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Unformatted text preview: The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 1 (Week 2) MATH2962: Real and Complex Analysis (Advanced) Semester 1, 2009 Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/ Lecturer: Daniel Daners Questions marked with * are more difficult questions. Questions to complete during the tutorial 1. Determine the supremum of the following sets. Determine whether it is a maximum or not. (a) A = { x Q : x 2 8 } Solution: sup A = 2 2 is not a maximum because 2 2 negationslash Q . (b) B = { 1 1 /n : n N } Solution: sup B = 1 is not a maximum as 1 1 /n < 1 for all n N (c) C = { (1 / 2 n ): n N } Solution: sup C = 1 = max C since (1 / 2) = 1 is an element of C . 2. Suppose A,B are nonempty subsets of R . We set A + B := { x + y : x A,y B } and A := { x : x A } . Prove the following statements. (a) sup( A ) = inf( A ); Solution: Set L := inf( A ). We want to show that sup A = L . We first show that L is an upper bound for A . Since L = inf( A ) we have L x for all x A and so x L for all x A . Hence L is an upper bound for A . Next we need to show that given an arbitrary upper bound m of A , it follows that L m . Since m is an upper bound for A we have x m and so m x for all x A . (b) sup( A + B ) = sup A + sup B ; Solution: The idea is to prove two inequalities: sup( A + B ) sup A + sup B , and sup( A + B ) sup A +sup B . Since sup A is an upper bound for A we have x sup A for all x A . Similarly, y sup B for all y B . Hence x + y sup A + sup B for all x A and y B , that is, sup A + sup B is an upper bound for A + B . Hence by definition of a supremum sup( A + B ) sup A + sup B . We next show the opposite inequality. By....
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