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  "body": "\n## Proof\n\nWe take the supremum of the set where $f$ is below $y$.\n\n1. Let $a \u003c b$, let $f \\colon [a, b] \\to \\R$ be [[continuous|0P27W9RW0G]], and let $y$ lie between $f(a)$ and $f(b)$.\n2. If $y = f(a)$, take $c = a$. If $y = f(b)$, take $c = b$. So we may assume that $y$ lies strictly between $f(a)$ and $f(b)$.\n3. If $f(a) \u003e y \u003e f(b)$, replace $f$ by $-f$ and $y$ by $-y$. Since $\\abs{-f(x) - (-f(z))} = \\abs{f(x) - f(z)}$, the [[epsilon-delta form|0W8YHHGCJG]] shows that $-f$ is continuous. A point with $-f(c) = -y$ has $f(c) = y$.\n4. So we may assume $f(a) \u003c y \u003c f(b)$. Let $S = \\set{x \\in [a, b] : f(x) \u003c y}$.\n5. The set $S$ contains $a$, and $b$ is an [[upper bound|0B5H2V4JW8]] of $S$. By the [[completeness axiom|0AXWET80YW]], $S$ has a [[supremum|0BTMQ04556]] $c$. Since $a \\in S$ and $b$ is an upper bound, $a \\le c \\le b$.\n6. Suppose that $f(c) \u003c y$. Then $c \\ne b$, since $f(b) \u003e y$. So $c \u003c b$.\n7. In this case, the epsilon-delta form with $\\varepsilon = y - f(c)$ gives $\\delta \u003e 0$ with $f(x) \u003c f(c) + \\varepsilon = y$ for all $x \\in [a, b]$ with $\\abs{x - c} \u003c \\delta$.\n8. Let $x = \\min(c + \\delta/2, b)$. Then $c \u003c x \\le b$ and $\\abs{x - c} \u003c \\delta$, so $f(x) \u003c y$ and $x \\in S$. This contradicts $x \u003e c = \\sup S$. Hence $f(c) \\ge y$.\n9. Suppose that $f(c) \u003e y$. The epsilon-delta form with $\\varepsilon = f(c) - y$ gives $\\delta \u003e 0$ with $f(x) \u003e y$ for all $x \\in [a, b]$ with $\\abs{x - c} \u003c \\delta$.\n10. By the [[approximation property of the supremum|0M8TGGP9JP]], there is $s \\in S$ with $c - \\delta \u003c s \\le c$. Then $s \\in [a, b]$ and $\\abs{s - c} \u003c \\delta$, so $f(s) \u003e y$.\n11. But $s \\in S$ means $f(s) \u003c y$. This contradiction shows $f(c) \\le y$.\n12. By steps 8 and 11, $f(c) = y$, with $c \\in [a, b]$.\n\n## Idea\n\nLet $c$ be the last point up to which $f$ can still be below $y$, namely the supremum of the set where $f \u003c y$. If $f(c) \u003c y$, continuity would push the set past $c$. If $f(c) \u003e y$, continuity would keep the set away from $c$. So $f(c) = y$, and completeness of $\\R$ is what makes $c$ exist.\n",
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