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Unformatted text preview: Math 598 Feb 22, 2005 1 Geometry and Topology II Spring 2005, PSU Lecture Notes 8 2.10 Measure of C 1 maps If X is a topological space, we say that A X is dense is X provided that A = X , where A denotes the closure of A . In other words, A is dense in X if every open subset of X intersects A . Theorem 1. Let f : M n N m be a C 1 map. Suppose that m > n . Then N f ( M ) is dense in M . To prove the above result we need to develop the notion of measure zero, which is defined as follows. We say that C R n is a cube of side length provided that C = [ a 1 , a 1 + ] [ a n , a n + ] , for some a 1 , . . . , a n R n . We define the mesure or volume of a cube of side length by ( C ) := n . We say a X R n has measure zero if for every > 0, we may cover X by a family of cubes C i , i I , such that i I ( C i ) . Lemma 2. A countable union of sets of measure zero in R n has measure zero. Proof. Let X i , i = 1 , 2 , . . . be a countable collection of subsets of R n with measure zero. Then we may cover each X i by a family C ij of cubes such that j C ij < / 2 i . Then X i =1 X j C ij = X i =1 2 i = . Since i X i ij C ij , it follows then that i X i has measure zero. 1 Last revised: February 27, 2005 1 Lemma 3. If U R n is open and nonemepty, then it cannot have measure zero. Proof. By definition, for evey point p U there exists r > 0 such that B r ( p ) U . Thus, since U 6 = , U contains a cube C (of side length 2 r/ n ). Suppose there is a covering of U by a family of cubes. Then, since C is compact, there exists a finite subcollection C i , i = 1 . . . , m which cover C . Let N be the number of integer lattice points (i.e., points with integer coecients) which lie in C , then ( max(0 , 1) ) n N ( + 1) n ....
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