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Unformatted text preview: + (1 % !) y) (xi % yi ) .
" xi
i=i Deﬁnition 23 Given E " RN , we say that
(i) E is disconnected if there exist two nonempty disjoint open sets U, V "
RN such that
E " U ' V, E ( U != ), E ( V != ), (ii) E is connected if it is not disconnected.
Theorem 24 Let U " RN be open and connected and let f : U # R be such
!f
that for all x $ U and all i = 1, . . . , N there exists ! xi (x) = 0. Then f is
constant.
Proof. Since all the partial derivatives are zero, in particular they are continuous. It follows from Theorem 16 that f is di"erentiable for all x $ U .
13 Step 1: Let x $ U . Since U is open, there exists B (x, r) " U . We claim that
f is constant in B (x, r). Fix y $ B (x, r). By the mean value theorem and
Remark 22, there exists ! $ (0, 1) such that
f (x) % f (y) = N
& "f
(!x + (1 % !) y) (xi % yi ) = 0.
" xi
i=i This proves the claim.
Step 2: Next ﬁx x0 $ U , let c := f (x0 ), and deﬁne the set
A := {x $ U : f (x) = c} .
We claim that A is open. Indeed, if x $ A, then by the previous step there
exists B (x, r) such that f is constant in B (x, r). Hence, f = c in B (x, r),
which implies that B (x, r) " A. Thus every point of A is an interior point and
so A is open. Moreover, A is nonempty since x0 $ A. Next consider the set
A1 = {x $ U : f (x) != c} = f !1 ((%*, c) ' (c, *)) .
Since (%*, c) ' (c, *) is open, f is continuous, and U is open, we have that
f !1 ((%*, c) ' (c, *)) is open. Thus, we c...
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 Spring '14

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