21-356 Lecture 6

# e v ii e is connected if it is not disconnected

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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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## This document was uploaded on 03/31/2014.

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