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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 each family $(U_i)_{i \\in I}$ of open subsets of $X$ with $K \\subseteq \\bigcup_{i \\in I} U_i$ has a finite subfamily $U_{i_1}, \\ldots, U_{i_n}$ with $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$.\n\n## Remarks\n\nProof: the open sets of the [[subspace topology|0D5YHQGDJY]] on $K$ are the sets $U \\cap K$ with $U$ open in $X$. So covers of $K$ by open sets of $K$ and covers by open sets of $X$ match each other.\nFor $K = X$, the statement is the definition of a [[compact space|0QPHW703PV]].\nThus compactness of a subset does not depend on the space around it. If $K \\subseteq Y \\subseteq X$, then $K$ is compact in $Y$ if and only if it is compact in $X$.\n\n## Explanations\n\n### L1 human\n\nTo test if a subset is compact, you can use open sets of the big space.\nEach time you cover the subset with open sets, finitely many of them must already be enough.\n",
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