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  "body": "\n## Definition\n\nLet $G = (V, E)$ be a finite [[graph|0JD5C82QMB]].\nA list assignment for $G$ is a map $L$ that assigns to each vertex $v$ a finite set $L(v)$ of colours.\nAn $L$-colouring of $G$ is a map $c$ on $V$ such that:\n\n1. $c(v) \\in L(v)$ for every vertex $v$.\n2. $c(u) \\ne c(v)$ for every edge $uv$.\n\nFor an integer $k \\ge 1$, the graph $G$ is $k$-choosable if it has an $L$-colouring for every list assignment $L$ with $\\abs{L(v)} \\ge k$ for all $v$.\nThe choice number $\\operatorname{ch}(G)$ is the least $k$ such that $G$ is $k$-choosable.\n\n## Remarks\n\nIf all lists equal $\\set{1, \\dots, k}$, an $L$-colouring is a proper [[colouring|0874138YP6]] with $k$ colours, so $\\operatorname{ch}(G) \\ge \\chi(G)$, where $\\chi$ is the [[chromatic number|0C5M7AM6FH]].\nExample: every tree is $2$-choosable.\nColour the vertices in an order in which each vertex after the first has exactly one earlier neighbour.\nEach vertex then picks a colour of its list other than the colour of that neighbour.\nNon-example: $K_{3,3}$ is not $2$-choosable, although $\\chi(K_{3,3}) = 2$.\nGive the vertices on each side the lists $\\set{1,2}$, $\\set{1,3}$ and $\\set{2,3}$.\nThe colours used on one side meet each of these three sets, so they include at least two colours.\nTwo of these colours form the list of some vertex on the other side, and that vertex fails clause 2.\nThomassen proved that every planar graph is $5$-choosable.\nChoice number is also called list chromatic number and written $\\chi_\\ell(G)$.\n",
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