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  "body": "\n## Definition\n\nLet $L$ be a [[complete lattice|D7A0ZGZT5S]].\nLet $J$ be a set, let $(K_j)_{j \\in J}$ be a family of sets, and let $x_{j,k} \\in L$ for $j \\in J$ and $k \\in K_j$.\nLet $F$ be the set of choice functions $f$ on $J$ with $f(j) \\in K_j$ for every $j \\in J$.\nThe lattice $L$ is *completely distributive* if for all such data\n$$\\bigwedge_{j \\in J} \\bigvee_{k \\in K_j} x_{j,k} = \\bigvee_{f \\in F} \\bigwedge_{j \\in J} x_{j, f(j)}.$$\n\n## Remarks\n\nExample: the [[power set|00E0YZCERH]] of a set $X$.\nA point $x$ lies in $\\bigcap_j \\bigcup_k x_{j,k}$ if and only if for each $j$ there is $k = f(j)$ with $x \\in x_{j, f(j)}$.\nThis is membership in the right side, where the axiom of choice picks $f$.\nNon-example: the lattice of open subsets of $\\R$ is a frame but not completely distributive.\nTake $J = \\N_{\\ge 1}$, $K_j = \\set{1, 2}$, $x_{j,1} = (-1/j, 1/j)$ and $x_{j,2} = \\R \\setminus \\set{0}$.\nThe left side is $\\R$.\nEach meet on the right is the interior of an intersection and never contains $0$, so the right side is $\\R \\setminus \\set{0}$.\nComplete distributivity is self-dual: the condition is equivalent to the same identity with $\\bigwedge$ and $\\bigvee$ exchanged.\nEvery completely distributive lattice is a [[distributive lattice|0HM0HQCK6N]].\n",
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