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Unformatted text preview: Honors Introduction to Analysis I Homework VII Solution March 30, 2009 Problem 1 Let f be a monotone function on an interval. Show that if the image of f is an interval, then f is continuous. Give an example of a nonmonotone function on an interval whose image is an interval, but is not continuous. Solution. Since it is defined on an interval I , by Theorem 4.2.6, f has at most countably many number of points where it has jump discontinuities and perhaps at the endpoints, it may have lim x → x + f ( x ) or lim x → x f ( x ) equal to ±∞ . Since the image is an interval, the last part is not possible, meaning the limits are finite for the endpoints too. Wlog assume f is increasing and that it has at least one dicontinuity, i.e. there is x ∈ I such that lim x → x f ( x ) = a and lim x → x + f ( x ) = b and wlog a < b . This means that any value c ∈ ( a,b ) is not in the image of f : if it were, then we would have c = f ( y ), for some y ∈ I for which we can assume y > x (the case y < x is similar). Take z ∈ ( x ,y ) and by monotonicity of f , lim x → x + f ( x ) ≤ f ( z ) ≤ f ( y ) so b ≤ f ( z ) ≤ f ( y ) = c , contradicting c < b . So ( a,b ) is not in the image, hence the image cannot be an interval. For the example, take f : [ 1 , 1] → [0 , 1] with f ( x ) = ( x x ∈ [0 , 1] 1 x ∈ [ 1 , 0) . The image of [ 1 , 1] is [0 , 1], but it is not continuous at 0. Problem 2 If f is a continuous function on a compact set, show that either f has a zero, or f is bounded away from zero (  f ( x )  > 1 /n for all x in the domain, for some 1 /n ). Solution. The image of a continuous function on a compact set is a compact set, which is closed and bounded. If there is 0 in the image, than we are done. Assume f ( x ) 6 = 0, for any x , but it is not bounded away from zero, i.e. ∀ n there is x n such that  f ( x n )  < 1 /n . This means that the sequence {...
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This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell University (Engineering School).
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