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text: 'The standard $\\leq$ relation on integers, the $\\subseteq$ relation on sets, and the $|$ (divisibility) relation on natural numbers are all examples of poset orders.\n\nInteger Comparison\n================\n\nThe 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 [540 totally ordered sets]. Here is its Hasse diagram.\n\n\n%%%comment:\ndot source (doctored in GIMP)\n\ndigraph G {\n node [width = 0.1, height = 0.1]\n edge [arrowhead = "none"]\n a [label = "-3"]\n b [label = "-2"]\n c [label = "-1"]\n d [label = "0"]\n e [label = "1"]\n f [label = "2"]\n g [label = "3"]\n rankdir = BT;\n a -> b\n b -> c\n c -> d\n d -> e\n e -> f\n f -> g\n}\n%%%\nPower sets\n=========\n\nFor 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 [5dy 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 [5lt 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 [573 transitivity] of $\\subseteq$ follows from the transitivity of $\\Rightarrow$.\n\nNote that the strict subset relation $\\subset$ is the strict ordering derived from the poset $\\langle \\mathcal{P}(X), \\subseteq \\rangle$.\n\nDivisibility on the natural numbers\n===========================\n\nLet [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 $a|c$.\n',
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