Poset: Examples

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 = 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 \(a|c\).