# Poset: Examples

https://arbital.com/p/poset_examples

by Kevin Clancy Jun 8 2016 updated Dec 6 2016

The standard $\leq$ relation on integers, the $\subseteq$ relation on sets, and the $|$ (divisibility) relation on natural numbers are all examples of poset orders.

# Integer Comparison

The set $\mathbb Z$ of integers, ordered by the standard "less than or equal to" operator $\leq$ forms a poset $\langle \mathbb Z, \leq \rangle$. This poset is somewhat boring however, because all pairs of elements are comparable; such posets are called chains or totally ordered sets. Here is its Hasse diagram.

%%%comment:
dot source (doctored in GIMP)

digraph G {
node [width = 0.1, height = 0.1]
edge [arrowhead = "none"]
a [label = "-3"]
b [label = "-2"]
c [label = "-1"]
d [label = "0"]
e [label = "1"]
f [label = "2"]
g [label = "3"]
rankdir = BT;
a -> b
b -> c
c -> d
d -> e
e -> f
f -> g
}
%%%

# Power sets

For any set $X$, the power set of $X$ ordered by the set inclusion relation $\subseteq$ forms a poset $\langle \mathcal{P}(X), \subseteq \rangle$. $\subseteq$ is clearly reflexive, since any set is a subset of itself. For $A,B \in \mathcal{P}(X)$, $A \subseteq B$ and $B \subseteq A$ combine to give $x \in A \Leftrightarrow x \in B$ which means $A = B$. Thus, $\subseteq$ is antisymmetric. Finally, for $A, B, C \in \mathcal{P}(X)$, $A \subseteq B$ and $B \subseteq C$ give $x \in A \Rightarrow x \in B$ and $x \in B \Rightarrow x \in C$, and so the transitivity of $\subseteq$ follows from the transitivity of $\Rightarrow$.

Note that the strict subset relation $\subset$ is the strict ordering derived from the poset $\langle \mathcal{P}(X), \subseteq \rangle$.

# Divisibility on the natural numbers

Let [45h $\mathbb N$] be the set of natural numbers including zero, and let $|$ be the divides relation, where $a|b$ whenever there exists an integer $k$ such that $ak=b$. Then $\langle \mathbb{N}, | \rangle$ is a poset. $|$ is reflexive because, letting k=1, any natural number divides itself. To see that $|$ is anti-symmetric, suppose $a|b$ and $b|a$. Then there exist integers $k_1$ and $k_2$ such that $a = k_1b$ and $b = k_2a$. By substitution, we have $a = k_1k_2a$. Thus, if either $k$ is $0$, then both $a$ and $b$ must be $0$. Otherwise, both $k$'s must equal $1$ so that $a = k_1k_2a$ holds. Either way, $a = b$, and so $|$ is anti-symmetric. To see that $|$ is transitive, suppose that $a|b$ and $b|c$. This implies the existence of integers $k_1$ and $k_2$ such that $a = k_1b$ and $b = k_2c$. Since by substitution $a = k_1k_2c$, we have $a|c$.