The doors are open; and the surfeited grooms Do mock their charge with snores:
I have drugg’d their possets,
That death and nature do contend about them Macbeth, Act II, Scene II
William Shakespeare A poset or partially ordered set is a set A with a relation R ⊂ A × A on the elements of A which we will typically write as a ≤ b instead of (a, b) ∈ R, such that for all a, b, c ∈ A:
(reflexive) a ≤ a.
(antisymmetric) a ≤ b and b ≤ a imply that a = b.
(transitive) a ≤ b and b ≤ c imply that a ≤ c.
For example, A could be the set of subsets of a particular set, and ≤ with be the
“subset or equal” relation.
A convenient way do describe a poset for a finite set A is by its Hasse-diagram.
Say that a covers b if a ≥ b, a /= b and there is no a /= c =/= b with a ≥ c ≥ b. The Hasse diagram of the poset is a graph in the plane which connects two vertices a and b only if a covers b, and in this case the edge from b to a goes upwards.
Because of transitivity, we have that a ≤ b if and only if one can go up along edges from a to reach b.
Figure III.2 gives a number of examples of posets, given by their Hasse dia-grams, including all posets on 3 elements.
An isomorphism of posets is a bijection that preserves the ≤ relation.
An element in a poset is called maximal if there is no larger (wrt. ≤) element, minimal is defined in the same way. Posets might have multiple maximal and min-imal elements.
III.2. PARTIALLY ORDERED SETS AND LATTICES 37 Linear extension
My scheme of Order gave me the most trouble Autobiography Benjamin Franklin A partial order is called a total order, if for every pair a, b ∈ A of elements we have that a ≤ b or b ≤ a.
While this is not part of our definition, we can always embed a partial order into a total order.
Proposition III.3: Let R ⊂ X × X be a partial order on X. Then there exists a total order (called a linear extension) T ⊂ X × X such that R ⊂ T.
To avoid set acrobatics we shall prove this only in the case of a finite set X.
Note that in Computer science the process of finding such an embedding is called a topological sorting.
Proof: We proceed by induction over the number of pairs a, b that are incompara-ble. In the base case we already have a total order.
Otherwise, let a, b such an incomparable pair. We set (arbitrarily) that a < b.
Now let
L = {x ∈ X ∣ x ≤R a}, U = {x ∈ X ∣ b ≤R x}
We claim that S = R ∪ {(l, u) ∣ l ∈ L, u ∈ u} is a partial order. As (a, b) ∈ S it has fewer incomparable pairs, this shows by induction that there exists a total order T ⊃ S ⊃ R, proving the theorem.
Since R is reflexive, S is. For antisymmetry, suppose that for x /= y we have that (x, y), (y, x) ∈ S. Since R is a partial order, not both can be in R. Suppose that
(x, y), (y, x) ∈ S ∖ R = {(l, u) ∣ l ∈ L, u ∈ u}.
This implies that x ≤R a and b ≤R x, thus by transitivity b ≤R a, contradicting the incomparability.
If (x, y) ∈ R, (y, x) ∈ S ∖R, we have that b ≤R x ≤R y ≤Ra, again contradicting incomparability.
For transitivity, suppose that (x, y) ∈ S ∖ R and (y, z) ∈ S. Then (y, z) ∈ R, as otherwise b ≤R y ≤R a. But then b ≤R y ≤R z, implying that (x, z) ∈ S. The other
case is analog. ◻
This theorem implies that we can always label the elements of a countable poset with positive integers, such that the poset ordering implies the integer ordering (but this is in general not unique).
Lattices
Definition III.4: Let A be a poset and a, b ∈ A.
• A greatest lower bound of a and b is an element c ≤ a, b which is maximal in the set of elements with this property.
• A least upper bound of a and b is an element c ≥ a, b which is minimal in the set of elements with this property.
A is a lattice if any pair a, b ∈ A have a unique greatest lower bound, called the meet and denoted by a ∧ b; as well as unique least upper bound, called the join and denoted by a ∨ b.
Amongst the Hasse diagrams in figure III.2, e,j,k,l) are lattices, while the others are not. Lattices always have unique maximal and minimal elements, sometimes denoted by 0 (minimal) and 1 (maximal).
Other examples of lattices are:
1. Given a set X, let A = P(X) = {Y ⊆ X} the power set of X with ≤ defined by inclusion. Meet is the intersection, join the union of subsets.
2. Given an integer n, let A be the set of divisors of n with ≤ given by “divides”.
Meet and join are gcd, respectively lcm.
3. For an algebraic structure S, let A be the set of all substructures (e.g. group and subgroups) of S and ≤ given by inclusion. Meet is the intersection, join the substructure spanned by the two constituents.
4. For particular algebraic structures there might be classes of substructures that are closed under meet and join, e.g. normal subgroups. These then form a (sub)lattice.
Using meet and join as binary operations, we can axiomatize the structure of a lattice:
Proposition III.5: Let X be a set with two binary operations ∧ and ∨ and two distinguished elements 0, 1 ∈ X. Then (X, ∧, ∨, 0, 1) is a lattice if and only if the following axioms are satisfies for all x, y, z ∈ X:
Associativity: x ∧ (y ∧ z) = (x ∧ y) ∧ z and x ∨ (y ∨ z) = (x ∨ y) ∨ z;
Commutativity: x ∧ y = y ∧ x and x ∨ y = y ∨ x;
Idempotence: x ∧ x = x and x ∨ x = x;
Inclusion: (x ∨ y) ∧ x = x = (x ∧ y) ∨ x;
Maximality: x ∧ 0 = 0 and x ∨ 1 = 1.
Proof: The verification that these axioms hold for a lattice is left as exercise to the reader.
III.2. PARTIALLY ORDERED SETS AND LATTICES 39 Vice versa, assume that these axioms hold. We need to produce a poset structure and thus define that x ≤ y iff x ∧ y = x. Using commutativity and inclusion this implies the dual property that x ∨ y = (x ∧ y) ∨ y = y.
To show that ≤ is a partial order, idempotence shows reflexivity. If x ≤ y and y ≤ x then x = x ∧ y = y ∧ x = y and thus antisymmetry. Finally suppose that x ≤ y and y ≤ z, that is x = x ∧ y and y = y ∧ z. Then
x ∧ z = (x ∧ y) ∧ z) = x ∧ (y ∧ z) = x ∧ y = x
and thus x ≤ z. Associativity gives us that x ∧ y ≤ x, y if also z ≤ x, y then z ∧ (x ∧ y) = (z ∧ x) ∧ y = z ∧ y = z
and thus z ≤ x ∧ y, thus x ∧ y is the unique greatest lower bound. The least upper bound is proven in the same way and the last axiom shows that 0 is the unique
minimal and 1 the unique maximal element. ◻
Definition III.6: An element x of a lattice L is join-irreducible (JI) if x /= 0 and if x = y ∨ z implies that x = y or x = z.
For example, figure III.3 shows a lattice in which the black vertices are JI, the others not.
When representing elements of a finite lattices, it is possible to do so by storing the JI elements once and representing every element based on the JI elements that are below. This is used for example in one of the algorithms for calculating the subgroups of a group.
Product of posets
The cartesian product provides a way to construct new posets (or lattices) from old ones: Suppose that X, Y are posets with orderings ≤X, ≤Y, we define a partial order on X × Y by setting
(x1, y1) ≤ (x2, y2) if and only if x1≤Xx2 and y1≤Y≤ y2.
Proposition III.7: This is a partial ordering, so X × Y is a poset. If furthermore both X and Y are lattices, then so is X × Y.
The proof of this is exercise ??.
This allows us to describe two familiar lattices as constructed from smaller pieces (with a proof also delegated to the exercises):
Proposition III.8: a) Let ∣A∣ = n and P(A) the power-set lattice (that is the subsets of A, sorted by inclusion). Then P(A) is (isomorphic to) the direct product of n copies of the two element lattice {0, 1}.
b) For an integer n = ∏ri=1peii > 1 written as a product of powers of distinct primes, let D(n) be the lattice of divisors of n. Then D(n) ≅ D(p1e1) × ⋯ × D(perr).
Figure III.3: Order-ideal lattice for the “N” poset.