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hw6sol

# hw6sol - Honors Introduction to Analysis I Homework VI...

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Honors Introduction to Analysis I Homework VI Solution March 23, 2009 Problem 1 Let f be a function defined on a closed domain D . Show that f is continous if and only if the inverse image of every closed set is a closed set. Solution. “= ” Assume f is continuous. Let A be a closed set in the codomain. Then A c is open, which means that f - 1 ( A c ) is also open (since f is continuous), so ( f - 1 ( A c )) c is closed. But f - 1 ( A ) = D ( f - 1 ( A c )) c , an intersection of closed sets, so it’s closed. =” Assume that if A is closed, then f - 1 ( A ) is closed. For any x 0 D and any n N , consider the set A = ( -∞ , f ( x 0 ) - 1 /n ] [ f ( x 0 ) + 1 /n, ) = ( f ( x 0 ) - 1 /n, f ( x 0 ) + 1 /n ) c , which is closed. Hence f - 1 ( A ) is also closed. By the construction f ( x 0 ) / A , so x 0 / f - 1 ( A ). Since f - 1 ( A ) is closed, x 0 is not a limit point, so there is a neighborhood of x 0 not contained in f - 1 ( A ), i.e. there is some interval ( x 0 - 1 /m, x 0 + 1 /m ) such that ( x 0 - 1 /m, x 0 + 1 /m ) f - 1 ( A ) = . Then f (( x 0 - 1 /m, x 0 + 1 /m )) ( f ( x 0 ) - 1 /n, f ( x 0 ) + 1 /n ) = A c . We have that n , there is an m such that if | x - x 0 | < 1 /n , then | f ( x ) - f ( x 0 ) | < 1 /m , hence f is continuous. Problem 2 Let A be the set defined by the inequalities f 1 ( x ) 0 , f 2 ( x ) 0 ,..., f n ( x ) 0 , where f 1 , f 2 , ..., f n are continuous functions defined on the whole line. Show that A is closed. Show that the set defined by f 1 ( x ) > 0 , f 2 ( x ) > 0 ,..., f n ( x ) > 0 is open. Solution. A = { x R | f k ( x ) 0 , k = 1 , ..., n } = n \ k =1 { x R | f k ( x ) 0 } = n \ k =1 f - 1 k ([0 , )) Since [0 , ) = R \ ( -∞ , 0) is closed, by the previous problem, f - 1 k ([0 , )) is closed, for each k . So A

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