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  "body": "\n## Problem\n\nLet $X$ be a [[topological space|0CQHFPGJ45]] and let $A \\subseteq X$.\nApply to $A$, again and again and in any order, the two operations of taking the [[closure|0QEAJB91GZ]] and taking the [[complement|0RKF1ZT5JJ]] in $X$.\nHow many distinct subsets of $X$ can arise in this way, at most?\n\n## Remarks\n\nThe problem is solved: at most $14$ distinct sets arise, counting $A$ itself, and the bound $14$ is attained in $\\R$.\nThe [[interior|0RZH6W98HK]] of $A$ is the complement of the closure of the complement of $A$.\nWith $k$ for closure and $i$ for interior, every set that arises is one of $A, kA, iA, ikA, kiA, kikA, ikiA$ or the complement of one of them.\nThe key identity is $kikik A = kik A$, which stops the list.\n\nIn $\\R$, the set\n$$A = (0, 1) \\cup (1, 2) \\cup \\set{3} \\cup ([4, 5] \\cap \\Q)$$\ngives $14$ distinct sets.\nFor it, $kA = [0, 2] \\cup \\set{3} \\cup [4, 5]$, $iA = (0, 1) \\cup (1, 2)$, $ikA = (0, 2) \\cup (4, 5)$, $kiA = [0, 2]$, $kikA = [0, 2] \\cup [4, 5]$ and $ikiA = (0, 2)$.\nThese seven sets are distinct and bounded, and their seven complements are unbounded, so all $14$ are distinct.\n\n## Explanations\n\n### L1 human\n\nStart with a set and keep pressing two buttons: \"close it\" and \"take what is outside\".\nOne might expect to get infinitely many different sets this way.\nIn fact the process always cycles back after a short while, and one can never get more than $14$ different sets.\nA clever subset of the real line, made of intervals, a single point and some rational numbers, shows that all $14$ can really appear.\n",
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      "first_stated": "1922",
      "posed_by": "Kazimierz Kuratowski"
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    "name": "Kuratowski's closure-complement problem",
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  "name": "Kuratowski's closure-complement problem",
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