21-356 Lecture 6

e v ii e is connected if it is not disconnected

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online