# poset

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A poset, or partially ordered set, consists of a set $P$ and a binary relation $\leq$ on $P$ which satisfies the following properties:

• $\leq$ is reflexive (http://planetmath.org/Reflexive), so $a\leq a$ always holds;

• $\leq$ is antisymmetric, so if $a\leq b$ and $b\leq a$ hold, then $a=b$; and

• $\leq$ is transitive (http://planetmath.org/Transitive3), so if $a\leq b$ and $b\leq c$ hold, then $a\leq c$ also holds.

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

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

• $<$ is transitive.

Since $\leq$ 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 $\leq$ 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 $\preceq\,=\{(a,a),(a,b),(b,a),(b,b)\}$. A diagram of this structure, omitting loops, is displayed below.

 $\xymatrix{b\ar@/^{/}[d]\\ a\ar@/^{/}[u]}$

Observe that the binary relation on $Q$ is reflexive and transitive, so $Q$ is a preorder. On the other hand, $a\preceq b$ and $b\preceq a$, while $a\neq b$. 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 $a\leq b$ or $b\leq a$. Let $P$ be the structure with ground set $\{a,b,c\}$ and binary relation $\leq\,=\{(a,a),(a,b),(a,c),(b,b),(c,c)\}$. A diagram of this structure, omitting loops, is displayed below.

 $\xymatrix{b&&c\\ &a\ar[ul]\ar[ur]&}$

Observe that the binary relation on $P$ is reflexive, antisymmetric, and transitive, so $P$ is a poset. On the other hand, neither $b\leq c$ nor $c\leq b$ 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 $a\leq b$ or $b\leq a$ holds in $P$, we say that $a$ and $b$ are comparable; otherwise, we say they are incomparable. We use the notation $a\shortparallel b$ to indicate that $a$ and $b$ are incomparable.

If $(P,\leq_{P})$ and $(Q,\leq_{Q})$ are posets, then a function $\varphi\colon P\to Q$ is said to be order-preserving, or monotone, provided that it preserves inequalities. That is, $\varphi$ is order-preserving if whenever $a\leq_{P}b$ holds, it follows that $\varphi(a)\leq_{Q}\varphi(b)$ also holds. The identity function on the ground set of a poset is order-preserving. If $(P,\leq_{P})$, $(Q,\leq_{Q})$, and $(R,\leq_{R})$ are posets and $\varphi\colon P\to Q$ and $\psi\colon Q\to R$ are order-preserving functions, then the composition $\psi\circ\varphi\colon P\to R$ is also order-preserving.

Posets together with order-preserving functions form a category, which we denoted by $\mathbf{Poset}$. 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 $\mathbb{N}$ be the set of natural numbers. Inductively define a binary relation $\leq$ on $\mathbb{N}$ by the following rules:

• for any $n\in\mathbb{N}$, the relation $0\leq n$ holds; and

• whenever $m\leq n$, the relation $m+1\leq n+1$ also holds.

Then $(\mathbb{N},\leq)$ is a chain, hence a poset. This structure can be naturally embedded in the larger chains of the integers, the rational numbers, and the real numbers.

The next example shows that nontrivial antichains exist.

###### Example 2.

Let $P$ be a set with cardinality greater than $1$. Let $\leq$ be the diagonal of $P$. Thus $\leq$ represents equality, which is trivially a partial order relation (which is also the intersection of all partial orderings on $P$). By construction, $a\leq b$ in $P$ if and only $a=b$. Thus no two elements of $P$ are comparable.

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 $A\leq B$ if and only if $A\subseteq B$.

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 $\leq$. The dual poset of $P$ is defined as follows: it has the same underlying set as $P$, whose order is defined by $a\leq^{\prime}b$ iff $b\leq a$. It is easy to see that $\leq^{\prime}$ is a partial order. The dual of $P$ is usually denoted by $P^{\partial}$.

## 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 $P$ and edge set $<$. For example, the following diagram displays the Boolean algebra $B_{2}$ as a directed graph.

 $\xymatrix{&\{0,1\}&\\ \{0\}\ar[ur]&&\{1\}\ar[ul]\\ &\emptyset\ar[ul]\ar[uu]\ar[ur]&}$

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 of $P$ has finite height. If $a$ and $b$ are elements of $P$, then we say that $b$ covers $a$ if $a and there are no elements of $P$ 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

 $a<:b\text{\ if and only if b covers a.}$

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 and the height of $[a,b]$ is 1, then $a<:b$.

Assume for induction that whenever $a and the height of $[a,b]$ is at most $n$, then $(a,b)$ is in the transitive closure of $<:$. Suppose that $a 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 induction assumption 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 and the height of $[a,b]$ is at most $n+1$, then $(a,b)$ is in the transitive closure of $<:$.

This completes the proof. ∎

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 $<:$ of $B_{2}$.

 $\xymatrix{&\{0,1\}&\\ \{0\}\ar[ur]&&\{1\}\ar[ul]\\ &\emptyset\ar[ul]\ar[ur]&}$

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

## References

 Title poset Canonical name Poset Date of creation 2013-03-22 11:43:41 Last modified on 2013-03-22 11:43:41 Owner mps (409) Last modified by mps (409) Numerical id 22 Author mps (409) Entry type Definition Classification msc 06A99 Synonym partially ordered set Related topic Relation Related topic PartialOrder Related topic Semilattice Related topic StarProduct Related topic HasseDiagram Related topic GreatestLowerBound Related topic NetsAndClosuresOfSubspaces Related topic OrderPreservingMap Related topic DisjunctionPropertyOfWallman Defines comparable Defines incomparable Defines cover Defines covering Defines order-preserving function Defines monotone Defines monotonic Defines order morphism Defines morphism of posets Defines dual poset Defines consecutive