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  "body": "\n## Proof\n\nWe pass to complements and use the definition of a Baire space.\n\n1. Let $X$ be a [[Baire space|0D5KR33H7X]], and let $U_n$ be open and [[dense|0D198GPMTJ]] in $X$ for each $n \\in \\N$. Let $F_n = X \\setminus U_n$, a [[closed set|0G5AV9Q4RR]].\n2. Fix $n$, and let $V$ be an open set with $V \\subseteq F_n$. Then $V \\cap U_n = \\emptyset$, so $U_n \\subseteq X \\setminus V$.\n3. The set $X \\setminus V$ is closed. Since the [[closure is the smallest closed superset|0SHC6Z304Z]], $\\overline{U_n} \\subseteq X \\setminus V$.\n4. Since $U_n$ is dense, $\\overline{U_n} = X$. So $X \\subseteq X \\setminus V$, which gives $V = \\emptyset$.\n5. The [[interior|0C38S0KMJK]] $\\operatorname{int} F_n$ is an open subset of $F_n$, because the [[interior is the largest open subset|0HAZEFH137]]. By steps 2 to 4, $\\operatorname{int} F_n = \\emptyset$.\n6. Since $X$ is a Baire space, $\\operatorname{int} \\bigcup_n F_n = \\emptyset$.\n7. Let $D = \\bigcap_n U_n$. A point lies outside $D$ if and only if it lies outside some $U_n$. So $X \\setminus D = \\bigcup_n F_n$.\n8. Let $\\overline{D}$ be the [[closure|0DRKT1QJR7]] of $D$. By the same statement as in step 3, it is closed and contains $D$. So $X \\setminus \\overline{D}$ is open.\n9. Since $D \\subseteq \\overline{D}$, we get $X \\setminus \\overline{D} \\subseteq X \\setminus D = \\bigcup_n F_n$.\n10. By steps 8 and 9, and since the interior is the largest open subset, $X \\setminus \\overline{D} \\subseteq \\operatorname{int} \\bigcup_n F_n = \\emptyset$.\n11. So $\\overline{D} = X$, and $\\bigcap_n U_n$ is dense in $X$.\n\n## Idea\n\nThe complement of a dense open set is a closed set with empty interior.\nThe Baire property says a countable union of such sets has empty interior. Taking complements again, the intersection of the open sets is dense.\n",
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