104-so4 - Math 104 - Lecture 1 - Problem Set 4 Solutions...

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Unformatted text preview: Math 104 - Lecture 1 - Problem Set 4 Solutions Thursday, July 15, 2010 1.40 (5 points) Let > be given, and defined A ( ) and A as in the problem. By definition, a A ( ) for every , since the only point in [ a,a ] is a , and | f ( a )- f ( a ) | = 0 < . So A is not empty. Also, every u A is in [ a,b ] , so b is an upper bound of A . Thus sup A exists. Suppose c = sup A < b . Apply the continuity of f at c using error of 2 . So there is some c so that if | x- c | < c then | f ( x )- f ( c ) | < 2 . Let c 1 = max ( a,c- 2 c 3 ) , c 2 = max ( a,c- c 3 ) , and c 3 = min ( b,c + c 3 ) . Since a c 2 < c = sup A , c 2 A , so c 2 A ( 2 ) for some 2 > . Put = min ( c 3 , 2 ) > . We will prove c 3 A ( ) A . Let x,t [ a,c 3 ] with | x- t | < . Without loss of generality, assume x < t . So then also t < x + . If x < c 1 then c 1 6 = a , and so t < x + < c 1 + c 3 = c 2 . So t,x [ a,c 2 ] . Since c 2 A ( 2 ) and | x- t | <...
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104-so4 - Math 104 - Lecture 1 - Problem Set 4 Solutions...

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