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  "body": "\n## Statement\n\nLet $n \\ge 1$ be an integer and give $\\mathbb{R}^n$ the [[Euclidean metric|0E1MMCAY5B]].\nA subset $K$ of $\\mathbb{R}^n$ is [[compact|00R0X1WGPX]] if and only if $K$ is [[closed|0G5AV9Q4RR]] and [[bounded|0P953MG05E]].\n\n## Remarks\n\nThe direct part holds in each metric space: [[compact subsets of a metric space are bounded|0R7BTEWCJ2]], and they are closed.\nThe converse uses the completeness of $\\mathbb{R}$. A bounded set lies in a cube $[-M, M]^n$, which is compact. A closed subset of a compact set is compact.\nThe converse fails in general metric spaces. In the space of bounded real sequences with the sup metric, the closed unit ball is closed and bounded but not compact.\nMetric spaces in which closed bounded sets are compact are the [[proper metric spaces|0Z4V98VN5E]].\n\n## History\n\nDirichlet used finite subcovers of a closed interval in his 1852 lectures, to prove that continuous functions on such intervals are uniformly continuous.\nÉmile Borel stated and proved a version for countable covers in 1895. Cousin, Lebesgue and Schoenflies later extended it to arbitrary covers.\n\n## Explanations\n\n### L1 human\n\nIn $\\mathbb{R}^n$, you can recognize compact sets by two easy tests: they contain their limit points and they fit in a ball.\nFor example, the closed disc and the sphere are compact, but the open disc and the whole plane are not.\n",
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