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  "body": "\n## Statement\n\nLet $(X, d)$ be a [[metric space|063W5JMY85]] and let $K \\subseteq X$.\nThen $K$ is [[compact|00R0X1WGPX]] if and only if $K$ is [[sequentially compact|0CARW59WD2]].\n\n## Remarks\n\nOne proof of the hard direction uses the [[Lebesgue number lemma|040CFNJEAP]] and the fact that a sequentially compact metric space is [[totally bounded|0WE5BBHGM6]].\nBoth conditions are also equivalent to: $K$ is complete and totally bounded. This is the statement [[compact iff complete and totally bounded|054Q5PW5GX]].\nWithout a metric, the two notions differ. For example, the product $\\{0, 1\\}^{[0, 1]}$ is compact but not sequentially compact.\n\n## Explanations\n\n### L1 human\n\nIn a metric space, you can test compactness with sequences. A set is compact exactly when each sequence in it has a subsequence with a limit in the set.\nThis test is often easier than the test with open covers.\n",
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