Zero derivative on a connected open set implies constant

Statement

Let $U \subseteq \mathbb{R}^n$ be a connected open set and let $f \colon U \to \mathbb{R}^m$ be differentiable on $U$. Suppose that $Df(x) = 0$ for each $x \in U$. Then $f$ is constant on $U$.

Remarks

The proof fixes $a \in U$ and shows that the set $S = \{x \in U : f(x) = f(a)\}$ is open and closed in $U$. It is open by the mean value inequality with $M = 0$ on small balls, and closed because $f$ is continuous. As $U$ is connected and $a \in S$, $S = U$. Connectedness is needed: on $U = (0, 1) \cup (2, 3)$, the function that is $0$ on the first interval and $1$ on the second has derivative $0$ but is not constant. For real-valued $f$, the condition $Df = 0$ means that all partial derivatives of $f$ vanish on $U$.

Explanations

L1 human

Suppose that $f$ does not change in any direction at any point, and that the domain is in one piece. Then $f$ cannot change at all. On a domain in two pieces, $f$ can take a different constant value on each piece.