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# solHMK3 - Section 2.4#4(a We prove that sup(aS = a sup S...

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Section 2.4 #4) (a) We prove that sup( aS ) = a sup S . The proof for inf( aS ) is similar. Let x aS . Then x/a S . So sup S x/a since sup S is an upper bound for S . Therefore, a sup S x , and a sup S is an upper bound for aS . Now let v be any upper bound for aS . If y S , then v ay . So v/a y . Thus, v/a is an upper bound for S , and so v/a sup S . Therefore, v a sup S , and so a sup S = sup( aS ) as desired. (b) We prove that inf( bS ) = b sup S . The proof for sup( bS ) is similar. Let x bS . Then x/b S . So sup S x/b since sup S is an upper bound for S . Therefore, b sup S x , and b sup S is a lower bound for bS . Now let v be any lower bound for bS . If y S , then v by . So v/b y . Thus, v/b is an upper bound for S , and so v/b sup S . Therefore, v b sup S , and so a sup S = inf( aS ) as desired. #6) We prove that sup( A + B ) = sup A + sup B . The proof of inf( A + B ) is similar. Let x A + B . Then x = a + b for some a A and some b B . Therefore, x = a + b sup A + sup B since sup A and sup B are upper bounds for A and B respectively. Thus, sup A + sup B is an upper bound for A + B . To prove it is the least upper bound, we show that for all > 0, there exists x A + B such that sup A + sup B - < x sup A + sup B . Let > 0. Choose a A such that sup A - 2 < a sup A , and choose b B such that sup B - 2 < b sup B . Then a + b A + B and sup A + sup B - < a + b sup A + sup B as desired. #7) We will prove the relation sup { f ( x ) + g ( x ) : x X } ≤ sup { f ( x ) : x X } + sup { g ( x ) : x X } . The corresponding relation for inf { f ( x ) + g ( x ) } is proved similarly. It suffices to prove the right hand side is an upper bound for { f ( x ) + g ( x ) : x X } , for then it must be at least as large as the supremum of the set. Let x 0 X be arbitrary. Then f ( x 0 ) + g ( x 0 ) sup { f ( x ) : x X } + sup { g ( x ) : x X } , and the right hand side is an upper bound as desired.

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