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Let $E$ and $F$ be Banach spaces and let $X \subset E$ be a connected open subset. I am trying to show that if that derivative function $\partial f: X \rightarrow \mathcal{L}(E, F)$ is identically $0$ on $X$ then $f$ must be constant.

Since $X$ is open, for any given $x_0 \in X$, we can find an open ball about $x_0$, $\mathbb{B}_{\epsilon}(x_0)$, such that $\mathbb{B}_{\epsilon}(x_0) \subset X$. Since the open ball is connected, we can apply the integral mean value theorem to conclude that

\|f(x) - f(x_0) \| = 0 \implies f(x) = f(x_0) \; \forall \; x \in \mathbb{B}_{\epsilon}(x_0)

It follows that $f$ is locally constant. Now, from this result we know that $f$ is constant on $X$. I am trying to follow an argument however that does not make use of this result. Let $y_0 = f(x_0)$. Then, since $\partial f$ exists, $f$ is continuous, and from this it follows that the (non-empty) fiber $f^{-1}(y_0)$ is closed. If we can show that $f^{-1}(y_0)$ is also open we can use the hypothesis that $X$ is connected to conclude that $f^{-1}(y_0) = X$ and so $f$ must be constant. I'm stuck on this point. Therefore, my question is:

Within the context of the preceding paragraph, how can we conclude that $f^{-1}(y_0)$ must be open?

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    Aiye. That's the answer.2011-07-27

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You use that $f$ is locally constant. =)

To show that a set is open, you need to show that every point in it is contained in an open ball. Analytically this is a local condition, since you can take the ball to be arbitrarily small. (Algebraically, however, one may choose a topology where open balls are no longer intuitively local.) So basically, you cannot get away from using some local property of $f$.