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  "body": "\n## Statement\n\nLet $X$ be a [[Hausdorff space|0MXB4CKHFW]] and let $K$ be a [[compact subset|00R0X1WGPX]] of $X$.\nThen $K$ is [[closed|0G5AV9Q4RR]] in $X$.\n\n## Remarks\n\nProof: fix $y \\notin K$. For each $x \\in K$, choose disjoint open sets $U_x \\ni x$ and $V_x \\ni y$. Finitely many $U_{x_1}, \\ldots, U_{x_n}$ cover $K$. Then $V_{x_1} \\cap \\cdots \\cap V_{x_n}$ is an open set that contains $y$ and misses $K$.\nThe Hausdorff condition is needed. With the [[indiscrete topology|0XJ02JBD6S]] on a set with two points, each one-point set is compact but not closed.\nEach metric space is Hausdorff, so compact subsets of metric spaces are closed.\n\n## Explanations\n\n### L1 human\n\nIn a Hausdorff space, you can separate a point outside $K$ from each point of $K$ by open sets.\nCompactness lets you do this with finitely many open sets at once, which gives an open set about the point that misses $K$.\n",
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