Notes for week of 11/19

MATH 681 Notes Combinatorics and Graph Theory IProof. Suppose C ⊂ P is a maximal chain, Since (C, ≼) is a finite nonempty poset in its own right,it has some maximal element x; since C is totally ordered, the nonexistence of any y in C such thatx ≼ y implies that y ≼ x for all y ∈ C, so x is in fact a greatest element of C (if not necessarily ofS). We have two possibilities to address: x may be maximal in S, or it may not. If it is maximal,our condition has been shown. If it is not maximal, there is a z ∈ S such that x ≼ z. Then, for ally ∈ C, y ≼ x ≼ z, so y ≼ z, so C ∪ {z} is totally ordered, contradicting C’s maximality. The prooffor minimal elements proceeds along similar lines.Proposition 2. The set of maximal elements of a finite poset S is a maximal antichain; likewise,the set of minimal elements of a finite poset S is a maximal antichain.Proof. Note that this is trivially true if S is empty; henceforth, we will consider the case where Shas at elast one element.Let A be the set of maximal elements of S. We shall first prove that A is an antichain, and thenthat any augmentation of A by adding an element is not an antichain.Consider distinct elements x, y ∈ A. Since x is a maximal element and y ≠ x, maximality guaranteesx y, Likewise, since y is maximal anddistinct from x, y x. Thus, x and y are incomparable.Since this is true of arbitrarily chosen distinct elements of A, all distinct elements of A are incomparableand A is an antichain.Now, consider z 0 ∈ S −A; that is, z 0 is a non-maximal element of S. We shall show that A∪{z 0 } isnot an antichain. Since z 0 is nonmaximal, there is some z 1 ≠ z 0 such that z 0 ≼ z 1 . If z 1 is maximal,it is in A; if it is nonmaximal, there is a z 2 ≠ z 1 such that z 1 ≼ z 2 . We continue along these linesuntil we get either a z k that is maximal, or an infinite ascending sequence z 0 ≼ z 1 ≼ · · · . Thelatter situation was shown last week to be impossible in a finite poset; thus some z k is maximal, soz k ∈ A and z 0 ≼ z 1 ≼ · · · ≼ z k gives z 0 ≼ z k by transitivity. Since z 0 and z k are comparable, andz k ∈ A, A ∪ {z 0 } is not an antichain.The proof for minimal elements proceeds along similar lines.1.2 Limits on width and height; Dilworth’s TheoremOne interesting result we see playing with maximum chains and antichains is that they can’t both bevery small. We can easily build posets with large height and small width (e.g. ({1, 2, 3, . . . , n}, ≤)),small height and large width ({1, 2, 3, . . . , n}, =), and even with large height and width ({a 1 , a 2 , a 3 , . . . , b 2 , b 3 , . . . b n },) with a i ≼ a j when i ≤ j, and all other terms incomparable. But keeping both the height andwidth of an arbitrarily large poset down seem hard (we can kind of do so, with {1, 2, 3, n} 2 subjectto (a, b) ≼ (c, d) if a ≤ c and b = d – this has n 2 elements and length and width of n.We shall show that this is in fact the best we can do.Proposition 3. Suppose P is partitioned into a finite set of chains C 1 , C 2 , . . . , C n . If A is anantichain, then there is at most one element of A in each C i ; thus n ≥ |A|.Proof. This result is actually quite trivial: suppose two distinct elements x and y of A were bothin the same chain C i . membership in A would force x and y to be incomparable; membership inC i would force them to be comparable. Thus, |A ∩ C i | ≤ 1 for all i.Page 2 of 5 November 19, 2009