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  "body": "\n## Proof\n\n1. Let $x, x' \\in I$ with $x \\ne x'$. One is smaller, so $f(x) \\ne f(x')$ since $f$ is [[strictly increasing|0TR9SR9FTB]]. So $f$ is injective, and $f^{-1} \\colon J \\to I$ is defined.\n2. Claim: $J$ is an [[interval|0TK8Y3J5RW]]. Let $y_1, y_2 \\in J$ and $y_1 \\le z \\le y_2$. If $z = y_1$ or $z = y_2$, then $z \\in J$.\n3. Otherwise $y_1 \u003c z \u003c y_2$. Write $y_1 = f(x_1)$ and $y_2 = f(x_2)$ with $x_1, x_2 \\in I$. If $x_2 \\le x_1$, then $y_2 \\le y_1$, which is false. So $x_1 \u003c x_2$.\n4. Since $I$ is an interval, $[x_1, x_2] \\subseteq I$, and $f$ is [[continuous|0P27W9RW0G]] on it. The [[intermediate value theorem|00KAFBR32E]] gives $c \\in [x_1, x_2]$ with $f(c) = z$. So $z \\in J$, and $J$ is an interval.\n5. Write $g = f^{-1}$. Claim: $g$ is strictly increasing. Let $y \u003c y'$ in $J$. If $g(y) \\ge g(y')$, then $y = f(g(y)) \\ge f(g(y')) = y'$, which is false. So $g(y) \u003c g(y')$.\n6. Now fix $y_0 \\in J$ and $\\varepsilon \u003e 0$, and put $x_0 = g(y_0)$. We look for $\\delta \u003e 0$ as in the [[epsilon-delta form of continuity|0W8YHHGCJG]].\n7. If $I$ has a point $p \u003c x_0$, put $x_- = \\max(p, x_0 - \\frac{\\varepsilon}{2})$. Then $p \\le x_- \u003c x_0$, so $x_- \\in I$. Also $x_- \u003e x_0 - \\varepsilon$. Put $\\delta_1 = y_0 - f(x_-)$, which is positive. If $I$ has no point below $x_0$, put $\\delta_1 = 1$.\n8. If $I$ has a point $q \u003e x_0$, put $x_+ = \\min(q, x_0 + \\frac{\\varepsilon}{2})$. Then $x_+ \\in I$ and $x_0 \u003c x_+ \u003c x_0 + \\varepsilon$. Put $\\delta_2 = f(x_+) - y_0 \u003e 0$. Otherwise put $\\delta_2 = 1$.\n9. Let $\\delta = \\min(\\delta_1, \\delta_2) \u003e 0$. Let $y \\in J$ with $\\abs{y - y_0} \u003c \\delta$, and put $x = g(y)$. If $y = y_0$, then $x = x_0$.\n10. Suppose $y \u003c y_0$. By step 5, $x \u003c x_0$, so $I$ has a point below $x_0$, and $x_-$ was defined in step 7. Then $y \u003e y_0 - \\delta_1 = f(x_-)$.\n11. By step 5, $x = g(y) \u003e g(f(x_-)) = x_-$. So $x_0 - \\varepsilon \u003c x_- \u003c x \u003c x_0$, and $\\abs{x - x_0} \u003c \\varepsilon$.\n12. Suppose $y \u003e y_0$. In the same way, $x \u003e x_0$, $x_+$ was defined, and $y \u003c f(x_+)$. So $x_0 \u003c x \u003c x_+ \u003c x_0 + \\varepsilon$.\n13. In all cases $\\abs{g(y) - g(y_0)} \u003c \\varepsilon$. So $g$ is continuous at $y_0$, for each $y_0 \\in J$.\n14. Hence $J$ is an interval, and $f^{-1} \\colon J \\to I$ is continuous and strictly increasing.\n\n## Idea\n\nThe image is an interval because the intermediate value theorem fills in every value between two attained values. For continuity of the inverse at $y_0 = f(x_0)$, take points $x_\\pm$ of $I$ within $\\varepsilon$ of $x_0$ on each side. Their images bracket $y_0$, and since $f^{-1}$ preserves order, every $y$ in that bracket goes back between $x_-$ and $x_+$.\n",
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