{
  "aliases": [
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  "body": "\n## Statement\n\nLet $X$ be a [[topological space|0CNQXVBHCX]] and let $K \\subseteq X$.\nThen $K$ is [[compact|00R0X1WGPX]] if and only if it has the property that follows.\nLet $(F_i)_{i \\in I}$ be a family of [[closed subsets|0G5AV9Q4RR]] of $X$ with $K \\cap F_{i_1} \\cap \\cdots \\cap F_{i_n} \\neq \\emptyset$ for all finitely many indices $i_1, \\ldots, i_n \\in I$.\nThen also $K \\cap \\bigcap_{i \\in I} F_i \\neq \\emptyset$.\n\n## Remarks\n\nA family of sets has the finite intersection property when each finite subfamily has a nonempty intersection.\nProof: take complements. The sets $X \\setminus F_i$ cover $K$ if and only if $K \\cap \\bigcap_i F_i = \\emptyset$. Then use [[compactness via open covers|0P6PMJB4MA]].\nFor $K = X$: a space is compact if and only if each family of closed sets with the finite intersection property has nonempty intersection.\n[[Cantor's intersection theorem|0EZBG8SKJM]] is a special case.\n\n## Explanations\n\n### L1 human\n\nThis is the dual form of compactness, with closed sets instead of open sets.\nSuppose that each finite subfamily of the closed sets has a common point in $K$. Then the whole family has a common point in $K$.\n",
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