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  "body": "\n## Proof\n\nWe combine the extreme value theorem with the intermediate value theorem.\n\n1. Let $a \\le b$ and let $f \\colon [a, b] \\to \\R$ be [[continuous|0P27W9RW0G]].\n2. By the [[extreme value theorem|0WS1A8E2P9]], there are $c, d \\in [a, b]$ with $f(c) \\le f(x) \\le f(d)$ for all $x \\in [a, b]$.\n3. Let $m = f(c)$ and $M = f(d)$. These are the minimum and the maximum of $f$ on $[a, b]$, and step 2 gives $f([a, b]) \\subseteq [m, M]$.\n4. Conversely, let $y \\in [m, M]$. If $c = d$, then $m = M$, so $y = f(c) \\in f([a, b])$.\n5. Otherwise, let $p = \\min(c, d)$ and $q = \\max(c, d)$. Then $a \\le p \u003c q \\le b$.\n6. Let $h$ be the restriction of $f$ to $[p, q]$. Let $z \\in [p, q]$ and let $V$ be an open interval containing $h(z)$.\n7. Continuity of $f$ at $z$ gives an open interval $U$ containing $z$ with $f(x) \\in V$ for all $x \\in U \\cap [a, b]$. Since $U \\cap [p, q] \\subseteq U \\cap [a, b]$, $h$ is [[continuous at|0EEE2CZB2A]] $z$.\n8. So $h$ is continuous. The numbers $h(p)$ and $h(q)$ are $m$ and $M$ in some order, so $y$ lies between $h(p)$ and $h(q)$.\n9. By the [[intermediate value theorem|00KAFBR32E]], there is $x \\in [p, q]$ with $f(x) = h(x) = y$. So $y \\in f([a, b])$.\n10. By steps 3, 4 and 9, $f([a, b]) = [m, M]$.\n\n## Idea\n\nThe extreme value theorem gives the smallest and largest values $m$ and $M$, both attained. The intermediate value theorem, used on the interval between a minimum point and a maximum point, fills in every value between them. Nothing outside $[m, M]$ can be hit, by the choice of $m$ and $M$.\n",
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