smt1 - Math 5615H. October 5, 2011. Midterm Exam 1....

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Math 5615H. October 5, 2011. Midterm Exam 1. Problems and Solutions. Problem 1. (10 points). Let A and B be nonempty bounded subsets of R . Show that sup( A B ) = sup { sup A, sup B } . Proof. Denote M 1 := sup( A B ) , M 2 := sup { sup A, sup B } . We have x sup A M 2 x A, and x sup B M 2 x B. Hence x M 2 x A B , which implies M 1 M 2 . On the other hand, since A A B and B A B , we also have sup A M 1 , sup B M 1 , so that M 2 M 1 . Together with the opposite inequality M 1 M 2 , this yields M 1 = M 2 . Problem 2. (10 points). Find an explicit expression for a one-to-one mapping f of A onto B , where A := [0 , 1] = { x R : 0 x 1 } , B := [0 , 1) = { x R : 0 x < 1 } . Hint: First define f on the sequence A 0 := { a n = 1 /n ; n = 1 , 2 , 3 ,... } . Solution. One can take f ( x ) = ( a n +1 if x = a n A 0 , x if x / A 0 . . Problem 3.
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smt1 - Math 5615H. October 5, 2011. Midterm Exam 1....

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