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  "body": "\n## Statement\n\nLet $a \\le b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be [[continuous|0P27W9RW0G]].\nThen $f$ is bounded. That is, there is $K$ with $|f(x)| \\le K$ for all $x \\in [a, b]$.\n\n## Remarks\n\nSuppose not. Then there are points $x_n \\in [a, b]$ with $|f(x_n)| \u003e n$. By the [[Bolzano–Weierstrass theorem|0KDJD249DT]], a subsequence converges to some $c \\in [a, b]$. Continuity at $c$ gives a contradiction.\nThe interval must be closed and bounded. The function $1/x$ on $(0, 1]$ is continuous but not bounded.\nMathlib states a form for compact sets, as `IsCompact.bddAbove_image`.\n\n## Explanations\n\n### L1 human\n\nOn a closed bounded interval, a continuous function cannot shoot off to infinity.\nIts graph stays between two horizontal lines.\n",
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  "name": "Boundedness theorem",
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