hw7-sol - Modern Analysis, Homework 7 Possible solutions 1....

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Unformatted text preview: Modern Analysis, Homework 7 Possible solutions 1. [12] (5 pts) Rudin, Ch 4, problem 2. Solution: Let f : X Y be a continuous mapping of metric spaces. Let y f ( E ). Then y = f ( e ) for some e E = E E . If e E , then y = f ( e ) f ( E ) f ( E ). If e E , then there exists a sequence e n E converging to e . Then since f is continuous, f ( e n ) f ( e ). If the set { f ( e n ) | n N } is infinite, then this shows that f ( e ) is a limit point of f ( E ), ie y f ( E ) . On the other hand, if the set { f ( e n ) | n N } is finite, then the sequence f ( e n ) must eventually stabilize (ie it is eventually constant) to some value. This value is the limit of the sequence, ie f ( e ). In this case it follows that y = f ( e ) = f ( e N ) for some sufficiently large N N , and that y f ( E ). Note that f ( E ) can be a proper subset of f ( E ): consider the continuous mapping f : R R given by f ( x ) = tan 1 ( x ) with E = R . Then f ( R ) = f ( R ) = ( 2 , 2 ) , but of course ( 2 , 2 ) = bracketleftbig 2 , 2 bracketrightbig . 2. [16] (5 pts) Suppose that A and B are connected subsets of a metric space ( X, d ) and that A B negationslash = {} . Prove that A B is connected. Solution: Suppose that A B = U V for a pair of disjoint, nonempty A B-opens U and V . Setting U A = U A , V A = V A , we see that A = U A V A . But both of these sets U A and V A are A-opens, and A is connected, so it must be the case that one of U A or V A is empty. Suppose V A = {} . Then A = U A . Using a similar argument, with similar notation, we have B = U B (and V B = {} ) or B = V B (and U B = {} ). Consider the case that B = U B . Then we have A U and B U , so A B U and V is empty. It follows that U, V is not a separation of A B . Now consider the case that A = U A and B = V B . Then we have A U V c and B V U c . In this case, we would have A B = {} , a contradiction! 3. [13] (5 pts) Let f : X Y be a continuous mapping of metric spaces and p be a point in X . Recall our definition of the oscillation of f at p : osc p f = inf > sup x N ( p ) d Y ( f ( x ) , f ( p )) . Prove that if osc p f = 0, then f is continuous at p . Solution: Let > 0. Then osc p f + = is not a lower bound for the set { sup x N ( x ) d Y ( f ( x ) , f ( p )) | > } , ie there exists > 0 such that sup x N ( x ) d Y ( f ( x ) , f ( y )) < . Suppose d X ( x, p ) < . Then x N ( x ) so d Y ( f ( x ) , f ( p )) sup x N ( x ) d Y ( f ( x ) , f ( y )) < , and were done. 4. [23] (5 pts) Suppose f : R R is uniformly continuous. Prove that there exist constants A, B > 0 such that for all x R we have | f ( x ) | A | x | + B....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.

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hw7-sol - Modern Analysis, Homework 7 Possible solutions 1....

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