Chain Partitions in Ordered Sets
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CHAPTER 1 Chain Partitions in Ordered Sets Editors' note: The background for this chapter is based on a lecture in tended for undergraduates. Background R. P. DILWORTH Let us consider the sets which can be made up from the numbers 1, 2, and 3. We have the singleton sets {1}, {2}, and {3} made up of one number each. Then there are the two-element sets {1,2}, {l, 3}, and {2,3}. Finally, there is the three element set {l, 2, 3}. It is customary to include the empty set 0 which has no numbers belonging to it. There is a natural relation between these sets, namely, {1} is a subset of {l, 2} and {l, 3} is a subset of {l, 2, 3}. If A and B denote two of these sets, we write A = B if A and B consist of the same numbers. Thus {1, 2} = {2,1}. If every member of A is also a member of B we write A ~ B and if, in addition, there are members of B which do not belong to A we write A < B. Thus {1, 2} < {1, 2, 3}. It is convenient to diagram the containing relations existing between these sets as shown in Figure 1. Each set is represented by a small circle in the plane. Larger sets lie above smaller sets and the circles representing two sets are joined by a line if one contains the other and there is no set lying strictly between the two. The subsets of a set also have an algebraic structure. If A and B are two sets, A V B (A join B) will denote the set made up of the objects of A together with the objects of B. A A B (A meet B) will denote the set of objects belonging to both A and B. These operations are commutative, associative, distributive and idempotent (A V A = A A A = A). There is also a unary operation of complementation. A' consists of the objects in the set which do not belong to A. In the example above, 1 o {1,2,3} {1,2} O~O~O {2,3} {I} oXO~O {3} ~/ Figure 1 we have {1,2} V {3} = {1, 2, 3} {1, 2} A {2, 3} = {2} {1,3}' = {2}. The subsets of the set {1, 2, 3} under the operations of join, meet, and complemen tation provide a simple example of a Boolean algebra-named after George Boole who studied such systems in the 1800's in connection with his work in symbolic logic. The structure of a finite Boolean algebra is completely determined once the number of atoms (singleton sets) is known, namely each member of Boolean algebra is the unique join of the singleton sets contained in it. Now a singleton set together with the empty set 0 make up a two-element Boolean algebra whose diagram is shown in Figure 2. Thus the basic structure result for finite Boolean algebras states that a finite Boolean algebra is a direct product of n two-element Boolean algebras where n is the number of atoms. Figure 2 An important generalization of a Boolean algebra is a distributive lattice which has the join and meet operations with the same properties as in the case of Boolean algebras but without the complementation operation. For example, the sets 0, {l}, {2}, {l,2}, {2,3}, {l,2,3} form a distributive lattice having the diagram shown in Figure 3. It is no longer true that every member of the lattice is a join of atoms, but it is true that every member is a join of join irreducibles. A join irreducible is an element of the lattice which cannot be represented as a join of elements distinct 2 THE DILWORTH THEOREMS o {1,2,3} {1,2} /~. {2,3} {l}./~.' ~.( Figure 3 from itself. In the example, 0, {I}, {2}, and {2,3} are join irreducibles. Although {I, 2, 3} can be represented as the join of {I}, {2}, and {2,3} and also as the join of {I} and {2,3} the latter representation which is minimal is unique. Indeed, in an abstract finite distributive lattice, the minimal representations as joins of join irreducibles are unique and associating each element of the lattice with the set of join irreducibles less than or equal to it gives a representation of the lattice as a lattice of sets of join irreducibles. Conversely, if a finite ordered set is given, the subsets of the ordered set which are such that with each element in the subset also all elements less than or equal to it also belong to the subset form a distributive lattice. For example, consider the ordered set diagrammed in Figure 4. boo d a./1/ Figure 4 The sets 0, {a}, {c}, {c,d}, {a,b,c}, {a,c,d}, {b,c,d}, {a,b,c,d} form a dis tributive lattice diagrammed in Figure 5. The join irreducibles {a}, {a,b,c}, {c}, {c, d} correspond to the elements a, b, c, d in the ordered set. The correspondence between finite ordered sets and finite distributive lattices is one-to-one. Thus, as is frequently the case, questions concerning the structure of distributive lattices can be formulated in terms of questions concerning ordered sets. We also note from this correspondence that any representation of a finite distributive lattice as a lattice of subsets of some set requires that the set have as many members as there are join irreducibles in the distributive lattice. In other terms, the minimum number of two-element Boolean algebras whose direct product has a sublattice isomorphic to a given finite distributive lattice is the number of join irreducibles in the lattice. When I completed my graduate work at Caltech and went to Yale University Chain Partitions in Ordered Sets 3 Figure 5 on a Sterling Research Fellowship, one of the most interesting problems in lattice theory at that time was the conjecture that a lattice in which every element had a unique complement was a Boolean algebra. The conjecture had been verified in several instances where one of a number of additional restrictions was imposed, namely modularity, orthocomplementation, and atomicity. The general conjecture was still open. Since the fellowship provided me with a full year in which to do research, I decided to concentrate on the conjecture. However, since it is difficult to work full time on one problem, I decided to fill the breaks by working on a problem concerning the representation of a distributive lattice as a sublattice of a direct product of totally ordered sets, i.e., chains. Since it was easy to determine the least number of two-element chains whose direct product could contain the given distributive lattice as a sublattice, it seemed natural to ask for the least number of chains of arbitrary length whose direct product could contain the given distributive lattice as a sublattice. Making use of the correspondence between finite distributive lattices and finite ordered sets it was quite straightforward to verify that imbedding a finite distributive lattice in a direct product of chains was equivalent to representing a finite ordered set as a set join of chains. One useful observation which turned up early in the investigation was the fact that the number of elements covering a given element in the distributive lattice was always a lower bound for the number of chains needed for a representation, since each covering element had to belong to a different chain. Again, making use of the correspondence between distributive lattices and ordered sets and examining many examples it became clear that the maximal number of elements covering an element in the distributive lattice was equal to the maximum number of non-comparable elements in the corresponding ordered set, i.e., to the maximal number of elements in an antichain of the ordered set. It was also clear that the maximal number of elements in an antichain was a lower bound to the minimum number of chains needed to represent the ordered set as a set join of chains. Thus the problem would be solved if it could be shown that these two numbers were 4 THE Oll...WORTH THEOREMS indeed equal. If n is the maximal number of elements in an anti chain of a finite ordered set, the most natural attack on the problem would be to take a maximal chain out of the ordered set and hope that the maximum size of an antichain in what is left is n - 1 or less. However, this doesn't work. In Figure 4, the maximum size of an anti chain is 2. The elements band c make up a maximal chain. The elements a and d which are left are an antichain of size 2. Obviously, a good place to begin would be the case n = 2, since for n = 1, the ordered set is already a chain. By starting at the bottom of the ordered set and making an inductive argument it was easy to see that the ordered set is a set join of two chains when n = 2. This special technique fails for n ~ 3 so a new approach was required. Concentrating on n = 3, I tried removing maximal chains having various special properties with the hope that the remaining elements would form an ordered set with n = 2. Although this approach worked for ordered sets of relatively small orders, the required properties became more and more complicated as the size of the ordered set increased. After several months I abandoned this approach and turned to trying inductive arguments.