This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Bopens 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 Aopens, 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....
View
Full
Document
This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
 Peters

Click to edit the document details