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Unformatted text preview: Math 8601: REAL ANALYSIS. Fall 2010 Homework #3. Problems and Solutions. #1. Let f be a real function on R 1 . The image and the inverse image of a subset A R 1 under f are correspondingly f ( A ) = { y : y = f ( x ) for some x A } , f 1 ( A ) = { x : f ( x ) A } . Show that f ( f 1 ( A )) A f 1 ( f ( A )) for arbitrary A R 1 . Give an example when f 1 ( f ( A )) 6 = A . Solution. (i). y f ( f 1 ( A )) = y = f ( x ) for some x f 1 ( A ) = y = f ( x ) A . Therefore, f ( f 1 ( A )) A . (ii). x A = y = f ( x ) f ( A ) = x f 1 ( f ( A )). Therefore, A f 1 ( f ( A )). (iii). For f ( x ) = sin x, A = { } , we have f 1 ( f ( A )) = f 1 (0) = { k : k = 0 , 1 , 2 , } 6 = A = { } . #2. Let f ( x ) be a continuous function on R 1 , and let B denote the Borel algebra on R 1 . Show that M := f 1 ( B ) := { E : E = f 1 ( F ) for some F B} is a algebra contained in B ....
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This note was uploaded on 02/15/2012 for the course MATH 8601 taught by Professor Staff during the Fall '08 term at Minnesota.
 Fall '08
 Staff
 Math

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