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  "body": "\n## Statement\n\nLet $(P, \u003c)$ be a finite [[strict partial order|60FNZ35H6R]].\nThen $P$ is an [[interval order|VSNQV29KFJ]] if and only if $P$ contains no $\\mathbf{2} + \\mathbf{2}$.\nBeing an interval order means that there are real intervals $[l_x, r_x]$, for $x \\in P$, with $x \u003c y$ if and only if $r_x \u003c l_y$.\nHere $P$ contains a $\\mathbf{2} + \\mathbf{2}$ when there are $a \u003c b$ and $c \u003c d$ in $P$ with $a$ incomparable to $d$ and $c$ incomparable to $b$.\n\n## Remarks\n\nOne direction is direct.\nIf intervals represent $P$ and $a \u003c b$, $c \u003c d$, then $r_a \u003c l_b$ and $r_c \u003c l_d$; if also $a \\not\u003c d$ and $c \\not\u003c b$, then $l_d \\le r_a$ and $l_b \\le r_c$, so $r_a \u003c l_b \\le r_c \u003c l_d \\le r_a$, a contradiction.\nFor the converse, the absence of $\\mathbf{2} + \\mathbf{2}$ means that the sets of predecessors $\\set{y : y \u003c x}$ are totally ordered by inclusion, and so are the sets of successors.\nList the distinct predecessor sets as $D_1 \\subsetneq D_2 \\subsetneq \\cdots \\subsetneq D_k$.\nLet $l_x$ be the index of the predecessor set of $x$, and let $r_x$ be the largest $j$ with $x \\notin D_j$.\nThen $x \u003c y$ means $x \\in D_{l_y}$, which happens exactly when $r_x \u003c l_y$.\n\nExample: the four elements $a \u003c b$, $c \u003c d$ with no other relations form a $\\mathbf{2} + \\mathbf{2}$, so they cannot be represented by intervals.\nThe order on three intervals $[0, 1]$, $[2, 3]$, $[0, 3]$ has the single relation $[0,1] \u003c [2,3]$ and contains no $\\mathbf{2} + \\mathbf{2}$.\n\n## Explanations\n\n### L1 human\n\nThink of events that each take up a time interval, with one event before another when it ends before the other starts.\nThe orders that arise this way are the interval orders.\nFishburn's theorem says that one small pattern is the only obstacle: two pairs $a$ before $b$ and $c$ before $d$, where $a$ overlaps $d$ and $c$ overlaps $b$.\nIntervals on a line cannot do this, and every order that avoids it can be drawn with intervals.\n\n### L2 human\n\nThe theorem is a forbidden-subposet characterization, like the characterization of semiorders by the absence of $\\mathbf{2} + \\mathbf{2}$ and $\\mathbf{3} + \\mathbf{1}$.\nIt shows that interval orders are exactly the posets whose strict down-sets form a chain under inclusion, which gives a fast test for being an interval order.\nInterval orders are the orders whose incomparability graphs are interval graphs, so the theorem connects order theory with the theory of interval graphs.\n",
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  "name": "Fishburn's theorem on interval orders",
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