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Unformatted text preview: MAT 257Y Solutions to Practice Term Test 2 (1) Let f : R n → R m be C 1 where n > m . Suppose [ df ( x )] has rank m . Show that there exists > 0 such that for any y ∈ B ( f ( x ) , ) there exists x ∈ R n such that f ( x ) = y . Solution By the Corollary to Implicit Function Theorem there exist an open set U ⊂ R n , an open set V ⊂ R n containing x and a diffeomorphism φ : U → V such that f ( φ ( x, y )) = y for any ( x, y ) ∈ U . Here x ∈ R n- m , y ∈ R m . Let p = ( a, b ∈ U be the preimage of x . i.e. φ ( p ) = x . Since U is open there is an > such that B ( p. ) ⊂ U . Then for any y ∈ B ( f ( x ) , ) there exists u ∈ U such that f ( φ ( u )) = y . (2) Let A be a rectangle in R n and let S ⊂ A be a set of measure zero which is rectifiable. Show that S has content zero. Hint: Use that R A χ S exists and must be equal to zero. Solution Let f = χ S . Then R A f exists and since f = 0 ex- cept on a set of measure 0, we must have R A f = 0 by a theorem from class. Leta theorem from class....
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- Winter '10
- Empty set, Open set, Fubini