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  "body": "\n## Statement\n\nLet $(P, \\le)$ be a finite [[partially ordered set|0QV2JS9RJ3]].\nLet $w$ be the largest size of an [[antichain|03HF498P1J]] in $P$.\nThen $P$ is the union of $w$ pairwise disjoint [[chains|0S8FE7ZMFM]], and $P$ is not the union of fewer than $w$ chains.\n\n## Remarks\n\nThe easy half: a chain meets an antichain in at most one element, so any cover of $P$ by chains uses at least $w$ chains.\nThe hard half can be proved by induction on $\\abs{P}$, by removing a maximal chain or by splitting $P$ along a maximum antichain.\nAnother proof builds a bipartite graph with two copies of $P$ and an edge from $x$ to $y$ when $x \u003c y$; then [[Kőnig's theorem|0JP0ZF57PN]] on matchings gives the chain partition.\nExample: in the set $\\set{1, 2, \\ldots, 10}$ ordered by divisibility, the set $\\set{6, 7, 8, 9, 10}$ is an antichain of size $5$, and $\\set{1, 2, 4, 8}$, $\\set{3, 6}$, $\\set{5, 10}$, $\\set{7}$, $\\set{9}$ are $5$ chains that cover it.\nThe number $w$ is called the width of $P$.\nThe theorem also holds for an infinite partially ordered set of finite width $w$.\nThe dual result, Mirsky's theorem, says that the largest size of a chain equals the least number of antichains that cover $P$.\n\n## Explanations\n\n### L1 human\n\nIn a partial order, some pairs of elements are comparable and others are not.\nA chain is a set of elements that can all be lined up, and an antichain is a set of elements no two of which are comparable.\nDilworth's theorem says that the least number of chains needed to cover everything equals the largest number of pairwise incomparable elements.\nSo a big antichain is the only obstacle to covering with few chains.\n\n### L2 human\n\nDilworth's theorem is one of the min-max theorems of combinatorics, like Kőnig's theorem and Menger's theorem, and each of these can be derived from the others.\nIt gives a quick proof of the [[Erdős–Szekeres theorem|0SNJM8HSD3]]: a sequence of $rs + 1$ distinct real numbers has an increasing subsequence of length $r + 1$ or a decreasing subsequence of length $s + 1$.\nIn algorithms, a minimum chain partition of a finite poset can be computed through a maximum matching in a bipartite graph.\n",
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