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  "body": "\n## Example\n\nLet $X$ be a [[topological space|0CQHFPGJ45]], and let $\\operatorname{RO}(X)$ be the set of its [[regular open sets|833XAA2Q65]], the open sets $U$ with $U = \\operatorname{int} \\overline{U}$.\nOrder $\\operatorname{RO}(X)$ by inclusion and define:\n\n1. $U \\wedge V = U \\cap V$;\n2. $U \\vee V = \\operatorname{int} \\overline{U \\cup V}$;\n3. $\\neg U = \\operatorname{int}(X \\setminus U)$;\n4. $0 = \\emptyset$ and $1 = X$.\n\nThen $\\operatorname{RO}(X)$ is a [[complete Boolean algebra|75BPV95AAS]], called the regular open algebra of $X$.\nThe supremum of a family $(U_i)$ is $\\operatorname{int} \\overline{\\bigcup_i U_i}$, and the infimum is $\\operatorname{int} \\bigcap_i U_i$.\n\nCheck of the key points: the intersection of two regular open sets is regular open.\nAlso $\\operatorname{int} \\overline{W}$ is regular open for every open $W$, so the operations stay in $\\operatorname{RO}(X)$.\nFor $U$ regular open, $\\neg U$ is regular open, $U \\cap \\neg U = \\emptyset$, and $\\operatorname{int} \\overline{U \\cup \\neg U} = X$, because $U \\cup \\neg U$ is dense in $X$.\nFor any family, $\\operatorname{int} \\overline{\\bigcup_i U_i}$ is regular open, contains every $U_i$, and lies in every regular open $V$ that contains all $U_i$, since then $\\overline{\\bigcup_i U_i} \\subseteq \\overline{V}$.\n\n## Remarks\n\nThe join is not the union: in $\\R$, $(0, 1) \\vee (1, 2) = (0, 2)$.\nFor $X = \\R$, the algebra has no atoms, since every nonempty regular open set contains two disjoint open intervals.\nEvery Boolean algebra $B$ embeds densely into the regular open algebra of its [[Stone space|ZV38HA5PAQ]], and this gives the completion of $B$.\nIn Mathlib, the regular elements of a Heyting algebra are `Heyting.Regular`; for the open sets of $X$ they are the regular open sets.\n\n## Explanations\n\n### L1 human\n\nThe open sets of a space are almost a Boolean algebra, but the complement of an open set is not open.\nThe fix is to keep only the regular open sets, which have no cracks, and to fill in the cracks after each operation.\nFor example, on the line, $(0, 1)$ \"or\" $(1, 2)$ is $(0, 2)$: the missing point $1$ is filled in.\nThe result is a Boolean algebra in which every family of elements has a least upper bound.\n",
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