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  "body": "\n## Proof\n\nWe show that $f$ is constant near each point, and then use connectedness.\n\n1. Let $U$ and $f$ be as in the statement. If $U$ is empty, there is nothing to prove. So fix $a \\in U$, and put $c = f(a)$.\n2. Let $z \\in U$. Since $U$ is [[open|01ZQQDCERP]], there is $\\rho \u003e 0$ such that the [[open ball|0DC3MPR7C9]] $B(z, \\rho)$ lies in $U$.\n3. Let $y \\in B(z, \\rho)$ and $0 \\le t \\le 1$. Then $\\|z + t (y - z) - z\\| = t \\|y - z\\| \u003c \\rho$. So the segment from $z$ to $y$ lies in $B(z, \\rho) \\subseteq U$.\n4. On this segment $Df = 0$, so $\\|Df\\| \\le 0$ there. The map $f$ is [[differentiable|0135XZZQ8M]] on $U$. By the [[mean value inequality|02CYB184RC]] with $M = 0$, $\\|f(y) - f(z)\\| \\le 0$. So $f(y) = f(z)$.\n5. Thus $f$ is constant on the ball $B(z, \\rho)$, for each $z \\in U$ and its $\\rho$ from step 2.\n6. Put $S = \\{x \\in U : f(x) = c\\}$ and $T = U \\setminus S$. Then $S \\cap T = \\varnothing$, $S \\cup T = U$, and $a \\in S$.\n7. The set $S$ is open. If $z \\in S$, then by step 5 each $y \\in B(z, \\rho)$ has $f(y) = f(z) = c$. So $B(z, \\rho) \\subseteq S$.\n8. The set $T$ is open. If $z \\in T$, then by step 5 each $y \\in B(z, \\rho)$ has $f(y) = f(z) \\ne c$. So $B(z, \\rho) \\subseteq T$.\n9. The sets $S$ and $T$ are open in $\\mathbb{R}^n$ and lie in $U$, so they are open in $U$ with the subspace topology.\n10. The set $U$ is [[connected|0R37DFJKFW]], so $U$ is not the union of two disjoint nonempty open subsets. This is the definition of a [[connected space|0D939CMME7]]. Since $S \\ne \\varnothing$ by step 6, steps 6 to 9 force $T = \\varnothing$.\n11. So $S = U$, that is, $f(x) = c$ for each $x \\in U$. Hence $f$ is constant on $U$.\n\n## Idea\n\nA zero derivative makes $f$ constant on each small ball, by the mean value inequality. So the set where $f$ takes the value $f(a)$ and the set where it does not are both open. A connected set cannot be split into two disjoint nonempty open pieces, so the second set is empty.\n",
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