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Home , Distributive lattice, Join and meet

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 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 . 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 -named after George Boole• who studied such systems in the 1800's in connection with his work in symbolic . The structure of a finite 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 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 which has the 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 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 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 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. Clearly, if the conjecture was true and a portion of the ordered set was split into n chains and x was an element not in any of the n chains, then it should be possible to include x and reorganize the chains so that the enlarged set was again a set join of n chains. Experimenting with a few examples made it clear that the reorganization should involve the portions of the chains lying above x and portions lying below x. For example, in Figure 4, the ordered set consisting of a, b, and c can be split into the two chains {a} and {b, c}. The portion of {b, c} lying below the element d is the element c. If c and d are put together as a chain, then the remaining elements a and b also form a chain, and the ordered set is represented as a join of two chains. If this method was to work in general, then it should be possible to select an upper portion of one of the chains and a lower portion of another chain so that when these elements are removed from the set join of the n chains, the remaining ordered set can be represented as set join of n - 1 chains. Assuming that this is not the case for each selection of an upper section and a lower section turns out to lead to a contradiction and thus the theorem follows by induction. While I was working on the finite case it became clear that the result must hold for infinite ordered sets as long as the size of remains bounded. However, by this time I had come back to Cal tech as an assistant professor and was winding up my work on the unique complementation problem. Since it was now in the middle of World War II, I had been asked by John Harlan, who was then with the DOD, to be a member of one of the small operations analysis units which were being set up at the headquarters of each of the three divisions of the Eighth Air Force. I felt that I could not refuse, so until the end of the war I was in England doing an entirely different kind of research. At the end of the war I returned to Caltech and eventually turned again to the decomposition problem for infinite ordered sets in which the size of antichains was bounded. It seemed me, at this time, that the problem now was more a set-

Chain Partitions in Ordered Sets 5 theoretic rather than a combinatorial problem. So again I went searching for chains which when removed from the ordered set left an ordered set in which the maximum size of an antichain was reduced. After several months of trying various restrictive conditions, I decided to go to the strongest restriction possible, namely, to define a chain to be 'strong' if its meet with any finite subset of the ordered set was a chain in some decomposition of the finite subset into a minimal number of chains. Since the theorem holds in the finite case, a chain consisting of a single element is 'strong' and since 'strongness' is a finite property, there exists a maximal 'strong' chain. When such a chain is removed from the ordered set, the size of a maximal antichain in the remaining ordered set is reduced and theorem follows by induction. The time had finally arrived when it was appropriate to write up the results and send the manuscript off for publication.

6 THE DILWORTH THEOREMS