 2 -free families in 2 [n] is 2 (1+o(1))( n ⌊n/2⌋ ) ; this result was further refined by Korshunov [16].More generally, a trivial lower bound on the number of P -free families in 2 [n] is 2 La(n,P ) .The following conjecture from [7] states that this lower bound is close to being tight.Conjecture 1.4 (Gerbner, Nagy, Patkós and Vizer [7]).The number of P -free families in 2 [n] is 2 (1+o(1))La(n,P ) .
Our first result asserts that if Conjecture 1.3 is true for a given tree poset P , and for a specific choice of some x ∈ P , then Conjecture 1.4 is true for this P .
Theorem 1.5.Suppose P is a tree poset of height h for which Conjecture 1.3 holds for some choice of x ∈ P .Then the number of P -free families in 2 [n] is In Section 3.1 we prove Theorem 1.5 via a variation on the standard graph container algorithm.The following result gives a class of tree posets P for which Conjecture 1.3 holds for some choices of x ∈ P .Theorem 1.6.Conjecture 1.3 holds for all tree posets P and any x ∈ P for which all elements of P are within distance at most 2 of x in the (undirected) Hasse diagram of P .
In Section 4 we discuss why it seems challenging to extend Theorem 1.6 to arbitrary tree posets.The next corollary follows immediately from Theorems 1.5 and 1.6.Here we say a tree poset P has radius at most t if there is some x ∈ P so that every vertex in the (undirected) Hasse diagram is at distance at most t from x.
Corollary 1.7.Given any tree poset P of height h ≤ 5 and radius at most 2, the number of P -free families in 2 [n] is Notice that every tree poset P of radius at most 2 has height at most 5; we only include the h ≤ 5 condition in Corollary 1.7 (and in the statement of other results below) to make this restriction on the height explicit.Note that Corollary 1.7 resolves Conjecture 1.4 for various natural posets P including all monotone trees of height at most 3 and the poset N (see Figure 2).We also prove the following special version of Theorem 1.6 that ensures a larger blow-up of P .Lemma 1.8.Let P be any tree poset of height h ≤ 5 and radius at most 2. Let x ∈ P such that every element of P is of distance at most two in the (undirected) Hasse diagram of P .Given any ε > 0, the following holds for any sufficiently large n ∈ N. Consider the n 1.9 -blow-up P (x, n 1.9 ) of P rooted at x.If F ⊆ 2 [n] is such that |F | ≥ 4(h − 1 + ε) n ⌊n/2⌋ , then F contains a copy of P (x, n 1.9 ).
This lemma is a crucial tool in the proof of Theorem 1.10.Note that we have not tried to optimise the lower bound on |F | in Lemma 1.8 as the result suffices for our purposes.
Our other application of Conjecture 1.3 is to the random version of the P -free problem.Let P(n, p) be obtained from 2 [n] by selecting each element of 2 [n] independently at random with probability p.The model P(n, p) was first investigated by Rényi [18] who determined the probability threshold for the property that P(n, p) is not itself an antichain, thereby answering a question of Erdős.There has also been interest in the problem of determining when the largest P -free family of P(n, p) has size p • (1 + o(1))La(n, P ) with high probability (w.h.p.). 2 For example, the next result resolves this problem when P = C 2 .Theorem 1.9 (Balogh, Mycroft and Treglown [1] and Collares Neto and Morris [4]).For any ε > 0 there exists a constant C such that if p > C/n then w.h.p. the largest C 2 -free family (i.e., antichain) in P(n, p) has size (1 ± ε)p n ⌊n/2⌋ .
Osthus [17] had earlier proven Theorem 1.9 in the case when pn/ log n → ∞.Moreover, Osthus showed that, for a fixed c > 0, if p = c/n then with high probability the largest antichain in P(n, p) has size at least (1 + o( 1))(1 + e −c/2 )p n ⌊n/2⌋ .So the bound on p in Theorem 1.9 is best-possible up to the constant C. The results in [1,4] actually yield an analogous result for C t -free families (for any fixed t ∈ N).Hogenson [13] also adapted the proof of Theorem 1.9 from [1] to obtain an analogous result for the so-called t-fork.In [7, Conjecture 7] a general conjecture was made on the threshold for the P -free problem in P(n, p) and the corresponding problem for induced P was also considered.Our next result resolves the random version of the P -free problem for all tree posets of radius at most 2.
Theorem 1.10.Let P be any tree poset of height h ≤ 5 and radius at most 2. Given any ε > 0, there exists C = C(ε, P ) > 0 such that the following holds.If p > C/n then w.h.p. the largest P -free family in P(n, p) has size (h − 1 ± ε)p n ⌊n/2⌋ .
Notice that Theorem 1.10 implies the random version of Sperner's theorem (Theorem 1.9).Furthermore, the lower bound on p in Theorem 1.10 is essentially tight.Indeed, for such P , [7, Corollary 6] implies that if p = o(1/n) then w.h.p. the largest P -free family in P(n, p) has size at least (h − o(1))p n ⌊n/2⌋ .The proof of Theorem 1.10 applies Theorem 1.6, Lemma 1.8 and the graph container method.As noted by one of the referees of the paper, it may be possible to prove Theorem 1.10 via the hypergraph container method.In fact, the analogous result for C t -free families was proven using the hypergraph container method in [4].Our proofs of Theorems 1.5 and 1.10 show that variants of the graph container algorithm have the potential to attack problems that cannot be naturally stated in the language of independent sets in graphs.This philosophy is perhaps the most important message for the reader to take from our work.
The paper is organised as follows.In Section 2 we introduce some useful results.The proofs of Theorems 1.5, 1.6 and 1.10 and Lemma 1.8 are presented in Section 3. In Section 4 we provide some further discussion and make a few simple observations.Notation.Throughout this paper we omit floors and ceilings whenever this does not affect the argument.Given a set S and t ∈ N we write S ≤t to denote the set of all subsets of S of size at most t, and let |S| ≤t := S ≤t .Given any poset P , when considering P as a Hasse diagram we will often refer to the elements of P as vertices.Consider a tree poset P with root x ∈ P .Suppose u ∈ P is of distance ρ from x in the undirected Hasse diagram of P , and let v ∈ P be a neighbour of u of distance ρ + 1 from x in the undirected Hasse diagram.Then we call u the parent of v and v a child of u.Note that every vertex other than x has precisely one parent, though every vertex can have multiple children.

Preliminaries
In this section, we introduce a few tools that will be used in the proofs of some of our main results. 2 Here, by with high probability we mean with probability tending to 1 as n tends to infinity.
. This is the expected number of sets in F that are contained in a maximum chain C in 2 [n] chosen uniformly at random among all such chains.Let C n denote the set of all maximal chains in 2 [n] .Given a family F ⊆ 2 [n] , the min-partition of C n is As λ n (F ) is the expected number of sets of F in a randomly selected C ∈ C n , and λ n−|F | (U(F )) is the expected number of sets of F in a randomly selected chain C ∈ C F , we obtain the following proposition.
The proof of the next lemma is essentially identical to that of Theorem 3.1 in [9].Let where here we used properties (i) and (ii).Propositions 2.2 and 2.3 complete the proof.
Propositions 2.2 and 2.3 complete the proof.

The proofs of our main results
3.1.Proof of Theorem 1.5.To prove Theorem 1.5 we apply the following graph container result.
Lemma 3.1.Suppose P is a tree poset of height h for which Conjecture 1.3 holds for some choice of x ∈ P .Given any ε > 0, let δ > 0 be as in Conjecture 1.3 and suppose n ∈ N is sufficiently large.Then there exists a function f : ) such that for any P -free family F in 2 [n] there is a subfamily Given any H ∈ 2 [n]   ≤|P |2 n /δn we refer to the family H ∪ f (H) produced by Lemma 3.1 as a container.We call H the fingerprint of the container H ∪ f (H).With Lemma 3.1 at hand, Theorem 1.5 now follows easily.
Proof of Theorem 1.5.Suppose P is a tree poset of height h for which Conjecture 1.3 holds for some choice of x ∈ P .By Theorem 1.1, there exists a P -free family F in 2 [n] of size (h − 1 + o(1)) n ⌊n/2⌋ .By considering all subfamilies of F we see that there are at least 2 (h−1+o( 1))( n ⌊n/2⌋ ) P -free families in 2 [n] .To prove the theorem, note that it therefore suffices to prove the following: given any ε > 0, if n ∈ N is sufficiently large then there are at most 2 (h−1+3ε)( n ⌊n/2⌋ ) P -free families in 2 [n] .Given any ε > 0 let δ > 0 be as in Conjecture 1.3 and suppose n ∈ N is sufficiently large.Let C be the collection of all the containers H ∪ f (H) (for all H ∈ 2 [n]   ≤|P |2 n /δn ) obtained when applying Lemma 3.1.Thus, where the last inequality follows as n is sufficiently large.Moreover, by Lemma 3.
for every C ∈ C. Since every P -free family in 2 [n] is a subset of some container C ∈ C, we immediately obtain that there are at most ) P -free families in 2 [n] , as desired.
It remains to prove Lemma 3.1.The proof uses a modified version of the graph container algorithm.In particular, there are some specific subtleties we need to take care of.
In the proof of Lemma 3.1 we will be constructing our containers through a process of analysing copies of t-blow-ups P (x, t) in 2 [n] .For this, it will be helpful to consider the following process of 'constructing' a copy of P in P (x, t) with root x.Suppose the children of x in P are u 1 , . . ., u s .Then for each of these children u i there is a set U i of size t in P (x, t) where each x i ∈ U i can 'play the role' of u i in a copy of P rooted at x.That is, there is an edge between x i and x in the Hasse diagram of P (x, t) and it is oriented in the 'same' direction as the edge between u i and x in the Hasse diagram of P .Fix some choice of x i ∈ U i for each i ∈ [s]; so x i will play the role of u i in the copy of P we are constructing.Fix any i ∈ [s], and let u 1,i , . . ., u si,i denote the children of u i in P .Again, for each of these children u j,i there is a set U j,i of size t in P (x, t) where each x j,i ∈ U j,i can play the role of u j,i in a copy of P rooted at x that contains x 1 , . . ., x s .In particular, every vertex x j,i ∈ U j,i is adjacent to x i in the Hasse diagram of P (x, t) and the edge between x j,i and x i is oriented in the 'same' direction as the edge between u j,i and u i in the Hasse diagram of P .For each i ∈ [s] and j ∈ [s i ], fix some x j,i ∈ U j,i to play the role of u j,i in the copy of P we are constructing.We repeat this process to construct a copy of P rooted at x in P (x, t): at every step we consider the children u of every vertex selected in the previous step; we have t vertices in P (x, t) that are able to play the role of u in the copy of P we are constructing, so we can continue this process to obtain our desired copy of P .
Suppose now F is a P -free family in 2 [n] and consider a copy of P (x, t) in 2 [n] for some t ∈ N and x ∈ P .Run the process described in the previous two paragraphs, however, now at every step insist that any vertex in P (x, t) selected must be in F .Since F is a P -free family this process cannot finish and produce a copy of P .That is, either the root of our copy of P (x, t) does not belong to F , or at some step of the process we must encounter a set U of t vertices that corresponds to some u ∈ P and where U ∩ F = ∅.As we will now see, this simple observation is crucial to the success of running our version of the graph container algorithm.
Proof of Lemma 3.1.Suppose P is a tree poset of height h for which Conjecture 1.3 holds for some choice of x ∈ P .Set p := |P |.Given any ε > 0, let δ > 0 be as in Conjecture 1.3 and suppose n ∈ N is sufficiently large.
Fix a total order O 2 [n] of the elements of 2 [n] .Let O P be a total order of the vertices of P such that the first vertex is x; the next vertices are the children of x; the next vertices are those vertices of distance two from x in the undirected Hasse diagram of P , and so forth.We write y 1 , . . ., y p for the vertices of P ordered as in O P ; so y 1 = x.Further, let P blow denote the set of all copies of P (x, t) in 2 [n] for all t ∈ N. Let O blow be a total order of the elements of P blow .
We now run our modified version of the graph container algorithm.The input of the algorithm is a P -free family F ⊆ 2 [n] .The algorithm will output a fingerprint H and a container H ∪ f (H) where H ⊆ F ⊆ H ∪ f (H).
Initially we set G 0 := 2 [n] and H 0 := ∅.We will add vertices from 2 [n] to H 0 and remove vertices from G 0 through the following iterative process, beginning at Step 1.
At Step i, let P 1 (x, t) ∈ P blow be a copy of P (x, t) in G i−1 ⊆ 2 [n] where we choose t to be as large as possible.If there is more than one copy of P (x, t) in G i−1 for this choice of t we choose P 1 (x, t) to be the copy of P (x, t) appearing earliest in the total order O blow .Let x 1 be the vertex of P 1 (x, t) that plays the role of x.
• Suppose x 1 ∈ F and t ≥ δn.We will update G i−1 and H i−1 in several subphases to obtain G i and H i respectively.
-Subphase 1. Delete x 1 from G i−1 and add it to H i−1 .Proceed to Subphase 2.
-Subphase 2. Recall that we write y 1 , . . ., y p for the vertices of P ordered as in O P .So y 1 = x and y 2 is a child of x in P .In P 1 (x, t) there is a set of t vertices U 2 ⊆ G i−1 so that each x 2 ∈ U 2 is adjacent to x 1 in the Hasse diagram of P 1 (x, t) and further the edge between x 1 and x 2 is oriented in the same direction as the edge between y 1 (= x) and y 2 in the Hasse diagram for P .Look through the vertices in U 2 ⊆ 2 [n] one by one, following the total order O .So here we have deleted between 1 and t elements of G i−1 , and of these only x 2 lies in F .Proceed to Subphase 3. -Subphase j (for 3 ≤ j ≤ p).In the previous subphases we have defined x 1 , . . ., x j−1 ∈ P 1 (x, t) corresponding to y 1 , . . ., y j−1 ∈ P .By definition of O P , the unique parent y k of y j in P must be one of y 1 , . . ., y j−1 .Furthermore, by definition of P (x, t), there is a set of t vertices U j ⊆ G i−1 in P 1 (x, t) corresponding to y j so that every x j ∈ U j is adjacent to x k in the Hasse diagram of P 1 (x, t), and moreover, the edge between x k and x j is oriented in the same direction as the edge between y k and y j in the Hasse diagram for P .Look through the vertices in U j ⊆ 2 [n] one by one, following the total order O 2 [n] .If we run through the entire set U j without finding an element from F then we delete all t vertices in U j from G i−1 and stop Step i.
Otherwise, let x j ∈ U j denote the very first element from F that we discover in U j .Add x j to H i−1 and remove from G i−1 both x j and all elements of U j that occur before x j in the total order O 2 [n] .So here we have deleted at most t elements of G i−1 , of which only x j lies in F .Proceed to Subphase j + 1.At the end of these subphases relabel G i−1 as G i and H i−1 as H i ; then proceed to Step i + 1.Since F is P -free, for some j ∈ [p], Subphase j must consider a set U j where U j ∩ F = ∅.Thus, in Step i we have deleted at least t vertices from G i−1 to obtain G i and have added at most p − 1 vertices to H i−1 (at most one in each of the first p − 1 subphases) to obtain H i .
• Suppose x 1 ∈ F and t < δn.Then define H := H i−1 ∪ {x 1 } and f (H) := G i−1 \ {x 1 } and terminate the algorithm.Note that in every step of the algorithm we only add elements to H i−1 that lie in F .So certainly H ⊆ F .Similarly, by construction, F ⊆ H ∪ f (H).In every step of the algorithm, except the final step, if we add (at most p − 1) elements to H i−1 we delete at least δn elements from G i−1 ⊆ 2 [n] .Therefore, Furthermore, by construction of the algorithm, the family f (H) ⊆ 2 [n] does not contain a copy of the δn blow-up P (x, δn).Thus, by the assumption of the lemma, we have that .
Therefore all that remains to check is that the function f is well-defined.That is, suppose F 1 and F 2 are P -free families in 2 [n] such that, on input F i , our algorithm outputs the container H i ∪ f (H i ) for each i ∈ [2]; if H 1 = H 2 then we require that f (H 1 ) = f (H 2 ).This follows though immediately from the definition of the algorithm.Indeed, suppose one is only presented with the fingerprint that is outputted by the algorithm.
Then one can completely identify every action taken during the algorithm.Thus, two applications of the algorithm yielding the same fingerprint must also yield the same container.3.2.Proofs of Theorem 1.6 and Lemma 1.8.For the proofs, we need to define five specific posets.See their Hasse diagrams in Figure 3. Let • be the three element poset on {a, x, c} with a, x < c, and a, x unrelated; • Y be the four element poset on {a, x, c, d} with a, x > c > d being all its cover relations; • S be the five element poset on {a, b, x, d, e} with a > b < x < d > e being all its cover relations; • S + be the six element poset on {a, b, x, d, e, f } with a > b < x < d < f and d > e being all its cover relations; • S ++ be the seven element poset on {a, b, x, d, e, f, g} with a > b, g < b < x < d < f , and d > e being all its cover relations.
Proof of Theorem 1.6.Let ε > 0; note that it suffices to prove the theorem under the assumption that ε is sufficiently small and n ∈ N is sufficiently large.Observe that if x ∈ P is within distance at most 2 of all other elements of P , then a connected component of P \ {x} is either an isolated vertex or contains a vertex u that is contained in all edges of the Hasse diagram of that component, and where u is adjacent to x in the Hasse diagram of P .This and Remark 1 imply that it is enough to prove the statement for the posets , Y d , S, S + , S ++ and x as given in their definition and in Figure 3.For example, if P is of height 2 and γ 1 > 0, then there exists a γ 2 > 0 such that there is a choice of y ∈ P so that P (y, Consider first the poset ; its height is 2. It suffices to consider F ⊆ 2 [n] of size exactly (1 + ε) n ⌊n/2⌋ .Let δ := ε 2 (1+ε)120 .As n is large enough, we can and will assume (here and in the cases of the four other posets as well) that all sets in F have size between n/2 − n 2/3 and n/2 + n 2/3 .Indeed, there are o( n ⌊n/2⌋ ) sets in 2 [n] that do not have size between n/2 − n 2/3 and n/2 + n 2/3 , so we can ignore any such sets in F .We say that F ∈ F is of • Type 1 if there exist at least εn/30 sets G ∈ F with |G| = |F | − 1 and G ⊂ F ; • Type 2 if, for some j ≥ 2, there exist at least εn 2 /30 sets G ∈ F with |G| = |F | − j and G ⊂ F ; • Type 3 otherwise.Let F * denote the subfamily of F of sets of type 1 or 2. Applying Lemma 2.4 'upside down', we obtain that the subfamily F \ F * of sets of type 3 has size at most (1 + ε/2) n ⌊n/2⌋ .Therefore, |F * | ≥ ε n ⌊n/2⌋ /2.Consider the bipartite graph B with parts F and F * , where there is an edge between G ∈ F and F ∈ F * if and only if G ⊂ F .By definition of the types, the number of edges in B is at least εn|F * |/30 ≥ ε 2 n n ⌊n/2⌋ /60; so there exists a set F x ∈ F with degree at least 2δn in B. F x will play the role of x in the copy of P (x, δn) that we will construct.At least δn of F x 's neighbours in B are of the same type (i.e., all of type 1 or all of type 2); they will play the role of the many copies of c in P (x, δn).Let F c ⊆ F * denote a family of such sets so that |F c | = δn.
Suppose first that all sets in F c are of type 1.Each F ∈ F c has at least εn/30 subsets in F of size |F | − 1; so there are at least εn/30 − |F c | − 1 ≥ εn/60 such subsets of F that do not belong to F c ∪ {F x }.If F, F ′ ∈ F c have different sizes, then the subsets of F and F ′ that we consider are clearly distinct.If F, F ′ have the same size, then they share at most one common subset of size |F | − 1.Therefore, for all F ∈ F c we can pick εn/60 − |F c | ≥ δn distinct subsets of F to obtain a copy of (x, δn) in F .
Suppose next that F c consists only of sets of type 2. Then each of them contains at least εn 2 /30−|F c |−1 ≥ δn|F c | subsets that do not lie in F c ∪ {F x }.So we can pick greedily δn distinct subsets for each F ∈ F c to obtain a copy of (x, δn).
Consider next the poset Y d .In this case it suffices to consider F ⊆ 2 [n] of size exactly (2 + ε) n ⌊n/2⌋ .Let δ 1 be the output of the theorem on input ε/2, and x; thus, δ 1 = ε 2 /(480(1 + ε/2)).For Y d we show that we can take δ := δ 1 /2.Now we say that F ∈ F is of .By the previous case, we obtain a copy of (x, δ 1 n) in F * .Let F x be the set corresponding to x, F c be the family of sets corresponding to the copies of c, and F a be the family of sets corresponding to copies of a. Suppose that there exist F c ∈ F c , F a ∈ F a with F a ⊃ F c .Then we can exchange the roles of F a and F c and we still have a copy of (x, δn).As after such a swap, the number of pairs F a ⊃ F c decreases, after a finite number of changes, we can assume that no F a ∈ F a contains any F c ∈ F c .Now we are ready to use our copy of (x, δ 1 n) to obtain a copy of Y d (x, δ 1 n/2) by adding δ 1 n/2 distinct supersets of each F c ∈ F c .We distinguish two cases according to the type of sets in F c ; at the price of working with (x, δ 1 n/2) rather than (x, δ 1 n), we can assume that all sets in F c are of the same type.
Suppose first that all sets in F c are of type 1.Then, by our assumption that no F a ∈ F a contains any F c ∈ F c , any F c ∈ F c has at least (ε/30 − δ 1 /2)n supersets of size |F c | + 1 in F that are not used in our copy of (x, δ 1 n/2).If F c and F ′ c have different sizes, then these supersets are distinct because they are also of different size.If F c , F ′ c are of the same size, then they can share at most one common superset of size |F c | + 1; thus, as δ 1 is much smaller than ε, we can pick δ 1 n/2 distinct supersets of each F c ∈ F c to obtain Y d (x, δ 1 n/2).Suppose next that all F c ∈ F c are of type 2. Then as εn 2 /30 > δ 1 n|F c |, we can pick the supersets of each F c greedily to obtain a copy of Y d (x, δ 1 n/2).
Finally, we can consider the posets S, S + , and S ++ .As the proofs are almost identical, we only show the statement for S + .Let F ⊆ 2 [n] be a family of size exactly (3 + ε) n ⌊n/2⌋ .Let δ 1 be the value of δ that we showed to exist for ε/3, and x, and let δ 2 be the value of δ that we showed to exist for ε/3, Y d and x.Set δ * := min{δ 1 , δ 2 }.Let F 1 be the subfamily of F consisting of those sets F ∈ F for which there exists no copy of (x, δ * n) ⊂ F in which F plays the role of x.(The poset is the dual of .)Let F 2 be the subfamily of F consisting of those sets F ∈ F for which there exists no copy of Y d (x, δ * n) ⊂ F in which F plays the role of x.By definition of δ 1 , δ 2 and δ * , we have . By definition of F 1 and F 2 , F x plays the role of x both in a copy G 1 of (x, δ * n) and in a copy G 2 of Y d (x, δ * n).In G 1 ∪ G 2 , the containment relations are as required as in a copy of S + (x, δ * n) with F x playing the role of x; the only problem that can occur is that G 1 and G 2 might overlap.We overcome this problem by decreasing the value of δ * and filtering out sets from G 1 ∪ G 2 .
For y ∈ {d, e, f } let G y ⊂ G 2 denote the family of sets corresponding to y in G 2 and similarly for y ∈ {a, b} let G y ⊂ G 1 denote the family of sets corresponding to y in We may therefore throw away exactly half of E(G d ) ∩ G e and exactly half of We now must have that any overlaps between G 1 and G 2 occur between G a and G d ∪G e ∪G f .Note that currently we have that Remove from G 2 (and thus G d and G e ∪ G f respectively) all elements of G ′ d as well as all elements from {E(G ′ d ) : We now must have that any overlaps between G 1 and G 2 occur between G a and G e ∪ G f .Note that currently we have that This is a contradiction though as Remove all destroyed G b from G b and also remove from We have ensured that G a and G e ∪ G f are disjoint, and thus G 1 and G 2 are disjoint.Therefore, G 1 ∪ G 2 contains a copy of S + (x, δn) where δ := δ * /200, as desired.
Proof of Lemma 1.8.The proof closely follows that of Theorem 1.6, but instead of using Lemma 2.4 we apply Lemma 2.5.Again, as in the case of Theorem 1.6, by Remark 1, it is enough to only prove the result for P = , Y d , S, S + , S ++ , but with the slightly stronger conclusion that we seek a copy of P (x, n 1.91 ).As x is at distance at most 2 from any element of P , every element y ∈ P has at most n 3.82 ≪ n 4 sets corresponding to y in a copy of P (x, n 1.91 ).This combined with the fact that we now consider families F that are four times as large as those in Theorem 1.6 means the proof is actually cleaner than that of Theorem 1.6.As such, we only explicitly show the case of .
Let 0 < ε < 1/4 be fixed, n be sufficiently large and . Let F − be the family of sets F ∈ F that do not contain εn 4 /500 other sets of F of size |F | − j for any j ≥ 4. By Lemma 2.5, |F − | ≤ (4+ 4ε 5 ) n ⌊n/2⌋ and thus F + := F \F − has size at least 3ε n ⌊n/2⌋ .Consider the bipartite graph B with parts F + and F such that F + ∈ F + and F ∈ F are joined by an edge if and only if F ⊂ F + .By definition of F + , the number of edges in B is at least εn 4 |F + |/500 ≥ ε 2 n 4 n ⌊n/2⌋ /200; therefore there exists F x ∈ F with degree at least ε 2 n 4 /1000 in B. F x will play the role of x in the copy of (x, n 1.91 ) that we will construct, and n 1.91 of its neighbours F c in B will play the role of x's neighbours in (x, n 1.91 ).For each of these n 1.91 neighbours F c of F x , we greedily pick n 1.91 distinct subsets of F c to obtain a copy of (x, n 1.91 ) in F .
The proof of Lemma 3.2 closely follows that of Lemma 3.1 except that we analyse the container algorithm over two separate stages.The idea of a multi-stage analysis of the graph container algorithm was first introduced in [1].In particular, as in the proof of Theorem 1.9 presented in [1], using a two-stage container algorithm will allow us to employ more careful calculations in the proof of Theorem 1.10, thereby allowing us to deal with probabilities p 'close' to 1/n.
Proof of Lemma 3.2.Let P be as in the statement of the lemma and set p := |P |.Let x ∈ P be of distance at most two from every other element in the Hasse diagram of P .Given any 0 < ε < 1/5, let δ > 0 be as in Conjecture 1.3.Suppose n ∈ N is sufficiently large and let m := n ⌊n/2⌋ .Fix a total order O 2 [n] of the elements of 2 [n] .Let O P be a total order of the vertices of P such that the first vertex is x; the next vertices are the children of x; the next vertices are those vertices of distance two from x in the undirected Hasse diagram of P , and so forth.We write y 1 , . . ., y p for the vertices of P ordered as in O P ; so y 1 = x.Further, let P blow denote the set of all copies of P (x, t) in 2 [n] for all t ∈ N. Let O blow be a total order of the elements of P blow .
We now run our modified version of the graph container algorithm.The input of the algorithm is a P -free family F ⊆ 2 [n] .The algorithm will output fingerprints H 1 , H 2 and a container . We proceed in two stages.
Stage 1: Initially we set G 0 1 := 2 [n] and H 0 1 := ∅.We will add vertices from 2 [n] to H 0 1 and remove vertices from G 0 1 through the following iterative process, beginning at Step 1.At Step i, let P 1 (x, t) ∈ P blow be a copy of P (x, t) in G i−1 1 ⊆ 2 [n] where we choose t to be as large as possible.If there is more than one copy of P (x, t) in G i−1 1 for this choice of t we choose P 1 (x, t) to be the copy of P (x, t) appearing earliest in the total order O blow .Let x 1 be the vertex of P 1 (x, t) that plays the role of x.
• Suppose x 1 ∈ F .Then define • Suppose x 1 ∈ F and t ≥ n 1.9 .We will update G i−1 1 and H i−1 1 in several subphases to obtain G i 1 and H i 1 respectively.
-Subphase 1. Delete x 1 from G i−1 1 and add it to H i−1 1 .Proceed to Subphase 2. -Subphase 2. Recall that we write y 1 , . . ., y p for the vertices of P ordered as in O P .So y 1 = x and y 2 is a child of x in P .In P 1 (x, t) there is a set of t vertices U 2 ⊆ G i−1 1 so that each x 2 ∈ U 2 is adjacent to x 1 in the Hasse diagram of P 1 (x, t) and further the edge between x 1 and x 2 is oriented in the same direction as the edge between y 1 (= x) and y 2 in the Hasse diagram for P .Look through the vertices in U 2 ⊆ 2 [n] one by one, following the total order O 2 [n] .If we run through the entire set U 2 without finding an element from F then we delete all t vertices in U 2 from G i−1 1 and stop Step i.Otherwise, let x 2 ∈ U 2 denote the very first element from F that we discover in U 2 .Add x 2 to H i−1 1 and remove from G i−1 1 both x 2 and all elements of U 2 that occur before x 2 in the total order O 2 [n] .So here we have deleted between 1 and t elements of G i−1 1 , and of these only x 2 lies in F .Proceed to Subphase 3.
In every step of Stage 2 except the final step, if we add (at most p − 1) elements to H i−1 2 we delete at least δn elements from Combining this with (1) we see that condition (i) of the lemma holds.Note that g(H 1 ∪H 2 ) ⊆ 2 [n] is defined so that it does not contain a copy of P (x, δn).Thus, by Theorem 1.6, Therefore all that remains to check is that the functions f and g are well-defined.That is, if for two P -free families F and F ′ the algorithm outputs the same H 1 , then f (H 1 ) is defined the same, and furthermore, if H 1 ∪ H 2 is the same for both runs of the algorithm then so is g(H 1 ∪ H 2 ).This follows though immediately from the definition of the algorithm.Indeed, suppose one is only presented with the fingerprint H 1 that is outputted by Step 1 of the algorithm.Then one can completely identify every action taken during Step 1 of the algorithm.Similarly, if one is further given the second fingerprint H 2 , then one can completely identify every action taken during Step 2 of the algorithm.With Lemma 3.2 at hand, we can now prove Theorem 1.10.
Apply Lemma 3.2 with ε 1 playing the role of ε.Suppose for a contradiction that P(n, p) does contain some P -free family F with |F | > (h − 1 + ε)pm.Let H 1 and H 2 denote the fingerprints for F given by Lemma 3.2.Then since Note that the number of possibilities for H 1 is ≤|P |n −1.9 2 n , and for each possibility the probability that H 1 ⊆ P(n, p) is p |H1| .For any fixed H 1 we have |f (H 1 )| ≤ 4(h − 1 + ε 1 )m ≤ (4h − 3)m and H 2 ⊆ f (H 1 ), so the number of possibilities for H 2 is at most ≤|P |(4h−3)m/δn , and for each possibility the probability that H 2 ⊆ P(n, p) is p |H2| .Furthermore, for any fixed H 1 and H 2 we have g(H 1 ∪ H 2 ) ≤ (h − 1 + ε 1 )m = (h − 1 + ε/4)m, so the expected number of elements of g(H 1 ∪ H 2 ) selected for P(n, p) is at most (h − 1 + ε/4)pm.By the Chernoff bound for the binomial distribution, the probability that at least (h − 1 + ε/2)pm elements of g(H 1 ∪ H 2 ) are selected for P(n, p) is therefore at most e −ε 2 pm/(100h 2 ) .
Proof.The upper bound follows from the fact that a ♦-free family F ⊆ P(n, p) cannot contain a 4-chain which is a special copy of ♦.Therefore F can be partitioned into three antichains F 1 , F 2 , F 3 ; thus the upper bound on |F | follows from Theorem 1.9 applied to all F i s.
To see the lower bound, observe that the expected number E 1 of sets in the middle three layers of P(n, p) is (3 ± ε/4)p n ⌊n/2⌋ .Also, the expected number E 2 of copies of ♦ in the middle three layers of P(n, p) is (1±ε/4)p 4 n  As Theorem 1.5 articulates, Conjecture 1.3 has implications to the counting problem for tree posets P .The resolution of Conjecture 1.3 would also likely allow one to generalise Theorem 1.10 to all tree posets P of height h, not just those of radius 2. For this one would also need an analogue of Lemma 1.8.
One might wonder why our proof of Theorem 1.6 does not work for posets of radius larger than 2, and why Bukh's original argument [3] cannot be applied to prove Theorem 1.6 for arbitrary tree posets and arbitrary element x.Let us consider first the latter question.Bukh's proof uses some preliminary structural results on tree posets.Namely, he shows that any tree poset of height h is a subposet of a height h saturated tree poset, i.e., one in which all maximal chains contain h elements.Then he shows that any saturated tree poset T of height h can be obtained from C h by 'adding intervals' I 1 , I 2 , . . ., I m , i.e., branches of the Hasse diagram.This enables him to obtain an inductive proof by showing that T \ I m can be embedded not only into any large enough family F , but to nice, extendable sets of F , where nice means that a constant number of sets cannot ruin the extending property.There are two points of this proof that we do not see how to save to generalise Theorem 1.6.As a height h tree poset can have arbitrary radius, it is not enough to extend the embedding of the blow-up by one interval, but by a polynomial (of arbitrary high degree!)number of blow-ups of intervals, and now the extension property should not be ruined by a polynomial number of sets (rather than a constant number).Even more importantly, we do not know how to start the induction.Bukh's result for chains (even with the extension property) is immediate from any supersaturation result for h-chains, but we are unable to prove Theorem 1.6 for C 4 and x being its minimal element.
This brings us to the other question: why our proof does not work for the blow-up of trees with larger radius.Getting the result for and is basically an exercise if one is familiar with the notion of Lubellmass.Then for the poset Y , we try to apply some kind of an induction: we find lots of blow-ups of and then a blow-up of in the root of the blow-ups of the first round.All we need to take care of are the overlaps of these blow-ups.This is done via introducing type 1, 2, and 3 sets and observing that if we work with type 1 sets, then the pairwise overlaps are of size at most 1 and so at the price of shrinking δ, we get our pairwise disjoint blow-ups, while if we work with type 2 sets, then we need only a linear number of sets, while the blow-ups are quadratic.The latter could work for arbitrary long chains, but type 1 sets are problematic: type 1 copies of blow-ups of Y can intersect in a linear number of sets, and we were unable to handle how to get many pairwise disjoint copies of them.
Because of the above reasoning, the full resolution of each of Conjecture 1.3, Conjecture 1.4 and the random version of the P -free problem does currently seem out of reach.It would therefore be interesting to resolve these problems for other natural classes of posets P .For example, it would be interesting to extend our results to cover all tree posets P of height 2, particularly those posets whose (undirected) Hasse diagram is a path.In [11], La(n, D k ) was asymptotically determined for infinitely many choices of k.It would be interesting to resolve Conjecture 1.4 and the random version of the P -free problem for such D k .It is also natural to seek induced versions of our results, including Corollary 1.7 and Theorem 1.10.

Figure 2 .
Figure 2. Some posets of radius 2: the N poset, a monotone tree, and a more complicated one.

G 1 .
Therefore, G d , G e , G f are pairwise disjoint and so are G a , G b .Also, G b ∩ (G d ∪ G f ) = ∅ as all sets in G b are proper subsets of F x , while all sets in G d ∪ G f are proper supersets of F x .For a set G b ∈ G b , we write E(G b ) ⊂ G a for the family of sets that correspond to the component of G 1 \ {F x } containing G b , and for a set G d ∈ G d we write E(G d ) ⊂ G e ∪ G f for the family of sets that correspond to the component of G 2 \ {F x } containing G d .• Consider a subfamily G ′ b ⊂ G b with |G ′ b | = 1 2 |G b |.Remove from G 1 (and thus G b and G a respectively) all elements of G ′ b as well as all elements from {E(G ′ b
2 [n] .If we run through the entire set U 2 without finding an element from F then we delete all t vertices in U 2 from G i−1 and stop Step i.Otherwise, let x 2 ∈ U 2 denote the very first element from F that we discover in U 2 .Add x 2 to H i−1 and remove from G i−1 both x 2 and all elements of U 2 that occur before x 2 in the total order O 2 [n] • Type 1 if there exist at least εn/30 sets G ∈ F with |G| = |F | + 1 and G ⊃ F ; • Type 2 if, for some j ≥ 2, there exist at least εn 2 /30 sets G ∈ F with |G| = |F | + j and G ⊃ F ; • Type 3 otherwise.Applying Lemma 2.4, we obtain the subfamily of F of sets of type 3 has size at most (1 + ε/2) n ⌊n/2⌋ .Therefore, the subfamily F * ⊆ F of sets of type 1 or 2 has size at least (1 + ε/2) n ⌊n/2⌋