aa r X i v : . [ m a t h . C O ] J a n A Construction for Cube Ramsey
Tom Bohman ∗ Fei PengFebruary 2, 2021
Abstract
The (poset) cube Ramsey number R ( Q n , Q n ) is defined as theleast m such that any 2-coloring of the m -dimensional cube Q m admitsa monochromatic copy of Q n . The trivial lower bound R ( Q n , Q n ) ≥ n was improved by Cox and Stolee, who showed R ( Q n , Q n ) ≥ n + 1 for ≤ n ≤ and n ≥ using a probabilistic existence proof.In this paper, we provide an explicit construction that establishes R ( Q n , Q n ) ≥ n + 1 for all n ≥ . A central theme of combinatorics is the fact that large discrete systems oftencontain subsystems with a higher degree of organization than the original sys-tem. Results of this flavor appear in many areas. One example is the Erdős-Szekeres theorem, which states that every sequence of ab + 1 real numbershas a monotonously increasing subsequence of length a or a monotonouslydecreasing subsequence of length b . Another example is the Erdős-SzekeresConjecture, which asks for the minimum number n such that any n pointsin general position in the plane contain k points in convex position [Suk17].Graph Ramsey theory [GRS90] studies the graph-theoretic analog of thistheme. Ramsey’s Theorem for graphs states that for all k , there exists n such that any edge 2-coloring of a clique (complete graph) of size n contains ∗ This work was supported by a grant from the Simons Foundation (587088, TB) k . The least such n is denoted R ( k, k ) . Sig-nificant work has been done (see [Rad12], [CFS15]) to determine or boundthe Ramsey numbers for graphs.In this paper, we will focus on the Ramsey problem for poset cubes. Apartially ordered set, or a poset , is a set equipped with a partial relation ≤ that is reflexive, anti-symmetric and transitive. The cube Q n , which is thepower set of [ n ] equipped with the inclusion relation, plays an important rolein the theory of posets, and we consider the natural Ramsey question in thiscontext. We use the power set [ n ] itself to refer to the associated poset whenit is unambiguous. A poset embedding is an order-preserving (and nonorder-preserving) injection f from a poset P to another poset P ′ : that is, x ≤ P y ⇐⇒ f ( x ) ≤ P ′ f ( y ) . In the context of cubes, a function f : 2 [ n ] → [ m ] isan embedding if it is injective and satisfies S ⊆ T ⇐⇒ f ( S ) ⊆ f ( T ) . Inthis case, we say that the range of f – which is a subset of [ m ] – is a copy of [ n ] .This paper focuses on the following Ramsey theoretic quantity introducedby Axenovich and Walzer [AW17]. Given n ∈ N , define the (poset) cubeRamsey number R ( Q n , Q n ) to be the least integer m such that every 2-coloring of the elements of Q m admits a monochromatic copy of Q n . Note thatthe individual subsets of [ m ] are colored, instead of the inclusion relationsbetween them; in other words, we consider a vertex-coloring instead of anedge-coloring.It is not hard to see that R ( Q n , Q n ) ≥ n : that is, one can 2-colorthe cube Q n − without any monochromatic copy of Q n . The method is tosimply color all odd-sized subsets red and even-sized subsets blue. In fact,if we group subsets by their cardinality, we would split Q n − into n layers ,and any division of the layers into n totally-red layers and n totally-bluelayers will suffice.The best lower bound on R ( Q n , Q n ) to date is slightly better than n .Cox and Stolee [CS18] established that R ( Q n , Q n ) ≥ n + 1 for ≤ n ≤ and n ≥ . Their argument for n ≥ relies on a family of special sets,which is shown to exist using the Lovász Local Lemma, and there is no knownexplicit construction of this family. Here, we exhibit an explicit 2-coloring of Q n without a monochromatic copy of Q n for all n ≥ which, when combinedwith the previously known construction for n = 3 [AW17], gives: This can be generalized to different size requirements for colors, more than 2 colors,and/or hypergraphs. heorem 1.1. ∀ n ≥ , R ( Q n , Q n ) ≥ n + 1 . The best known upper bound of the cube Ramsey number is R ( Q n , Q n ) ≤ n − n + 2 established in [LT], and a recent paper [FRMTZ20] contains someresults regarding cube Ramsey numbers of a random poset, and also obtains R ( Q , Q ) = 7 .We mention in passing the closely-related poset variant of the classicalTurán problem, which asks how many elements we can choose from [ n ] with-out seeing a copy of [ k ] . This is unsolved even when k = 2 [AMM12] (thatis, picking elements to not see a “diamond”), and many believe that the an-swer is (cid:0) n ⌊ n/ ⌋ (cid:1) + (cid:0) n ⌊ ( n − / ⌋ (cid:1) , witnessed by taking the two layers closest to themiddle. The best known upper bound is (( √ / o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) [GMT18]. We exhibit a 2-coloring of [2 n ] admitting no monochromatic copy of [ n ] . Todefine the coloring, we first group [2 n ] into n pairs: Definition 2.1. a, b ∈ [2 n ] are said to be in the same pair if ⌈ a/ ⌉ = ⌈ b/ ⌉ .So the pairs are { , } , { , } . . . . We say a set S has a pair if S containsthat pair, and S misses the pair if S contains neither element of the pair.The partner of a ∈ [2 n ] is the unique b such that { a, b } is a pair. An element x ∈ S is a single in S if the partner of x is not an element of S .We sometimes characterize a set by its number of pairs and singles: forexample, { , , , , , , } is said to have 3 pairs and 1 single. Definition 2.2.
A collection
R ⊆ [2 n ] is pair-enforcing if for every S ⊆ [2 n ] s.t. ⌈ n/ ⌉ ≤ | S | < n , S ∈ R only if S has a pair. Definition 2.3.
A collection
R ⊆ [2 n ] is miss-forbiding if for every S ⊆ [2 n ] s.t. n < | S | ≤ n + ⌊ n/ ⌋ , S ∈ R only if S doesn’t miss any pair. Definition 2.4.
A collection
R ⊆ [2 n ] is not-too-high if for every S ∈ R wehave | S | ≤ n + ⌊ n/ ⌋ . Definition 2.5.
A collection
R ⊆ [2 n ] is flip-susceptible if for all S , S ⊆ [2 n ] such that | S | = | S | = n , | S ∪ S | = n + 1 and neither S nor S hasany pair, at most one of S and S is in R .3t is named as such because S and S would be all the same except fora choice-flip about one specific pair. Definition 2.6.
A collection
R ⊆ [2 n ] is restrictive if it is pair-enforcing,miss-forbiding, not-too-high and flip-susceptible.Now we are ready to present the construction. Definition 2.7.
Define the coloring c : 2 [2 n ] → { R, B } as follows: ∀ S ⊆ [2 n ] , • If | S | < ⌈ n/ ⌉ , S is red.• If ⌈ n/ ⌉ ≤ | S | < n , S is red iff S has a pair.• If | S | = n , S is red iff the sum of its elements is odd.• If n < | S | ≤ n + ⌊ n/ ⌋ , S is red iff S doesn’t miss any pair.• If | S | > n + ⌊ n/ ⌋ , S is blue.Let R be the collection of subsets of [2 n ] that are colored red by c . Notethat R is restrictive. Note further that the collection of sets of the form ¯ S such that S is colored blue by c is also restrictive. Note further that if Q ⊂ [2 n ] is a copy of [ n ] then the collection { ¯ S : S ∈ Q} is also a copyof [ n ] . So, as [AW17] gives a construction that establishes R ( Q , Q ) ≥ ,Theorem 1.1 follows from the following result: Theorem 2.8. If n ≥ , a restrictive collection R ⊆ [2 n ] does not containa copy of [ n ] . We prove this in the next Section.
We begin with two preliminary observations about poset embeddings. Forthe next few results and notations, we fix a poset embedding f : 2 X → Y ,where X and Y are finite sets. Lemma 3.1. If A ⊆ B ⊆ X then | f ( B ) | − | f ( A ) | ≥ | B | − | A | . In particular(setting A = ∅ ), | f ( B ) | ≥ | B | . roof. Say | B | − | A | = k . Then there exists a chain A = S ( S ( · · · ( S k = B . Since f is an embedding, f ( A ) = f ( S ) ( f ( S ) ( · · · ( f ( S k ) = f ( B ) . Hence, | f ( B ) | − | f ( A ) | ≥ k . Definition 3.2.
With respect to the poset embedding f : 2 X → Y , the topelement refers to f ( X ) , and the top children refers to all immediate childrenof the top element, namely all sets of the form f ( X \ { a } ) ( a ∈ X ). Lemma 3.3.
If the top element has cardinality N , then the intersection ofany k top children ( ≤ k ≤ | X | ) has cardinality at most N − k ; that is, ∀ nonempty I ⊆ X , (cid:12)(cid:12)(cid:12)(cid:12) \ a ∈ I f ( X \ { a } ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ N − | I | . Proof.
First note that every top child has size at most N − , so the lemmais true when k = 1 . Then, ∀ nonempty I ⊆ X, ∀ j ∈ X \ I , \ a ∈ I f ( X \ { a } ) ) \ a ∈ I ∪{ j } f ( X \ { a } ) because f ( { j } ) is a subset of the former but not the latter. Thus, when I grows from a singleton to a k -element set, the size of the intersection decreasesto at most N − k .With these observations in hand, we are ready to consider restrictivecollections. Lemma 3.4.
Let R be a restrictive collection. For all S ∈ R , there is some S + ⊇ S, | S + | = n + ⌊ n/ ⌋ and some restrictive collection R + ⊇ R such that S + ∈ R + .Proof. Let S ∈ R and asssume for the sake of contradiction that S has morethan ⌊ n/ ⌋ pairs. Then | S | > n . Since R is miss-forbiding and not-too-high, S ∈ R only if S doesn’t miss any pair. Note that a set that misses no pairand has k pairs has cardinality exactly n + k , so we have | S | > n + ⌊ n/ ⌋ ,witnessing S / ∈ R . Hence, the assumption is false.As S has no more than ⌊ n/ ⌋ pairs, by adding elements as needed, onecan find S + ⊇ S that has exactly ⌊ n/ ⌋ pairs and doesn’t miss any pair.Such S + will have size n + ⌊ n/ ⌋ . The collection R + = R ∪ { S + } wouldcontain S + while continuing to be restrictive.5 efinition 3.5. A copy of [ n ] is maximal if its top element has size n + ⌊ n/ ⌋ . Corollary 3.6.
If a restrictive collection R contains a copy of [ n ] , thensome restrictive collection R + ⊇ R contains a maximal copy of [ n ] . By Corollary 3.6, it suffices to show that there does not exist a restrictivecollection that contains a maximal copy of [ n ] . In the remainder of this Section, we fix a restrictive collection
R ⊆ [2 n ] and assume for the sake of contradiction that R contains a maximal copy of [ n ] . Let this maximal copy be the image of the embedding f : 2 [ n ] → [2 n ] .By definition, the top element of this maximal copy has size n + ⌊ n/ ⌋ .By the miss-forbiding property, this element cannot miss any pair, so ithas exactly ⌊ n/ ⌋ pairs and n − ⌊ n/ ⌋ = ⌈ n/ ⌉ singles. Denote the pairsas p , . . . , p ⌊ n/ ⌋ and the singles s , . . . , s ⌈ n/ ⌉ . For ease of notation we let Π denote the collection of pairs p , . . . , p ⌊ n/ ⌋ , and we let Σ denote the collectionof singles s , . . . , s ⌈ n/ ⌉ . Claim 3.7.
In the maximal copy, if a top child misses a single in Σ then thetop child has size at most n and has all ⌊ n/ ⌋ pairs in Π . Proof.
Consider such a top child S that misses a single from Σ . This topchild misses the pair that single lies in, because even the top element doesn’thave the partner. Since R is miss-forbiding and not-too-high, | S | ≤ n .As S is a top child, we can write S = f ([ n ] \ { a } ) for some a ∈ [ n ] . Setsin the copy that are below or equal to S naturally form a copy of [ n ] \{ a } ,with S being the top element. There are n − top children in the smallercube; by Lemma 3.3, their intersection has size at most n − ( n −
1) = 1 .Suppose S has at most ⌊ n/ ⌋ − of the pairs in Π . For each pair it has,one of its immediate children doesn’t have it – this is because at most oneelement of [2 n ] is in the intersection of all immediate children of S . Thus,there is a collection of at most ⌊ n/ ⌋ − immediate children of S such thatfor every pair in Π that S has, one immediate child in the collection does nothave that pair.These immediate children are of the form f ([ n ] \ { a, b i } ) , where i rangesfrom 1 to k , and k ≤ ⌊ n/ ⌋ − . Consider the set S = f ([ n ] \ { a, b , . . . , b k } ) .By Lemma 3.1, its size is at least | [ n ] \ { a, b , . . . , b k }| = n − ( k + 1) ≥ ⌈ n/ ⌉ ,and since it is below S , its size is smaller than n . But on the other hand, S is a subset of every f ([ n ] \ { a, b i } ) , so it does not have any pair. As R is pair-enforcing, S cannot be in R . But S is in the copy. To resolve thecontradiction, S must have all the ⌊ n/ ⌋ pairs.6 laim 3.8. The maximal copy has at most ⌈ n/ ⌉ − ⌊ n/ ⌋ + 1 top children(i.e., one if n is even and two if odd) which miss at least one single in Σ . Proof. If n is even, this is clear: by Claim 3.7, such a top child has size atmost n but has all ⌊ n/ ⌋ = n/ pairs in Π , so it must consist of exactly thepairs in Π and nothing else.If n is odd, the pairs are determined for the same reason, but there couldbe an extra single in the top child. Assume for the sake of contradiction thatthree such top children exist. Since they are mutually incomparable, they allhave an extra single from Σ . Say the top children are f ([ n ] \ { a } ) = p ⊔ · · · ⊔ p ⌊ n/ ⌋ ⊔ s i ,f ([ n ] \ { b } ) = p ⊔ · · · ⊔ p ⌊ n/ ⌋ ⊔ s j ,f ([ n ] \ { c } ) = p ⊔ · · · ⊔ p ⌊ n/ ⌋ ⊔ s k . Then, f ( { a } ) ⊆ f (([ n ] \ { b } ) ∩ ([ n ] \ { c } )) ⊆ f ([ n ] \ { b } ) ∩ f ([ n ] \ { c } )= p ⊔ · · · ⊔ p ⌊ n/ ⌋ ⊆ f ([ n ] \ { a } ) . This is a contradiction.
Claim 3.9. If n ≥ then there are b , . . . , b ⌊ n/ ⌋ and ˆ b , distinct elements of [ n ] , such that f ([ n ] \ { b i } ) doesn’t have p i for all i ∈ [ ⌊ n/ ⌋ ] , and f ([ n ] \ { ˆ b } ) doesn’t have p j for some j ∈ [ ⌊ n/ ⌋ ] . Proof.
By Claim 3.8, at least n − ( ⌈ n/ ⌉ − ⌊ n/ ⌋ + 1) top children have all thesingles in Σ . Construct a bipartite graph with bipartition Π ⊔ C where C isthe collection of the top children that have all singles in Σ , and p ∈ Π , S ∈ C are connected iff S does not have the pair p .We claim that ∀ P ⊆ Π , | N ( P ) | ≥ | P | . If P consists of x > pairs and | N ( P ) | ≤ x − , then at least n − ( ⌈ n/ ⌉ − ⌊ n/ ⌋ + 1) − ( x − top childrenhave all the x pairs in P . By Lemma 3.3, their intersection has size at most ( n + ⌊ n/ ⌋ ) − ( n − ( ⌈ n/ ⌉−⌊ n/ ⌋ +1) − ( x − ⌈ n/ ⌉ + x . But s , . . . , s ⌈ n/ ⌉ and the x pairs in P are already in the intersection, which implies that theintersection has size at least ⌈ n/ ⌉ + 2 x . Because x > , this is impossible.7y Hall’s Theorem, there is a Π -saturating matching. That means, thereare b , . . . , b ⌊ n/ ⌋ all distinct such that for every i , f ([ n ] \ b i ) doesn’t have p i .pdfFinally, since n − ( ⌈ n/ ⌉ − ⌊ n/ ⌋ + 1) > ⌊ n/ ⌋ when n ≥ , let f ([ n ] \ { ˆ b } ) be an unused top child in C . Since it has all the singles, it cannothave all the pairs (or it would be the top element), so there is some p j it doesnot have.We are now ready to complete the proof of Theorem 2.8. Consider b , . . . , b ⌊ n/ ⌋ and ˆ b from Claim 3.9. Say ˆ b doesn’t have p j and consider thesets S = f ([ n ] \ { b , . . . , b ⌊ n/ ⌋ } )= f ([ n ] \ { b , . . . , b j − , b j , b j +1 , . . . , b ⌊ n/ ⌋ } ) ,S = f ([ n ] \ { b , . . . , b j − , ˆ b, b j +1 , . . . , b ⌊ n/ ⌋ } ) ,S ∨ = f ([ n ] \ { b , . . . , b j − , b j +1 , . . . , b ⌊ n/ ⌋ } ) ,S ∪ = S ∪ S . Note that S , S , S ∨ ∈ R . Because S lies in every f ([ n ] \ { b i } ) , it doesn’thave any pair. Since R is pair-enforcing, to have S ∈ R we must have either | S | < ⌈ n/ ⌉ or | S | ≥ n . Because | S | ≥ | [ n ] \ { b , . . . , b ⌊ n/ ⌋ }| = ⌈ n/ ⌉ , wehave | S | ≥ n . Analogously, | S | ≥ n . Since S = S , | S ∪ | ≥ n + 1 .On the other hand, by Lemma 3.1, | f ([ n ]) | − | S ∨ | ≥ |{ b , . . . , b j − , b j +1 , . . . , b ⌊ n/ ⌋ }| = ⌊ n/ ⌋ − . Thus, | S ∨ | ≤ | f ([ n ]) | − ( ⌊ n/ ⌋ − n + ⌊ n/ ⌋ ) − ( ⌊ n/ ⌋ − n + 1 . Note that S , S ⊆ S ∨ , so S ∪ ⊆ S ∨ . Thus, | S ∪ | = n + 1 and | S | = | S | = n . Note again that neither S nor S has any pair. Because R isflip-susceptible, at most one of S and S is in R . This is a contradiction.Hence, no restrictive collection can contain a maximal copy of [ n ] . In viewof Corollary 3.6, the proof is complete. Remark . The condition n ≥ in Theorem 2.8 is necessary. Indeed, theconstruction we present here does have monochromatic copies of [3] = 2 [ n ] in8he case n = 3 . It turns out that the construction for the n = 3 case given in[AW17] is similar to our construction as the only difference is in the coloringof sets S with | S | = n = 3 . References [AMM12] Maria Axenovich, Jacob Manske, and Ryan Martin. Q -freefamilies in the Boolean lattice. Order , 29(1):177–191, 2012.[AW17] Maria Axenovich and Stefan Walzer. Boolean lattices: Ramseyproperties and embeddings.
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