The Boolean Rainbow Ramsey Number of Antichains, Boolean Posets, and Chains

Motivated by the paper of Axenovich and Walzer [2], we study the Ramsey-type problems on the Boolean lattices. Given posets $P$ and $Q$, we look for the smallest Boolean lattice $\mathcal{B}_N$ such that any coloring on elements of $\mathcal{B}_N$ must contain a monochromatic $P$ or a rainbow $Q$. This number $N$ is called the Boolean rainbow Ramsey number of $P$ and $Q$ in the paper. Particularly, we determine the exact values of the Boolean rainbow Ramsey number for $P$ and $Q$ being the antichains, the Boolean posets, or the chains. From these results, we also give some general upper and lower bounds of the Boolean rainbow Ramsey number for general $P$ and $Q$ in terms of the poset parameters.


Introduction
A poset P = (P, P ) is a set P equipped with a partial order P . For any two elements a, b ∈ P if either a P b or b P a holds, then a and b are comparable, otherwise they are incomparable. In this paper, we study the Ramsey-type problems of the wellknown poset, Boolean lattice, B n whose underlying set is the collection of all subsets of [n] := {1, 2, . . . , n} ([0] = ∅ as a convention) and the partial order is the inclusion relation on sets. The k-th level of B n is the collection of all k-subsets of [n], denoted by [n] k . A family F of subsets is isomorphic to a poset P if there exists an order-preserving bijection φ between P and F, i.e., P φ ←→ F such that for any x 1 , x 2 ∈ P , x 1 < P x 2 if and only if φ(x 1 ) ⊂ φ(x 2 ). If a family F contains a subfamily G isomorphic to P , we will say F contains P as a strong (or induced) subposet (F contains P , for short), or say the subsets in G form a copy of P . A coloring (k-coloring) of B n is a mapping c from B n to a set of positive integers (to [k]). Given a coloring c of B n , we say B n contains a monochromatic P under c if there is a family of subsets of the same color containing P .
In the literature on Ramsey theory, the Ramsey problems have been greatly studied on the set systems (hypergraphs), the graphs, the planes, and the general posets. For example, see [1,5,6,9,13,18]. The following type of Ramsey problems on Boolean lattices has been recently studied by Axenovich and Walzer [2]: Question 1. Given posets P 1 , . . . , P k , can one find the least integer n such that for any k-coloring of B n , it always contains a monochromatic P i of color i for some i?
In their paper, they called such a number the poset Ramsey number. Since there are many Ramsey properties investigated on the families of posets with various parameters, to make it more precise, we suggest the name the Boolean Ramsey number and use it in this paper. Let us define the term formally below: Definition 2. Given posets P 1 , . . . , P k , the Boolean Ramsey number R(P 1 , . . . , P k ) is the minimum integer n such that for any k-coloring of B n , it always contains a monochromatic P i of color i for some i. Moreover, if P i = P for all 1 i k, then we use R k (P ) to denote R(P 1 , . . . , P k ).
Including the above theorem, Axenvoich and Walzer obtained a series of results on the Boolean Ramsey number for the target posets being the Boolean lattices in [2], which are improved by Lu and Thompson [15] very recently.
Given a coloring c of B n , we say B n contains a rainbow P if it contains P and all the subsets forming P are of distinct colors under c. Imitating the rainbow Ramsey number in graph theory, we give the following definition: Definition 4. Given two posets P and Q, the Boolean rainbow Ramsey number RR(P, Q) is the minimum integer n such that for any coloring c of B n , it always contains either a monochromatic P or a rainbow Q as a subposet.
A result by Johnston, Lu, and Milans [12] implies that RR(P, Q) is finite for any P and Q. They proved that for n sufficiently large, any coloring of B n contains either a monochromatic Boolean algebra B alg(r) or a rainbow B alg(s) for any given positive integers r and s. A Boolean algebra B alg(r) in B n is such a family {S 0 ∪ (∪ i∈I S i ) | I ⊂ [r]}, where S 0 , S 1 , . . . , S r are pairwise disjoint subsets of [n] with S i = ∅ for i 1.
Containing the Boolean algebra B alg(r) implies containing B r since B alg(r) is isomorphic to B r . So RR(P, Q) is finite when P and Q are both Boolean lattices. In addition, Trotter [17] introduced the 2-dimension of a poset P , dim 2 (P ), the minimum number n for which B n contains P and proved that dim 2 (P ) is finite for any poset P . As a consequence, RR(P, Q) exists for any P and Q.
Analogous to [2], Cox and Stolee [7] studied the existence of the monochromatic weak subposets in the Boolean lattices. Here a weak subposet means an injection from the poset to the Boolean lattice, which preserves the inclusion relation but not necessary the non-inclusion relation. A very recent paper [4] by the third author and others presents the results on the Ramsey properties of both types of subposets in the Boolean lattices.
In this paper, we study the strong version of the Boolean rainbow Ramsey number and the relations between it and the Boolean Ramsey number. We determine the exact values of RR(P, Q) for specific posets and use the results to derive the upper and lower bounds for RR(P, Q) for general P and Q. We focus on the antichains A n , the Boolean posets B n , and the chains C n , where A n is a set of n pairwise incomparable elements, B n is isomorphic to B n , and C n is a set of n mutually comparable elements. In the paper, the scribe B n refers to the underlying Boolean lattice we color, and the capital B n refers to the desired monochromatic or rainbow Boolean posets. Table 1 is the summary of our results on RR(P, Q) when P and Q are one of the three types of posets.
C m A n n + 2, m = 2 and n 3 (m − 1)(n − 1) + 2, m = n = 2, or m 3 and n 2 Table 1: RR(P, Q) of antichains, Boolean posets, and chains The organization of the remaining sections in the paper is the following. In Section 2, we first determine the values of the Boolean rainbow Ramsey numbers for antichains or chains, and deduce some lower bounds from those results. In Section 3, we give an upper bound on RR(B m , B n ) for general m and n, and determine the values of RR(B 1 , B n ) and RR(B m , B 1 ) as listed in Table 1. Using the idea of determining RR(B 1 , B n ), we manage to solve RR(C m , B n ), which is also demonstrated in this section. The last section contains discussions for the relations of R k (P ), RR(P, Q), and the forbidden subposet problems.
2 Chains, antichains, and the general lower bounds When P and Q are both chains or both antichains, the Boolean rainbow Ramsey numbers can be determined by simple arguments. Proposition 6. For the chains, we have RR(C m , C n ) = (m − 1)(n − 1).
Proof. For N < (m−1)(n−1), we give a coloring c to B N by coloring the subsets with i in the consecutive m − 1 levels Since a chain C m in the Boolean lattices consists of m subsets of distinct sizes, B N does not contain a monochromatic C m under c. In the coloring c, the number of color classes is at most n − 1, so B N does not contain a rainbow C n as well. When N (m − 1)(n − 1), consider any coloring of the chain ∅ ⊂ By the pigeonhole principle, it contains at least n subsets of distinct colors or at least m subsets of the same color, as desired.
Let N m,n be the minimum integer such that Nm,n Nm,n/2 The height and width of a poset P , h(P ) and w(P ), are the sizes of a maximum chain and antichain of P , respectively. It is clear RR(P, Q) RR(P , Q ) if P and Q are subposets of P and Q, respectively. The following bounds on RR(P, Q) for general P and Q are the consequences of Proposition 6 and Proposition 7.
For P and Q being different types of posets, the problem becomes much harder and more interesting. we manage to determine RR(C m , A n ) in Theorem 9 and give an upper bound for RR(A m , C n ) in Theorem 11. Theorem 9. For the chains and antichains, we have RR(C m , A n ) = n + 2, m = 2 and n 3, (m − 1)(n − 1) + 2, m = n = 2, or m 3 and n 2.
To prove Theorem 9, we need more preparations. Two families F and G of subsets are said to be incomparable if for any F ∈ F and G ∈ G, neither F ⊆ G nor G ⊆ F . The following structure in the Boolean lattices will help us to determine RR(C m , A n ).
Lemma 10. Let m 2 and n 2 be integers. For m 3, we can find n pairwise incomparable chains C 1 , . . . , When m = 2, this also holds for n = 2. For m = 2 and n 3, we can find such a collection of chains in B n+2 .
Proof. We prove the theorem by induction on n and constructing the families of chains recursively. Fix m 3. For n = 2, we take the two chains . Also, (m − 1)n + 2 is in every set in C n+1 but not in any set in C i for 1 i n. Therefore, C 1 , . . . , C n+1 are incomparable. For m = n = 2, the above construction consists of {{2}} and {{3}, {1, 3}}, which are incomparable. Now for m = 2 and n 3, we first take C 1 = {{1, 2}} and C 2 = {{4}, {1, 4}} as the desired incomparable chains in B 4 . Suppose we already have pairwise incomparable chains C 1 , . . . , C n in B n+2 for some n 3. In B n+3 , take C i = {[n + 2] − X | X ∈ C i }, for 1 i n, and C n+1 = {{n + 3} ∪ [j] | 0 j n}. Similar to the case of m 3, C 1 , . . . , C n are pairwise incomparable. The argument of incomparability of C n +1 and C i , i n is slightly different from the case of m 3: we have n + 2 in each set in C i for 1 i n − 1 but not in any set in C n+1 , n + 1 in each set in C n but not in any set in C n+1 , and n + 3 in each set in C n+1 but not in any set in C i for 1 i n. Hence C 1 . . . , C n+1 are pairwise incomparable.
Proof of Theorem 9. We use the constructions in Lemma 10 to show RR(C 2 , A n ) n + 2 for n 3, and RR(C m , A n ) (m − 1)(n − 1) + 2 for m = n = 2, or for m 3.
Given m, n 2, we consider an arbitrary coloring on We have a family of pairwise incomparable chains C 1 , . . . , C n in B n as described in Lemma 10. In both cases of N , for i 2, there exist at least i colors on the subsets in C i , otherwise it contains a monochromatic C m . However, if there are i colors on the subsets in C i , then we can pick a set from each chain C 1 , C 2 , . . . , C n in order so that all sets are of distinct colors. The n sets form a rainbow A n since all the chains are incomparable.
Let us establish the lower bound RR(C 2 , A n ) > n + 1 for n 3. For 0 i n + 1, we color every subset in [n+1] i with i. Clearly, there is no monochromatic C 2 under this coloring. If it contains rainbow A n , then neither ∅ nor [n + 1] can be in the rainbow antichain. So the rainbow A n must contain a set X i ∈ [n+1] i for 1 i n. However, if X 1 ⊂ X n , then X n = [n + 1] − X 1 and any other X i either contains X 1 or is contained in X n . As a consequence, the rainbow A n does not exist.
Next, we show RR(C m , A n ) > (m − 1)(n − 1) + 1 for m = n = 2, or for m 3 and n 2. Let N = (m − 1)(n − 1) + 1. For each X ∈ B N , color X with 1 + |X| m−1 . This gives an (n + 1)-coloring of B N such that no color appears on m subsets of distinct sizes. So, there is no monochromatic C m under the coloring. Meanwhile, colors 1 and n + 1 appear only on ∅ and [N ], respectively. Any family containing subsets of n distinct colors must contain at least one of the two sets, and thus cannot be an antichain A n .
Recall N m,n is the minimum integer such that Nm,n Nm,n/2 (m − 1)(n − 1) + 1. Because |S| k and there is no monochromatic A m under c , we can find a rainbow chain C n in B S . Note that a rainbow chain cannot contain both S and ∅. If this chain does not contain ∅, then it together with [k + m − 1] form a rainbow C n+1 under c. Else the chain contains ∅, then replace ∅ with S. The new chain together with [k + m − 1] form a rainbow C n+1 under c. Therefore RR(A m , C n+1 ) k +m−1 = RR(A m , C n )+m−1 as we claimed. Repeatedly using the last inequality n−2 times and the fact RR(A m , C 2 ) = N m,2 we obtain the inequality in the theorem.
Finally, let us show the equality for m = 2. By simple calculation we have (m − 1)(n − 1) + N m,2 = n when m = 2. It suffices to show RR(A 2 , C n ) n. For N < n, we color the subsets ∅ and [N ] in B N with the same color, and each of the remaining subsets with distinct colors. Clearly, there does not exist monochromatic A 2 , and every maximal rainbow chain has size N n − 1. Therefore, the lower bound holds and the proof is completed.
Since the exact value of RR(C m , A n ) is known, we also have the following lower bound on RR(P, Q) in terms of h(P ) and w(Q).

Results related to Boolean posets
Recall that the 2-dimension of a poset, dim 2 (P ), is the minimum number d such that the Boolean lattice B d contains P as a subposet. Thus, the following upper bound on RR(P, Q) for general P and Q is immediate.
From the above inequality, it turns out that determining RR(B m , B n ) is a more fundamental task. In the following, we first give an upper bound on RR(B m , B n ) using the ordinary Boolean Ramsey numbers R k (B m ), and then present the exact value of RR(B m , B n ) for m = 1 or n = 1.
. Consider a coloring c on B N . We assume that B N does not contain a monochromatic B m under c, and then prove there exists a rainbow B n .
Let us arrange the nonempty subsets of [n] in the order I 1 , . . . , whose color is different from c(Y 0 ). We pick such a set and call it Y 1 . Similarly, we can find a set in [Y 0 , X 2 ] whose color is different from c(Y 0 ) and c(Y 1 ), otherwise the sets in [Y 0 , X 2 ] are colored by only two colors, which implies [Y 0 , X 2 ] contains a monochromatic B m . We pick such a set and call it Y 2 . Since Y 1 ⊆ X 1 and Y 2 ⊆ X 2 , we have Y 1 ∩ Y 2 = ∅. Imitating the same operations, we can find pairwise disjoint Y 0 , . . . , Y n , Y i ⊂ X i for 1 i n, of all distinct colors. These sets play the roles of I 0 , I 1 , . . . I n of the rainbow B n . For n + 1 k n + n 2 , we have |I k | = 2. Suppose I n+1 = {i, j}. Then (Bm) . As before, we can find a set whose , and choose it as the set Y n+1 . In general, we can find Y k in [∪ i: if I i and I j are incomparable, then there exist some i ∈ I i \ I j and j ∈ I j \ I i . Thus, , which implies that Y i and Y j are incomparable as well. As a conclusion, these sets Y 0 , . . . , Y 2 n −1 form a rainbow B n in B N .
Remark. By Theorem 3 (Theorem 5, [2]), R k (P ) C · k, where C is a constant determined by P only. Thus, Theorem 14 is a proof of the existence of RR(P, Q) without using Theorem 5. We will have more discussions on this aspect in next section. The next is RR(B 1 , B n ) = 2 n − 1 for n 1. For N < 2 n − 1, we can color all subsets in the level [N ] i with i for 1 i N . Then every color class is an antichain. Moreover, to have a rainbow B n , we need at least 2 n different colors, which is greater than the number of colors in the coloring. Therefore, no rainbow B n under this coloring, and we have RR(B 1 , B n ) 2 n − 1.
By Theorem 14, RR(B 1 , B n ) 2 n−1 (2 n − 1), which is far from the lower bound 2 n − 1 in the last paragraph. We will use an idea analogous to the proof of Theorem 14 to obtain the desired upper bound. For N 2 n − 1, let S be the union of two disjoint sets {x 1 , . . . , x n } and [N − n]. Since B S is isomorphic to B N , we can consider colorings of B S . Assume B S does not contain a monochromatic B 1 under a given coloring c. Then we will prove that it contains a rainbow B n under c. Our strategy is to inspect the colors of the subsets of {x 1 , . . . , x n } one by one. If all the subsets of {x 1 , . . . , x n } are of different colors, then we are done. Once the color of a set X ⊂ {x 1 , . . . , x n } already appeared before, then we look for a new color in a chain which contains X as the minimal element.
Let us arrange all subsets of {x 1 , . . . , x n } in the order X 0 , X 1 , . . . , X 2 n −1 such that |X k | |X k | if k < k. First define Y 0 = X 0 = ∅ and m 0 = 0. For k 1, let us define m k N − n the smallest integer such that m k m k for any X k ⊂ X k and c(X k ∪ [m k ]) is different from any color in {c(Y 0 ), c(Y 1 ), . . . , c(Y k−1 )}. Once such an integer exists, then define Y k = X k ∪ [m k ]. By the definition of Y k and m k , Y k ⊂ Y k if and only if X k ⊂ X k . Hence the sets Y 0 , Y 1 . . . , Y 2 n −1 form a rainbow B n .
Claim: For 0 k 2 n − 1, m k exists. Suppose, on the contrary, some m k does not exist. Let k 0 be the smallest k such that m k does not exist. For any k < k 0 , we call a chain C k in B S a principal chain of Y k if it consists of |Y k | + 1 sets and the largest one is Y k such that all the colors of the sets in C k are in {c(Y 0 ), c(Y 1 ), . . . , c(Y k )}. We show the principal chain exists for all Y k with k < k 0 . It is clear C 0 = {∅} is the principal chain of Y 0 . For Y k with k 1, let k * = max{k | X k ⊂ X k }. By induction, the principal chain C k * of Y k * exists, and all the colors of the sets in C k * are in {c(Y 0 ), c(Y 1 ), . . . , c(Y k * )}. Moreover, by the definition of m k , the colors c( sets. This is the principal chain of Y k . Now since m k 0 does not exist, it means the colors c( In other words, the number of the colors of the subsets in the Note that any chain under the coloring c is a rainbow chain since we assume there is no monochromatic B 1 . So we have |Y k * * | + 1 + (N − n − m k * * + 1) k 0 . Since |Y k * * | = |X k * * | + m k * * , we have N n + k 0 − |X k * * | − 2. By the choice of k * * , we also have |X k * * | = |X k 0 | − 1. In addition, k 0 |X k 0 | i=1 n i according to the ordering of the X i 's. Then This contradicts our assumption of N . As a conclusion, we have RR(B 1 , B n ) 2 n −1.
The idea of the principal chains in the proof of Theorem 15 can be applied to determine the Boolean rainbow Ramsey number for mixed types of posets below: Proof. For N < (m−1)(2 n −1), we give a coloring of B N by assigning color i to the sets in the consecutive m−1 levels On the one hand, since a color class contains at most m − 1 subsets of different sizes, B N does not contain a monochromatic C m . On the other hand, the number of colors is at most 2 n − 1. Hence it does not contain a rainbow B n . Now for N (m − 1)(2 n − 1) we define B S = {x 1 , x 2 , . . . , x n } ∪ [N − n] as in the proof of Theorem 15, and study the colorings of B S . For a coloring c of B S , we prove that once there is no monochromatic C n , then there is a rainbow B n in B S under c. Again, we arrange all subsets of {x 1 , x 2 , . . . , x n } as X 0 , . . . , X 2 n −1 with |X k | |X k | if and only if k k. Let Y 0 = X 0 = ∅ and m 0 = 0. Then for all 1 k 2 n − 1, define Y k = X k ∪ [m k ], where m k N − n is the least integer such that m k m k whenever X k ⊂ X k and c(X k ∪ [m k ]) is a color different from c(Y 0 ), . . . , c(Y k−1 ). Now it suffices to show all Y k 's exist as before. If some Y k does not exist, and Y k exists for all k < k, then we build up the principal chains C k of Y k with the properties: |C k | = |Y k | + 1, the maximal set is Y k , and the color of each set in C Y k is in {c(Y 0 ), c(Y 1 ), . . . , c(Y k )}. The only difference is now the principal chain is not necessarily rainbow. Nevertheless, the number of subsets of the same color in each C k is at most m − 1. Thus, if Y k does not exist, then the chain This is again a contradiction. So B S must contain a rainbow B n under c.

Discussions
We gave an upper bound for RR(A m , C n ) in Theorem 11. A lower bound can be deduced in the following way: Recall that N m,n is the minimum integer N such that N N/2 (m− 1)(n−1)+1. Now consider the set S that is a union of disjoint sets X = {x 1 , . . . , x N m,2 −1 } and Y = {y 1 , . . . , y n−2 }. For each subset Z ⊆ S, color it according to Z ∩ Y . That is, we color Z 1 and Z 2 with the same color if and only if Z 1 ∩ Y = Z 2 ∩ Y . Thus, each color class is isomorphic to a Boolean poset B N m,2 −1 which does not contain A m . On the other hand, if Z 1 , . . . , Z k form a rainbow chain, then Z 1 ∩ Y, . . . , Z k ∩ Y also form a rainbow chain. However, any rainbow chain formed by subsets of Y has length at most n − 1. So there does not exist a rainbow C n . Therefore, we have RR(A m , C n ) n − 2 + N m,2 . Combing the upper bound in Theorem 11, we have n − 2 + N m,2 RR(A m , C n ) (m − 1)(n − 2) + N m,2 .
Obviously, there is a huge gap between the lower and upper bounds. It would be interesting if one can tighten the gap.
One of our main interests is to estimate RR(B m , B n ). Theorem 5 provides us an upper bound RR(B m , B n ) n2 (2n+1)2 m−1 −2 . We also have RR(B m , B n ) . In the following, we give an estimation of 2 n −1 i=1 R i (B m ) in terms of m and n. Recall that in the Remark after Theorem 14, we mentioned that R k (P ) C · k, where C is a constant determined by P . Indeed, this is a consequence from a result of Méroueh in [16]. Let us introduce the Lubell functions and forbidden subposet problems studied intensively recently. The Lubell functionh n (F) for a family F of subsets of [n] is defined as It is obvious thath n (B n ) = n+1. For the background of the Lubell functions and forbidden subposet problems, we refer the readers to a recent survey [11]. In the literature, Lu and Milans [14] proposed the conjecture: Conjecture 17. [14] Let λ n (P ) be the maximum value of the Lubell function for families of subsets of [n] not containing P as a subposet. Then lim sup λ n (P ) is finite.
Méroueh [16] verified this conjecture by showing that ifh n (F) > 1000m 7 16 m , then the family F must contain B m , as well as any poset whose 2-dimension is at most m. Thus, when N 1000m 7 16 m k, any k-coloring of B N must contain one color class F i of subsets of color i withh(F i ) > 1000m 7 16 m , and hence a monochromatic B m of color i. In other words, R k (B m ) 1000m 7 16 m k. Using his result, we conclude that This upper bound is a little better than that derived from Theorem 5 under certain circumstances. It will be a challenging problem to determine or give a better estimation of RR(B m , B n ) for both m 2 and n 2.