Infinitely Many Trees Have Non-Sperner Subtree Poset

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1 Order ( : DOI /s Infinitely Many Trees Have Non-Sperner Subtree Poset Andrew Vince Hua Wang Received: 3 April 2007 / Accepted: 25 August 2007 / Published online: 2 October 2007 Springer Science Business Media B.V Abstract Let C(T denote the poset of subtrees of a tree T with respect to the inclusion ordering. Jacobson, Kézdy Seif gave a single exaple of a tree T for which C(T is not Sperner, answering a question posed by Penrice. The authors then ask whether there exist an infinite faily of trees T such that C(T is not Sperner. This paper provides such a faily. Keywords Poset Lattice Sperner property Subtree 1 Introduction A ranked partially order set (poset is called Sperner if soe axiu antichain consists of eleents fro a single rank. The set C(T of subtrees of a tree T is a poset with respect to the inclusion ordering, the rank of a subtree being the cardinality of its vertex set. Moreover, it is shown in [3] that C(T is a lattice. Such lattices appear as early as [4]. Jacobson, Kézdy Seif [2] give a single exaple of a tree T for which C(T is not Sperner, answering in the negative a question posed by Penrice [5]. The authors [2] then ask whether there exists, in fact, infinitely any trees T for which C(T is not Sperner. Such an infinite faily is provided in this paper. The ideas are inspired by [2]; in fact, the sallest eber of our faily is the exaple in that paper as shown in Fig. 1. A. Vince H. Wang (B Departent of Matheatics, University of Florida, Gainesville, FL 32611, USA e-ail: hua@ath.ufl.edu A. Vince e-ail: vince@ath.ufl.edu

2 134 Order ( : Fig. 1 A non-sperner tree on 11 vertices Let T, 3, be the faily of trees shown in Fig. 2. The path fro v 1 to x has s = s( vertices, the path fro v 2 to x has t = t( vertices, where s( = N 0 2 2, t( = N 0 2, { ( k1 ( } N 0 = ax. (1 0 k< k i Note that N 0 ( 1 ( i = 2, which iplies that s( 2 2 t(. Inparticulars( 4 t( 3 since 3, equality is achieved for the tree in Fig. 1. The proof of the following theore appears in the next section. Theore 1 If is not a perfect square, then C(T is not Sperner. There are infinitely any for which is not a perfect square. For exaple, if 0 (od 3, then (od 3. So is not a perfect square if 0 (od 3 because 2 is not even a square (od 3. A siilar arguent shows that is not a perfect square for 0 or 2 (od 5 or for 2, 3 or 4 (od 7, etc.thevalues = 14 = 103 are the first two for which is a perfect square. 2 Proof of the Theore Let T := T be the subtree of T rooted at vertex x obtained by deleting vertex y fro T. Further let T 1 := T 1 ( T 2 := T 2 ( denote the two subtrees of T Fig. 2 The tree T

3 Order ( : Fig. 3 The rooted trees T 1 T 2 rooted at x that appear in Fig. 3. Note that T 1 has s 2 vertices, T 2 has t vertices, t = s 2 by the forulas 1. For any tree T rooted at vertex x,letc(t, x denote the poset of all subtrees of T containing x. (Moreover, as pointed out by a referee, C(T, x is a locally distributive lattice [1]. Let r (i denote the nuber of subtrees in C(T, x of rank i, letr 1 (i r 2 (i denote the nuber of subtrees in C(T 1, x C(T 2, x, respectively, of rank i. Then it is easy to see that r (i = i r 1 ( r 2 (i 1 (2 =1 where 2 if i = s 1 r 1 (i = 1 if 1 i s 2, i = s 1 0 otherwise 1 if 1 i < t ( r 2 (i = if t i t i t 0 otherwise. Our proof of Theore 1 depends on the fact, proved below, that r (i attains its axiu on exactly two values a b of i that a b differ by at least 2. Lea 1 With notation as in the introduction, let 0 k 0 < be such that ( k 0 k01 ( i = N0 let a = s 2 b = 2s 2 k 0.If is not a perfect square, then b a 2 r (a = r (b = N 0 r (i < N 0 if i = a, b.

4 136 Order ( : Proof Concerning the first stateent, it follows fro Eq. 1 the fact that s 2 = t that b a = s k 0 s 1 = t 1 2. Concerning the second stateent, forula 2 iplies t r (a = r (t = r 1 (i r 2 (t i 1 i=1 1 = r 1 (i r 2 (t i 1 i=1 1 ( = t = i 1 i=1 r (b = r (2t 2 2 k 0 = t t i=2 r 1 (i r 2 (t i 1 t = 2 t = N 0 i i=t 1 k 0 r 1 (i r 2 (2t 2 1 k 0 i k 01 ( ( k 01 = r 1 (i t 1 k 0 = = N 0. i i Letting (i := r (i 1 r (i it follows fro Eq. 2 that (i = r 2 (i s 1 r 2 (i s r 2 (i s 1 r 2 (i 1. (3 It is then easy to check that In addition we clai that k 0 (i >0 if 1 i < a (i <0 if a i < s t (i >0 if i = s t. (i >0 if s t < i < b (i <0 if b i, which would coplete the proof of the lea. To prove the clai, let i > s t. Since T 2 has t = s 2 < s t vertices, r 2 (i 1 = 0. Equation 3 then iplies that (i >0 if only if ( = r 2 (i s 1 >r 2 (i s r 2 (i s 1 i t s 1 = ( ( i t s i t s 1 Letting k = i (s t, the above inequality becoes ( ( ( >, k 1 k k 1.

5 Order ( : an easy calculation using the quadratic forula shows that this is equivalent to k >αwhere α = 1 2 ( ( Note that α is not an integer because is not a perfect square. This iplies that (i <0 if only if k <α. For i > s t the rank function of T as given in forula 2, after soe siplification, is r (k = k k1 =0. By the coents above, this function achieves a unique axiu at k = k 0 = α. Equivalently, this axiu is at i = s t k 0 = 2s 2 k 0 = b. Proof of Theore 1 Let T = T. It is first shown that C(T, x is not Sperner. A axiu antichain A is provided whose eleents are not all fro a single rank. With vertex y asshowninfig.2, integers a b as in Lea 1, S denoting the nuber of vertices in a subtree S, let A ={S C(T, x : ( S =a 1 y S or ( S =b y / S }. By Lea 1 we know that b > a 1, soa is an antichain in C(T, x, the eleents of A do not all have the sae rank. Moreover A =r (a r (b = 2N 0.On the other h, letting r x (i denote the nuber of eleents of rank i in C(T, x, we have by Lea 1 that r x (i = r (i r (i 1 <2N 0.HenceC(T, x is not Sperner. To prove that C(T is not Sperner, it will be shown that the nuber of eleents at any given rank in C(T is less than the nuber of eleents in the antichain A. Let r(i denote the nuber of (unrooted subtrees of C(T with i vertices. It is now sufficient to show that r(i < A =2N 0 for all i.ifi s 2 = t, then any subtree of T with i vertices ust contain x. Hence in this case r(i = r x (i <2N 0. It is easy to check that r(1 <2N 0. For 1 < i < s 2, letr (i r (i denote the nuber of subtrees with i vertices in C(T that do do not, respectively, contain vertex y.then i 1 r (i = s t i 2 =0 i 1 r (i = t 1 i t 1 =0 if i t if i > t.

6 138 Order ( : Then r(i = r (i r (i is axiized either when i = s 1 or when i = t.noting that s, t = s 2 N 0 = 2 t fro forula 1, we have r(s 1 = 2t =0 2 =0 = 2 1 2t 1 < 2(2 t = 2N 0. t 1 r(t = s t 1 2 s t 1 = 2N < 2N 0. =0 References 1. Dilworth, R.P., Crawley, P.: Decoposition theory for lattices without chain conditions. Trans. Aer. Math. Soc. 96, 1 22 ( Jacobson, M.S., Kézdy, A., Seif, S.: The poset on connected induced subgraphs of a graph need not be Sperner. Order 12, ( Klavzar, S., Petkovsek, M., Steelan, J.H.: Probles solutions, solutions of eleentary probles, E3281. Aer. Math. Monthly 97, ( Trotter, W.T., Moore, J.L.: Soe theores on graphs posets. Discrete Math. 15, ( West, D.B.: Open probles. In: Newsletter of the Sia Activity Group on Discrete Matheatics, Spring, vol. 2 series 2 (1992