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.

Truncated 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 $a|c$ .