poset (original) (raw)

\xyoption

all

The relation ≤ is called a partial order on P. In practice, (P,≤) is usually conflated with P; if a distinction is needed, P is called the ground set or underlying set of (P,≤). The binary relation < defined by removing the diagonal from ≤(i.e. a<b iff a≤b and a≠b) satisfies the following properties:

• •
  < is irreflexive, so if a<b holds, then b<a does not hold; and
• •
  < is transitive.

Since ≤ is reflexive, it can be uniquely recovered from < by adding the diagonal. For this reason, an irreflexive and transitive binary relation < (called a strict partial order) also defines a poset, by means of the associated relation ≤ described above (which is called weak partial order).

Since every partial order is reflexive and transitive, every poset is a preorder. The notion of partial order is stricter than that of preorder, Let Q be the structure with ground set Q={a,b} and binary relation ⪯={(a,a),(a,b),(b,a),(b,b)}. A diagram of this structure, omitting loops, is displayed below.

Observe that the binary relation on Q is reflexive and transitive, so Q is a preorder. On the other hand, a⪯b and b⪯a, while a≠b. So the binary relation on Q is not antisymmetric, implying that Q is not a poset.

Since every total order is reflexive, antisymmetric, and transitive, every total order is a poset. The notion of partial order is weaker than that of total order. A total order must obey the trichotomy law, which states that for any a and b in the order, either a≤b orb≤a. Let P be the structure with ground set {a,b,c} and binary relation ≤={(a,a),(a,b),(a,c),(b,b),(c,c)}. A diagram of this structure, omitting loops, is displayed below.

Observe that the binary relation on P is reflexive, antisymmetric, and transitive, so P is a poset. On the other hand, neither b≤c nor c≤b holds in P. Thus P fails to satisfy the trichotomy law and is not a total order.

The failure of the trichotomy law for posets motivates the following terminology. Let P be a poset. If a≤b or b≤a holds inP, we say that a and b are comparable; otherwise, we say they are incomparable. We use the notation a∥bto indicate that a and b are incomparable.

If (P,≤P) and (Q,≤Q) are posets, then a function φ:P→Q is said to be order-preserving, or_monotone_, provided that it preserves inequalities. That is,φ is order-preserving if whenever a≤Pb holds, it follows that φ⁢(a)≤Qφ⁢(b) also holds. The identity function on the ground set of a poset is order-preserving. If (P,≤P),(Q,≤Q), and (R,≤R) are posets and φ:P→Q andψ:Q→R are order-preserving functions, then thecomposition ψ∘φ:P→R is also order-preserving.

Posets together with order-preserving functions form a category, which we denoted by 𝐏𝐨𝐬𝐞𝐭. Thus an order-preserving function between the ground sets of two posets is sometimes also called a_morphism of posets_. The category of posets has arbitrary products (http://planetmath.org/ProductofPosets). Moreover, every poset can itself be viewed as a category, and it turns out that a morphism of posets is the same as a functor between the two posets.

Examples of posets

The two extreme posets are the chain, in which any two elements are comparable, and the antichain, in which no two elements are comparable. A poset with a singleton underlying set is necessarily both a chain and an antichain, but a poset with a larger underlying set cannot be both.

Example 1.

Let ℕ be the set of natural numbers. Inductively define a binary relation ≤ on ℕ by the following rules:

• •
  for any n∈ℕ, the relation 0≤n holds; and
• •
  whenever m≤n, the relation m+1≤n+1 also holds.

Then (ℕ,≤) is a chain, hence a poset. This structure can be naturally embedded in the larger chains of the integers, therational numbers, and the real numbers.

The next example shows that nontrivial antichains exist.

So far the only posets we have seen are chains and antichains. Most posets are neither. The following construction gives many such examples.

Example 3.

If X is any set, the powerset P=P⁢(X) of X is partially ordered by inclusion, that is, by the relation A≤B if and only ifA⊆B.

There are important structure theorems for posets concerning chains and antichains. One of the foundational results is Dilworth’s theorem. This theorem was massively generalized by Greene and Kleitman.

A final example shows that one can manufacture a poset from an existing one.

Example 4.

Let P be a poset ordered by ≤. The dual poset of P is defined as follows: it has the same underlying set as P, whose order is defined by a≤′biff b≤a. It is easy to see that ≤′ is a partial order. The dual of Pis usually denoted by P∂.

Graph-theoretical view of posets

Let P be a poset with strict partial order <. Then P can be viewed as a directed graph with vertex set the ground set of P andedge set <. For example, the following diagram displays the Boolean algebra B2 as a directed graph.

If P is a sufficiently complicated poset, then drawing all of the edges of P can obscure rather than reveal the structure of P. For this reason it is convenient to restrict attention to a subrelation of< from which < can be uniquely recovered.

We describe a method of constructing a canonical subgraph of P from which the partial order can be recovered as long as every interval ofP has finite height. If a and b are elements of P, then we say that b covers a if a<b and there are no elements ofP strictly larger than a but strictly smaller than b, that is, if[a,b]={a,b}. Two elements are said to be consecutive if one covers another. Define a binary relation <: on P by

By construction, the binary relation <: is a subset of <. Since< is transitive, the transitive closure (http://planetmath.org/ClosureOfASetViaRelations) of <: is also contained in <.

Proposition.

Suppose every interval of P has finite height. Then < is the transitive closure of <:.

Proof.

We prove this by induction on height. By definition of <:, if a<band the height of [a,b] is 1, then a<:b.

Assume for induction that whenever a<b and the height of [a,b] is at most n, then (a,b) is in the transitive closure of <:. Suppose that a<b and that the height of [a,b] is n+1. Since every chain in [a,b] is finite, it contains an element c which is strictly larger than a and minimal (http://planetmath.org/MaximalElement) with respect to this property. Therefore [a,c]={a,c}, from which we conclude that a<:c. Since the interval [c,b] is a proper subinterval of [a,b], it has height at most n, so by the inductionassumption we conclude that (c,b) is in the transitive closure of<:. Since (a,c) and (c,b) are in the transitive closure of<:, so is (a,b). Hence whenever a<b and the height of [a,b]is at most n+1, then (a,b) is in the transitive closure of <:.

In the same way we associated a graph to < we can associate a graph to <:. The graph is usually called the Hasse diagram of the poset. Below we display the graph associated to the cover relation <: ofB2.

For simplicity, the Hasse diagram of a poset is usually drawn as anundirected graph. Elements which are higher in the partial order relation are drawn physically higher. Since a strict partial order isacyclic, this can be done uniquely and the partial order can be recovered from the drawing.

References

• 1 Grätzer, G., General lattice theory, 2nd ed., Birkhäuser, 1998.
• 2 Stanley, R., Enumerative Combinatorics, vol. 1, 2nd ed., Cambridge University Press, Cambridge, 1996.