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  "body": "\n## Proof\n\nWe first show that $f$ is monotone on any three points, using the [[intermediate value theorem|00KAFBR32E]].\n\n1. Let $x \u003c y \u003c z$ be points of $I$. Since $I$ is an [[interval|0TK8Y3J5RW]], $[x, z] \\subseteq I$. The restriction of $f$ to any closed subinterval of $[x, z]$ is [[continuous|0P27W9RW0G]].\n2. The values $f(x)$, $f(y)$, $f(z)$ are distinct, since $f$ is [[injective|0PVG1CRAJN]].\n3. Claim: $f(y)$ lies strictly between $f(x)$ and $f(z)$. Suppose not. Then $f(y)$ is larger than both $f(x)$ and $f(z)$, or smaller than both.\n4. Case $f(y) \u003e f(x)$ and $f(y) \u003e f(z)$, with $f(x) \u003c f(z)$. Then $f(x) \u003c f(z) \u003c f(y)$. The intermediate value theorem on $[x, y]$ gives $t \\in [x, y]$ with $f(t) = f(z)$. But $t \\le y \u003c z$, which contradicts injectivity.\n5. Case $f(y) \u003e f(x)$ and $f(y) \u003e f(z)$, with $f(z) \u003c f(x)$. Then $f(z) \u003c f(x) \u003c f(y)$. The intermediate value theorem on $[y, z]$ gives $t \\in [y, z]$ with $f(t) = f(x)$. But $t \\ge y \u003e x$, which contradicts injectivity.\n6. Case $f(y)$ smaller than both. Then $-f(y)$ is larger than $-f(x)$ and $-f(z)$, and $-f$ is continuous and injective. Steps 4 and 5 for $-f$ give a contradiction.\n7. So the claim holds. Hence $f(y) - f(x)$ and $f(z) - f(y)$ are both positive or both negative.\n8. Now let $w_1 \u003c w_2 \u003c \\cdots \u003c w_m$ be finitely many points of $I$, with $m \\ge 2$. Put $d_i = f(w_{i+1}) - f(w_i)$ for $1 \\le i \u003c m$.\n9. Each $d_i$ is nonzero by step 2. By step 7 for the triple $w_i \u003c w_{i+1} \u003c w_{i+2}$, $d_i$ and $d_{i+1}$ have the same sign. By induction on $i$, all $d_i$ have the sign of $d_1$.\n10. If $d_1 \u003e 0$, then $f(w_1) \u003c f(w_2) \u003c \\cdots \u003c f(w_m)$. If $d_1 \u003c 0$, then $f(w_1) \u003e \\cdots \u003e f(w_m)$.\n11. If $I$ has fewer than two points, $f$ is [[strictly increasing|0TR9SR9FTB]] because there is nothing to check. Otherwise fix $p \u003c q$ in $I$. By step 2, $f(p) \u003c f(q)$ or $f(p) \u003e f(q)$.\n12. Suppose $f(p) \u003c f(q)$. Let $u \u003c v$ be points of $I$. List the set $\\set{p, q, u, v}$ in increasing order as in step 8.\n13. By step 10, $f$ is strictly increasing on this list or strictly decreasing on it. Since $p \u003c q$ and $f(p) \u003c f(q)$, it is strictly increasing. So $f(u) \u003c f(v)$.\n14. Thus $f$ is strictly increasing on $I$. If instead $f(p) \u003e f(q)$, the same argument shows that $f$ is strictly decreasing.\n15. In both cases $f$ is strictly monotone.\n\n## Idea\n\nIf the middle one of three points had the largest or smallest value, the intermediate value theorem would make $f$ take some value twice. So $f$ goes the same way along any three points, and hence along any finite list of points. Comparing an arbitrary pair with one fixed pair then fixes the direction everywhere.\n",
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