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Unformatted text preview: 1.3.7 . Prove that if a is an upper bound for A , and if a is also an element of A , then it must be that a = sup A . Since we are given that a is an upper bound for A , we need only check that part (ii) of De&nition 1.3.2, namely that if b is any upper bound for A , then a & b . But a is in A , so from the de&nition of ¡ b is an upper bound for A ¢, it follows that a & b . 1.3.8 . If sup A < sup B , then show that there exists an element b 2 B that is an upper bound for A . Because sup A is an upper bound for A , it su£ ces to &nd an element b 2 B that is greater than sup A . We know that sup A is not an upper bound for B , because sup B is the least upper bound for B , and sup A < sup B . But to say that sup A is not an upper bound for B , is to say that there is an element b 2 B such that sup A < b . Bingo. If you prefer, you can appeal to Lemma 1.3.7, taking " = sup B ¡ sup A . The rest of the translation is that the set A in Lemma 1.3.7 is the set B in this exercise, and the element s in Lemma 1.3.7 is the number sup B in the exercise. Soexercise....
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe

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