aa r X i v : . [ m a t h . L O ] A p r DISJOINT TYPE GRAPHS WITH NO SHORT ODD CYCLES
CHRIS LAMBIE-HANSON
Abstract.
In this note, we provide a proof of a technical result of Erd˝os andHajnal about the existence of disjoint type graphs with no odd cycles. We alsoprove that this result is sharp in a certain sense.
The purpose of this note is to provide a proof of a result of Erd¨os and Hajnalabout the existence of disjoint type graphs with no short odd cycles. As far as weknow, a proof of this result has never been published, though forms of it are statedin a number of publications (cf. [2, Theorem 7.4] and [3, Lemma 1.1(d)]). If κ is anuncountable cardinal, then graphs of this form provide, again as far as we know,the only known ZFC examples of graphs with size and chromatic number κ andarbitrarily high odd girth.Before we state and prove the main result, we need some definitions and conven-tions. First, if n is a positive integer, we will sometimes think of elements of [Ord] n as strictly increasing sequences of length n . So, for instance, if a ∈ [Ord] n and i < n , then a ( i ) is the unique element α ∈ a such that | a ∩ α | = i . All graphs con-sidered here will be simple undirected graphs. If G is a graph, then V ( G ) denotesits vertex set and E ( G ) denotes its edge set. Definition 1.
Let n be a positive integer. A disjoint type of width n is a function t : 2 n → | t − (0) | = | t − (1) | = n. If a, b ∈ [Ord] n are disjoint and a ∪ b is enumerated in increasing order as { α i | i < n } , then we say that the type of a and b is t , denoted tp( a, b ) = t , if a = { α i | i ∈ t − (0) } and b = { α i | i ∈ t − (1) } . Let ˆ t denote the disjoint type of width n denoted by letting ˆ t ( i ) = 1 − t ( i ) forall i < n . It is evident that, if a, b ∈ [Ord] n are disjoint and tp( a, b ) = t , thentp( b, a ) = ˆ t .A type t of width n will sometimes be represented by a binary string of length2 n in the obvious way. We will particularly be interested in the following family oftypes. Definition 2.
Let 1 ≤ s < n < ω . Then t ns is the disjoint type of width n whosebinary sequence representation consists of s copies of ‘0’, followed by n − s copiesof ‘01’, followed by s copies of ‘1’. More formally, t ns is defined by letting, for all Date : April 10, 2020. i < n , t ns ( i ) = i < s s ≤ i < n − s and i − s is even1 if s ≤ i < n − s and i − s is odd1 if i ≥ n − s. For example, t = 0001010111. Definition 3.
Suppose that n is a positive integer, β is an ordinal, and t is adisjoint type of width n . The graph G ( β, t ) is defined as follows. Its vertex set is V ( G ( β, t )) = [ β ] n . Given a, b ∈ [ β ] n , we put the edge { a, b } into E ( G ( β, t )) if andonly if a and b are disjoint and tp( a, b ) ∈ { t, ˆ t } .Before we get to our main result, we need a basic lemma. Given a function f from a natural number to Z , let max( f ) and min( f ) denote the maximum andminimum values attained by f , respectively. Lemma 4.
Suppose that k is a positive integer and f : k → Z is a function suchthat • f (0) = 0 and • | f ( i + 1) − f ( i ) | = 1 for all i < k .Then max( f ) − min( f ) < k .Proof. The proof is by induction on k . If k = 1, then max( f ) = min( f ) = f (0) = 0.Suppose that k > k −
1. Fix f : k → Z , andlet f − = f ↾ ( k − f ( k ) − f ( k −
1) = 1, then we have max( f ) ≤ max ( f − ) + 1and min( f ) = min( f − ), so, applying the induction hypothesis to f − , we obtainmax( f ) − min( f ) ≤ (cid:0) max( f − ) − min( f − ) (cid:1) < k −
1) = k. If f ( k ) − f ( k −
1) = −
1, then we have max( f ) = max( f − ) and min( f ) ≥ min( f − ) − f ) − min( f ) ≤ (cid:0) max( f ) − min( f − ) (cid:1) + 1 < ( k −
1) + 1 = k. (cid:3) We are now ready for the main result of this note. The proof is rather technical;we recommend that the reader first draw some pictures to convince themselves ofthe truth of the theorem in the special case s = 1, n = 3 (this pair does not satisfy n > s + 3 s + 1, but the conclusion of the theorem still holds). This will help thereader to get a feel for the problem and motivate the calculations in the proof. Wealso note that the lower bound of 2 s + 3 s + 1 is probably not optimal and canlikely be improved with a more careful analysis. Since a precise lower bound for n is not necessary for our desired applications (cf. [5]), the primary interest of theresult for us is the fact that such a lower bound exists at all. Theorem 5.
Suppose that s and n are positive integers with n > s + 3 s + 1 , andsuppose that β is an ordinal. Then the graph G ( β, t ns ) has no odd cycles of length s + 1 or shorter.Proof. Let t = t ns , V = [ β ] n , E = E ( G ( β, t )), and G = G ( β, t ) = ( V, E ). We beginby making some preliminary observations. If { a, b } ∈ E , then either tp( a, b ) = t ortp( a, b ) = ˆ t . If tp( a, b ) = t , then, for all i with s < i < n , we have b ( i − s − < a ( i ) < b ( i − s ) . ISJOINT TYPE GRAPHS WITH NO SHORT ODD CYCLES 3
If tp( a, b ) = ˆ t , then, for all i < n − s −
1, we have b ( i + s ) < a ( i ) < b ( i + s + 1) . Suppose that k is a positive integer and P = h a , . . . , a k i is a path of length k in G . For j ≤ k , let U j ( P ) = (cid:8) i < j | tp( a i , a i +1 ) = t (cid:9) , and D j ( P ) = (cid:8) i < j | tp( a i , a i +1 ) = ˆ t (cid:9) . Intuitively, U j ( P ) is the set of steps “up” in the path among the first j steps, and D ( P ) is the set of steps “down” among the first j steps. Then set u j ( P ) = | U j ( P ) | and d j ( p ) = | D j ( P ) | ; note that u j ( P ) + d j ( P ) = j for all j ≤ k . Claim 6.
Suppose that ≤ k ≤ s + 1 and P = h a , . . . , a k i is a path in G . Let u = u k ( P ) and d = d k ( P ) . Then there is i < n such that a k (cid:0) i − u ( s + 1) + ds (cid:1) < a (cid:0) i (cid:1) < a k (cid:0) i − us + d ( s + 1) (cid:1) . Remark 7.
Implicit in the statement of the claim is the assertion that0 ≤ i − u ( s + 1) + ds < i − us + d ( s + 1) < n, the truth of which will follow readily from the proof. Proof of Claim 6.
Define a function f : k + 1 → Z by letting f ( j ) = u j ( P ) − d j ( P )for every j ≤ k . Then f satisfies the hypotheses of Lemma 4, so, letting M =max( f ) and m = min( f ), we have M − m ≤ k ≤ s + 1.Let i = M ( s + 1). Note that M ( s + 1) ≤ (2 s + 1)( s + 1) = 2 s + 3 s + 1 , so we certainly have i < n . Subclaim 8.
For every < j ≤ k , we have a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) < a (cid:0) i (cid:1) < a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) . Remark 9.
Implicit in the statement of this subclaim is the assertion that, foreach 0 < j ≤ k , we have0 ≤ i − sf ( j ) − u j ( P ) < i − sf ( j ) + d j ( P ) < n. This will follow readily from the proof.
Proof of Subclaim 8.
We proceed by induction on j . We begin by proving thesubclaim for j = 1. Suppose first that tp( a , a ) = t , so f (1) = 1, u ( P ) = 1,and d ( P ) = 0. Then M ≥
1, so i ≥ s + 1. Therefore, since tp( a , a ) = t , thepreliminary observations at the beginning of the proof of the theorem imply that a ( i − s − < a ( i ) < a ( i − s ) , as desired.If, on the other hand, tp( a , a ) = ˆ t , and hence f (1) = − u ( P ) = 0, and d ( P ) = 1, then m ≤ −
1. Therefore, we have M ≤ s , so i = M ( s + 1) ≤ s + 2 s < n − s −
1. Therefore, since tp( a , a ) = ˆ t , the preliminary observationsat the beginning of the proof imply that a ( i + s ) < a ( i ) < a ( i + s + 1) , as desired. CHRIS LAMBIE-HANSON
Now suppose that 0 < j < k and we have established that a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) < a (cid:0) i (cid:1) < a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) . We will prove the corresponding statement for j + 1. Suppose to begin thattp( a j , a j +1 ) = t , so f ( j + 1) = f ( j )+ 1, u j +1 ( P ) = u j ( P )+ 1, and d j +1 ( P ) = d j ( P ).In this case, it follows that f ( j ) ≤ ( M −
1) and u j ( P ) ≤ ( M − i − sf ( j ) − u j ( P ) ≥ M ( s + 1) − s ( M − − ( M −
1) = s + 1 > s. Therefore, by the preliminary observations, we have a j +1 (cid:0) i − sf ( j ) − u j ( P ) − s − (cid:1) < a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) a j +1 (cid:0) i − sf ( j + 1) − u j +1 ( P ) (cid:1) < a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) and a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) < a j +1 (cid:0) i − sf ( j ) + d j ( P ) − s (cid:1) a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) < a j +1 (cid:0) i − sf ( j + 1) + d j +1 ( P ) (cid:1) . Combining these inequalities with the inductive hypothesis yields a j +1 (cid:0) i − sf ( j + 1) − u j +1 ( P ) (cid:1) < a ( i ) < a j +1 (cid:0) i − sf ( j + 1) + d j +1 ( P ) (cid:1) , as desired.On the other hand, suppose that tp( a j , a j +1 ) = ˆ t , so f ( j + 1) = f ( j ) − u j +1 ( P ) = u j ( P ), and d j +1 ( P ) = d j ( P ) + 1. In this case, it follows that f ( j ) ≥ ( m + 1) and d j ( P ) ≤ − ( m + 1). In particular, we have i − sf ( j ) + d j ( P ) ≤ i − s ( m + 1) − ( m + 1) = i − ( m + 1)( s + 1) . We know that M − m ≤ s + 1, so m + 1 ≥ M − s . As a result, the above inequalitybecomes i − sf ( j ) + d j ( P ) ≤ M ( s + 1) − ( M − s )( s + 1) = 2 s + 2 s < n − s − . Therefore, by the preliminary observations, we have a j +1 (cid:0) i − sf ( j ) − u j ( P ) + s (cid:1) < a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) a j +1 (cid:0) i − sf ( j + 1) − u j +1 ( P ) (cid:1) < a j (cid:0) i − sf ( j ) − u j ( P ) (cid:1) and a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) < a j +1 (cid:0) i − sf ( j ) + d j ( P ) + s + 1 (cid:1) a j (cid:0) i − sf ( j ) + d j ( P ) (cid:1) < a j +1 (cid:0) i − sf ( j + 1) + d j +1 ( P ) (cid:1) . Combining these inequalities with the inductive hypothesis yields a j +1 (cid:0) i − sf ( j + 1) − u j +1 ( P ) (cid:1) < a (cid:0) i (cid:1) < a j +1 (cid:0) i − sf ( j + 1) + d j +1 ( P ) (cid:1) , as desired, finishing the proof of the subclaim. (cid:3) Since f ( k ) = u k ( P ) − d k ( P ), we have i − u ( s + 1) + ds = i − sf ( k ) − u k ( P ) and i − us + d ( s + 1) = i − sf ( k ) + d k ( P ) . Therefore, the claim follows immediately from Subclaim 8. (cid:3)
ISJOINT TYPE GRAPHS WITH NO SHORT ODD CYCLES 5
Now suppose for sake of contradiction that G has an odd cycle of length 2 s + 1or shorter. In other words, there is a positive integer k ≤ s and a path C = h a , . . . , a k +1 i with a = a k +1 . Let u = u k ( C ) and d = d k ( C ). Note that u + d = 2 k + 1. Apply Claim 6 to find i < n such that a k +1 (cid:0) i − u ( s + 1) + ds (cid:1) < a ( i ) < a k +1 (cid:0) i − us + d ( s + 1) (cid:1) . Since a = a k +1 , this reduces to i − u ( s + 1) + ds < i < i − us + d ( s + 1) . Cancelling i from all three terms yields ds − u ( s + 1) < < d ( s + 1) − us. Since d and u are both non-negative integers, this implies that they are bothnonzero. Therefore, the left inequality gives us du < s + 1 s and the right inequality gives us ss + 1 < du , so we have ss + 1 < du < s + 1 s . In particular, du is close to 1. But we know that d + u = 2 k + 1; the assignments ofvalues to d and u subject to this constraint that put du closest to 1 are either d = k and u = k + 1 or vice versa. But k ≤ s , so, if d = k and u = k + 1, then du ≤ ss + 1and, if d = k + 1 and u = k , then du ≥ s + 1 s . Either possibility gives us a contradiction, so we are done. (cid:3)
We end this note by making a few further observations about these disjoint typegraphs. We first point out a minor error in the literature. In [1, Remark 1], theauthors write, using slightly different terminology, that, for any positive integer n ≥
3, the graph G ( β, t n ) has no odd cycles of length less than 2 ⌈ n/ ⌉ . This istrue for n = 3 but false for every larger value of n ; G ( β, t n ) always has a cycle oflength 5, as long as β is large enough to allow room for the cycle. In fact, we havethe following general result, showing that Theorem 5 is sharp in a sense. Proposition 10.
Suppose that < s < n < ω and β > ( n − s + 3) + (2 s + 1)(2 s + 2) . Then the graph G ( β, t ns ) has a cycle of length s + 3 . CHRIS LAMBIE-HANSON
Proof.
Let m = 2 s + 3. We will define a path h a , a , . . . , a m i in G ( β, t ns ) with a m = a . First define a = a m by letting a m ( i ) = im for all i < n . The definitionof each of the remaining elements of the cycle depends on the parity of its index.For j with 0 < j ≤ s + 1, define a j − by setting a j − ( i ) = ( i + s + j ) m − (2 j − i < n , and define a j by setting a j ( i ) = ( i + j ) m − j for all i < n . The following facts are easily verified and left to the reader. • For all j ≤ s , tp( a j , a j +1 ) = t ns . • For all j ≤ s , tp( a j +1 , a j +2 ) = ˆ t ns . • tp( a s +2 , a m ) = ˆ t ns . • The largest element of any of the vertices in the cycle is a s +1 ( n −
1) = ( n − s + 3) + (2 s + 1)(2 s + 2) . Therefore, h a , a , . . . , a m i forms a cycle of length 2 s + 3 in G ( β, t ns ). (cid:3) We conclude by noting the following result, which is one of the primary reasonsfor interest in disjoint type graphs. The result is due to Erd˝os and Hajnal [2]; thespecial case t = t is due to Erd˝os and Rado [4]. A proof of the full result can befound in [1, Theorem 2.1]. Theorem 11.
Suppose that n is a positive integer and t is a disjoint type of width n . For every infinite cardinal κ , the graph G ( κ, t ) has chromatic number κ . References [1] Christian Avart, Tomasz Luczak, and Vojtˇech R¨odl. On generalized shift graphs.
Fund. Math. ,226(2):173–199, 2014.[2] P. Erd˝os and A. Hajnal. On chromatic number of graphs and set-systems.
Acta Math. Acad.Sci. Hungar , 17:61–99, 1966.[3] P. Erd˝os, A. Hajnal, and E. Szemer´edi. On almost bipartite large chromatic graphs. In
Theoryand practice of combinatorics , volume 60 of
North-Holland Math. Stud. , pages 117–123. North-Holland, Amsterdam, 1982.[4] P. Erd˝os and R. Rado. A construction of graphs without triangles having pre-assigned orderand chromatic number.
J. Lond. Math. Soc. , s1-35(4):445–448, 1960.[5] Chris Lambie-Hanson. On the growth rate of chromatic numbers of finite subgraphs.
Adv.Math. , 2020. To appear.
Department of Mathematics and Applied Mathematics, Virginia Commonwealth Uni-versity, Richmond, VA 23284, United States
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