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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 there are $c, d \\in [a, b]$ with $f(c) \\le f(x) \\le f(d)$ for all $x \\in [a, b]$.\n\n## Remarks\n\nBy the [[boundedness theorem|0ATVXC3NCC]], $M = \\sup f([a, b])$ is finite. If $f(x) \u003c M$ for all $x$, then $1/(M - f(x))$ is continuous and unbounded, which is a contradiction.\nEach hypothesis is needed. The function $x$ on $[0, 1)$ has no maximum, and $x$ on $[0, \\infty)$ is not bounded above.\nMathlib states it for compact sets, as `IsCompact.exists_isMinOn` and `IsCompact.exists_isMaxOn`.\n\n## History\n\nBernard Bolzano proved the theorem in the 1830s. His proof stayed unpublished until 1930.\n\n## Explanations\n\n### L1 human\n\nOn a closed bounded interval, a continuous function has a highest value and a lowest value.\nIts graph has a top point and a bottom point.\n",
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