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  "body": "\n## Statement\n\nLet $X$ be a set and let $\\le$ be a [[partial order|0QV2JS9RJ3]] on $X$.\nThen there is a [[total order|0FRXBVVR03]] $\\preceq$ on $X$ such that $x \\le y$ implies $x \\preceq y$ for all $x, y \\in X$.\nIn other words, every partial order has a [[linear extension|W8DZ5KGGHN]].\n\n## Remarks\n\nFor a finite set, take a minimal element of $X$, put it first, and repeat with the remaining elements; this is topological sorting.\nFor an infinite set, apply [[Zorn's lemma|04RXTQA5ZA]] to the set of partial orders on $X$ that contain $\\le$, ordered by inclusion.\nA maximal such order is total: if $a$ and $b$ were incomparable in it, adding the pairs $(u, v)$ with $u \\preceq a$ and $b \\preceq v$ would give a strictly larger partial order.\nThe same argument shows more: if $a$ and $b$ are incomparable for $\\le$, then some linear extension puts $a$ before $b$.\nHence $\\le$ is the intersection of all its linear extensions.\nExample: the divisibility order on $\\set{1, 2, 3, 6}$ has the linear extensions $1, 2, 3, 6$ and $1, 3, 2, 6$, and $2$ and $3$ are incomparable because they appear in both relative orders.\n\n## Explanations\n\n### L1 human\n\nA partial order says that some things come before others, but leaves some pairs undecided.\nThe theorem says that one can always decide the remaining pairs and line up all the elements in one row without breaking any of the given rules.\nFor a list of tasks with prerequisites, this is a schedule that does the tasks one at a time.\n\n### L2 human\n\nThe theorem is what makes the order dimension of a poset well defined: the set of all linear extensions realizes the order.\nFor finite posets the proof is an algorithm, and counting the linear extensions is a central problem of order combinatorics.\nFor infinite sets the proof uses the axiom of choice through Zorn's lemma.\n",
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