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Let $U $ be an open subset of $ \mathbb R^n$ ; then how to prove that $U$ is connected iff for every differentiable

function $f:U \to \mathbb R$ , $\nabla f(x)=0 \implies f $ is constant on $U$ ?

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  • $\begingroup$ Can you prove that any connected open set is path-connected? $\endgroup$ Aug 26, 2015 at 15:58
  • $\begingroup$ @MiloBrandt : yes , that I can , actually I'm having more trouble with the converse part that is the condition on maps implies connected .. $\endgroup$
    – user228168
    Aug 26, 2015 at 16:07
  • $\begingroup$ @ShaunDev Ah. It might be easier to think of the converse in terms of the following equivalent statement: "Any disconnected space has a non-constant $f$ with zero gradient." $\endgroup$ Aug 26, 2015 at 16:11
  • $\begingroup$ Suppose $U = (0,1) \cup (2,3)$ in $\mathbb R ^1.$ Can you find a non constant $f$ on $U$ such that $f'\equiv 0?$ $\endgroup$
    – zhw.
    Aug 26, 2015 at 16:13
  • $\begingroup$ @MiloBrandt , zhw : yes , that I can do if $n=1$ , but I'm having trouble for higher dimensions , please help $\endgroup$
    – user228168
    Aug 26, 2015 at 16:40

1 Answer 1

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Hint: Suppose that $U$ is open and disconnected. Then, we can write $U = U_1 \cup U_2$ where each $U_i$ is open and non-empty, and $U_1 \cap U_2 = \emptyset$.

Define your function by $$ f(x) = \begin{cases} 1 & x \in U_1\\ 0 & x \notin U_1 \end{cases} $$ Why is this function differentiable?

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