Proposition 27.3.6.label Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and connected, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$ with $Df = 0$, then $f$ is constant.
Proof. Let $x \in V$, then for any $U \in \cn(0)$ circled with $U + x \subset V$ and $y \in U + x$,
by the Mean Value Theorem.
Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by Proposition 10.1.11, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by Lemma 5.4.3.
For any $y \in \ol W \cap V$, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. As $y \in \ol W \cap V$, $U \cap W \ne \emptyset$. Thus $f(y) = f(x)$, $y \in W$, and $W$ is relatively closed.
Since $V$ is connected and $W \subset V$ is both open and closed, $W = V$.$\square$
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