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  "body": "\n## Statement\n\nLet $X$ be a [[Baire space|0D5KR33H7X]] and let $(U_n)_{n \\in \\mathbb{N}}$ be a sequence of open subsets of $X$, each [[dense|0D198GPMTJ]] in $X$.\nThen $\\bigcap_{n} U_n$ is dense in $X$.\n\n## Remarks\n\nProof: let $F_n = X \\setminus U_n$. Each $F_n$ is closed, and $\\operatorname{int} F_n = X \\setminus \\overline{U_n} = \\emptyset$. Thus $\\operatorname{int} \\bigcup_n F_n = \\emptyset$. Its complement $\\bigcap_n U_n$ therefore has closure $X$.\nThe converse also holds, so this property characterises Baire spaces.\nIn $\\mathbb{R}$, for an enumeration $q_1, q_2, \\ldots$ of $\\mathbb{Q}$, the sets $U_n = \\mathbb{R} \\setminus \\{q_n\\}$ are open and dense. Their intersection is the set of irrational numbers, which is dense.\nThe intersection need not be open.\n\n## Explanations\n\n### L1 human\n\nEach dense open set leaves out only a thin closed set.\nIn a Baire space, removing countably many thin sets still leaves a set that is everywhere.\n",
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