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Unformatted text preview: Math 301/ENAS 513: Homework 9 Solution Triet M. Le 1. Page 89: 8, 10, 11, 13. 15. 2. Page 91: 3,6. 3. Page 94: 2, 3, 6. Page 89: 8: B S being not closed implies B does not contain all its limit points. Let a S be a limit point of B such that a / B . Now, consider f : B R , with f ( x ) = 1 d ( x,a ) > 0, for all x B . To show f is continuous: For any x, y B ,  f ( x ) f ( y )  = 1 d ( x, a ) 1 d ( y, a ) = d ( y, a ) d ( x, a ) d ( x, a ) d ( y, a ) Suppose d ( y, a ) d ( x, a ), then  d ( y, a ) d ( x, a )  = d ( y, a ) d ( x, a ) d ( y, x ) + d ( x, a ) d ( x, a ) = d ( x, y ) On the other hand if d ( x, a ) d ( y, a ), then  d ( y, a ) d ( x, a )  = d ( x, a ) d ( y, a ) d ( x, y ) + d ( y, a ) d ( y, a ) = d ( x, y ) Thus, in either cases,  f ( x ) f ( y )  d ( x, y ) d ( x, a ) d ( y, a ) , x, y B. Let x = d ( x, a ) / 2 > 0. Pick any > 0, we have  f ( x ) f ( y )  d ( x, y ) d ( x, a ) d ( y, a ) d ( x, y ) 2 x < , whenever d ( x, y ) < x = min { x , 2 x } . Hence f is continuous at any x B . To show f is not uniformly continuous: Suppose f is uniformly continuous. I.e. for each > 0, there exsists > 0 such that  f ( x ) f ( y )  < whenever d ( x, y ) < . Since a is a 1 limit point of B , then for each (0 , 1), we can find x, y B such that d ( y, a ) < and d ( x, a ) < d ( y, a ) / 2. Note f ( x ) > f ( y ) and d ( x, y ) < . On the other hand, f ( x ) f ( y ) = 1 d ( x, a ) 1 d ( y, a ) 2 d ( y, a ) 1 d ( y, a ) = 1 d ( y, a ) 1 as . (1) This is a contradition. Page 89: 10: Let a R be any irrational number such that a > 0, and define B = [0 , a ] Q . It is clear that B contains all its limit points in Q . Hence it is closed in Q . Define f : B R with f ( x ) = 1 / ( x a ). By the above problem, f is continuous on B . However,  f ( x )  as x a and x B , which shows that f is unbounded. Page 89: 11: Let ( S, d S ) and ( T, d T ) be metric spaces with d S being discrete. Let f : S T be any function. To show f is uniformly continuous: For any > 0. Let = 1 / 2. Then N ( x ) = { x } for all x S . Hence  f ( x ) f ( y )  = 0 < , for all y N ( x ) ....
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This note was uploaded on 07/18/2008 for the course MATH 301 taught by Professor Trietle during the Fall '07 term at Yale.
 Fall '07
 TrietLe
 Math

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