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.

digraph G { node = 0.1, height = 0.1 edge = “none” a = “-3″ b = “-2” c = “-1″ d = “0” e = “1″ f = “2” g = “3″ rankdir = BT; a → b b → c c → d d → e e → f f → g } <div>

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 $\mathbb N$ 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 $a|c$.