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  "body": "\n## Statement\n\nLet $A$ be a set and let $C$ be a [[closure operator|2HJGRABSPZ]] on $A$.\nAssume that $C$ is algebraic: $C(X) = \\bigcup \\set{C(Y) : Y \\subseteq X,\\ Y \\text{ finite}}$ for every $X \\subseteq A$.\nThen there is an [[algebra|QJGPT2W0FT]] $\\mathbf{A}$ with universe $A$ and finitary operations with the following property.\nA set $X \\subseteq A$ is a subuniverse of $\\mathbf{A}$ if and only if $C(X) = X$.\n\n## Remarks\n\nProof idea.\nFor each finite set $B = \\set{b_1, \\dots, b_n} \\subseteq A$ with $n$ distinct elements and each $c \\in C(B)$, add an $n$-ary operation.\nIt sends $(b_1, \\dots, b_n)$ to $c$ and sends every other tuple to its first entry; for $n = 0$ it is the constant $c$.\nA closed set is closed under these operations because $C(B) \\subseteq C(X)$ when $B \\subseteq X$.\nA set closed under them contains $C(B)$ for each finite $B$ inside it, so it is closed because $C$ is algebraic.\n\nConsequence: every [[algebraic lattice|3J0BTXSCVN]] is isomorphic to the [[subalgebra lattice|T0TNG7AGXT]] of some algebra.\nIndeed, an algebraic lattice is isomorphic to the lattice of closed sets of some algebraic closure operator.\n\nThe hypothesis that $C$ is algebraic is necessary.\nFor any algebra with finitary operations, the union of a chain of subuniverses is a subuniverse.\nOn $A = \\N$, let $C(X) = X$ for finite $X$ and $C(X) = \\N$ for infinite $X$.\nThis is a closure operator, and it is not algebraic.\nThe closed sets $\\set{0, 2, \\dots, 2k}$ form a chain whose union, the set of even numbers, is not closed.\nSo no algebra on $\\N$ has exactly the closed sets of $C$ as its subuniverses.\n\n## Explanations\n\n### L1 human\n\nA closure operator takes a set of elements and adds everything that the set \"forces\".\nIt is algebraic when whatever a set forces is already forced by some finite part of it.\nThe theorem says that every such operator comes from an algebra: the closed sets are exactly the subsets closed under some operations.\nSo the abstract idea of \"generating\" is no more general than generating with operations.\n\n### L2 human\n\nThe theorem is the converse of a basic fact: in any algebra, the subuniverse operator $\\operatorname{Sg}$ is an algebraic closure operator.\nTogether they show that algebraic closure systems, subuniverse lattices and algebraic lattices describe the same objects.\nA similar but harder result for congruences is the theorem of Grätzer and Schmidt, which says that every algebraic lattice is a [[congruence lattice|REE07CSXHR]].\n",
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