Solutions to HW 3

Solutions to HW 3 - Math 350 Homework Solutions 3 # 1:...

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Math 350 Advanced Calculus I Spring 2008 Homework Solutions 3 # 1: Supplementary Problem A Let S R be a nonempty and bounded set. Let m = sup S . Prove: If m 6∈ S , then S has no maximum. This result can be used in Problem 12.3. Solution We prove the contrapositive: If S has a maximum x , then x has to be sup S . That is, sup S S . Let x = max S . Then for any s S , s x . Hence x is an upper bound of S . For any y IR such that y < x , since x S and x > y , y is not an upper bound of S . Hence by the definition of supremum, x = sup S . Since x S , we have sup S S . # 2: Supplementary Problem B Indicate the logical error in the following proof: “For S = (0 , 4), since all x such that x > 4 are upper bounds of S , sup S = 4.” Solution The reasoning misses half of the proof. It only proves that sup S 4 (as the least upper bound), but does not prove that sup S = 4. That is, it does not show for any x such that x < 4, x is not a upper bound. 12.3 For each subset of IR , give its supremum and its maximum, if they exist. For each party, you must verify your results. That is, if you are claiming that a number a is the supremum, you must show that it is using the definition. (d) (0 , 4) Solution sup S = 4, No maximum We want to prove that sup S = 4. To show this, we need to show (1) 4 is an upper bound for S : Suppose s is an arbitrary element of S . Then 0 < s < 4, so s 4, as required. Therefore 4 is an upper bound for S . (2) For all m IR , if m < 4, then m is not an upper bound for S : Suppose m is an arbitrary real number and m < 4. We need to show that m is not an upper bound for S . To show this, we need to show there exists an element s S such that s > m . In the case that m 0, choose s = 1, then clearly s = 1 S and s = 1 > m . In the case that m > 0, choose s to be the midpoint of m and 4, so s = m + 4 2 = m + 4 - m 2 . Then by the first part of the equation, 0 < s < 4, so s S . By the second part of the equation, since 4 - m > 0, we have s > m . Therefore m is not an upper bound for S . Thus 4 is the least upper bound of S . Since sup S = 4 and 4 6∈ S , by Problem #1, S has no maximum. (f) ' 1 - 1 n | n IN Solution sup S = 1, No maximum We want to prove that sup S = 1. To show this, we need to show (1) 1 is an upper bound for S : 1
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Suppose s is an arbitrary element of S . Then s = 1 - 1 n for some n IN . Since 1 /n > 0, then s < 1, as required. Therefore 1 is an upper bound for S . (2) For all
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Solutions to HW 3 - Math 350 Homework Solutions 3 # 1:...

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