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  "body": "\n## Definition\n\nLet $X$ be a set and let $\\mathcal{P}(X)$ be its [[power set|00E0YZCERH]].\nA map $c \\colon \\mathcal{P}(X) \\to \\mathcal{P}(X)$ satisfies the *Kuratowski closure axioms* if, for all subsets $A, B \\subseteq X$:\n\n1. $c(\\emptyset) = \\emptyset$.\n2. $A \\subseteq c(A)$.\n3. $c(c(A)) = c(A)$.\n4. $c(A \\cup B) = c(A) \\cup c(B)$.\n\nSuch a map is a *closure operator* on $X$.\n\n## Remarks\n\nGiven a closure operator $c$, the sets $F$ with $c(F) = F$ are the [[closed sets|0ZCK8DQYN1]] of a unique topology on $X$.\nThe [[closure|0QEAJB91GZ]] of each $A$ in this topology is $c(A)$.\nConversely, the closure in any [[topological space|0CQHFPGJ45]] satisfies the four axioms.\nSo closure operators on $X$ and topologies on $X$ correspond one to one.\n\nAxiom 4 implies that $c$ is monotone: if $A \\subseteq B$, then $c(A) \\subseteq c(B) = c(A) \\cup c(B \\setminus A)$.\n\nExample: on $\\R$, let $c(A)$ be the set of limits of convergent sequences of points of $A$.\nThis $c$ is the usual closure, and it satisfies the four axioms.\n\nNon-example: fix a point $x_0$ of a nonempty set $X$ and let $c(A) = A \\cup \\set{x_0}$.\nAxioms 2, 3 and 4 hold, but $c(\\emptyset) = \\set{x_0} \\ne \\emptyset$, so axiom 1 fails.\n",
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