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  "body": "\n## Statement\n\nLet $X$ be a [[topological space|0CQHFPGJ45]], and let $\\mathcal S$ be a [[subbasis|09S8DV34Y0]] of its topology.\nSuppose that every [[cover|049PXYHW1T]] of $X$ by members of $\\mathcal S$ has a finite subcover.\nThen $X$ is [[compact|0TVJZYB6SH]].\n\n## Remarks\n\nSuppose $X$ is not compact.\nBy [[Zorn's lemma|04RXTQA5ZA]], there is an open cover $\\mathcal C$ without finite subcover that is maximal for inclusion among such covers.\nFor $x \\in U \\in \\mathcal C$, choose $S_1, \\ldots, S_n \\in \\mathcal S$ with $x \\in S_1 \\cap \\cdots \\cap S_n \\subseteq U$.\nIf no $S_i$ were in $\\mathcal C$, maximality would give finite sets $\\mathcal C_i \\subseteq \\mathcal C$ with $S_i \\cup \\bigcup \\mathcal C_i = X$, and then $U$ with all $\\mathcal C_i$ would be a finite subcover.\nSo $\\mathcal C \\cap \\mathcal S$ covers $X$, and by hypothesis it has a finite subcover, a contradiction.\n\nExample: the sets $[0, c)$ and $(c, 1]$ for $0 \u003c c \u003c 1$ form a subbasis of $[0, 1]$.\nA cover by such sets contains some $[0, c)$ and some $(c', 1]$ with $c' \u003c c$, by a supremum argument, so $[0, 1]$ is compact.\nThe theorem also gives a direct proof of [[Tychonoff's theorem|0JTRSJKCFB]].\n\nThe subbasis hypothesis is needed.\nOn $\\R$, the family $\\set{\\R}$ has the finite subcover property, but it is not a subbasis of the usual topology, and $\\R$ is not compact.\n\n## Explanations\n\n### L1 human\n\nTo show that a space is compact, one must handle every open cover.\nThe theorem says that it is enough to handle covers made of the simplest open sets, those of a chosen subbasis.\nThis cuts down the work a lot when the subbasis is easy to describe.\n\n### L2 human\n\nThe product topology has a natural subbasis: the preimages of open sets under single projections.\nWith the theorem, compactness of a product of compact spaces reduces to covers by such sets, which is a short argument; this is one standard proof of Tychonoff's theorem.\nThe proof uses Zorn's lemma, and the theorem is equivalent to the ultrafilter lemma over the axioms of set theory without choice.\n",
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