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  "body": "\n## Proof\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $I$ be a nonempty set, and let $(F_i)_{i \\in I}$ be a family of [[closed subsets|0G5AV9Q4RR]] of $X$.\n2. Put $F = \\bigcap_{i \\in I} F_i$, the [[intersection|0839XNQXDK]] of the family.\n3. We claim that $X \\setminus F = \\bigcup_{i \\in I} (X \\setminus F_i)$, a [[union|05GK32H72H]] of [[complements|0B1FSS1VZW]].\n4. Let $x \\in X$. Then $x \\notin F$ if and only if $x \\notin F_j$ for some $j \\in I$.\n5. This holds if and only if $x \\in X \\setminus F_j$ for some $j \\in I$, that is, $x \\in \\bigcup_{i \\in I} (X \\setminus F_i)$. This proves the claim.\n6. Each $F_i$ is closed, so each $X \\setminus F_i$ is open, by the definition of a closed set.\n7. By [[union of open sets is open|0HEF46RPEK]], the set $\\bigcup_{i \\in I} (X \\setminus F_i)$ is open. By step 3, $X \\setminus F$ is open.\n8. By the definition of a closed set, $F$ is closed in $X$.\n\n## Idea\n\nComplements turn intersections into unions. A union of open sets is open, so the complement of the intersection is open. That is what it means for the intersection to be closed.\n",
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