The standard $\le$ 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 $Z$ of integers, ordered by the standard "less than or equal to" operator $\le$ forms a poset $⟨Z,\le ⟩$. 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.

Power sets

For any set $X$, the power set of $X$ ordered by the set inclusion relation $\subseteq$ forms a poset $⟨P\left(X\right),\subseteq ⟩$. $\subseteq$ is clearly reflexive, since any set is a subset of itself. For $A,B\in P\left(X\right)$, $A\subseteq B$ and $B\subseteq A$ combine to give $x\in A\iff x\in B$ which means $A=B$. Thus, $\subseteq$ is antisymmetric. Finally, for $A,B,C\in P\left(X\right)$, $A\subseteq B$ 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 $⟨P\left(X\right),\subseteq ⟩$.

Divisibility on the natural numbers

Let $N$ 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 $⟨N,|⟩$ 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 $a=k_{1}b$ and $b=k_{2}a$. By substitution, we have $a=k_1{k}_{2}a$. Thus, if either $k$ is $0$, then both $a$ and $b$ must be $0$. Otherwise, both $k$'s must equal $1$ so that $a=k_1{k}_{2}a$ 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 $a=k_{1}b$ and $b=k_{2}c$. Since by substitution $a=k_1{k}_{2}c$, we have $a|c$.