Sec.%2012%20The%20Completeness%20Axiom

# Sec.%2012%20The%20Completeness%20Axiom - Math 350 Advanced...

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Unformatted text preview: Math 350 Advanced Calculus I Fall 2009 Homework Solutions Section 12 THE COMPLETENESS AXIOM 12.3 For each of the following subsets of R , give its supremum and its maximum, if they exist. For each part, 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) S = (0 , 4) 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 . To show this, we need to show ( ∀ s ∈ S ) [ s ≤ 4] . 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 ∈ R , if m is an upper bound for S , then 4 ≤ m . To prove this, we use a proof by contrapositive. The contrapositive statement is ( ∀ m ∈ R ) [ m < 4 → ( 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 . Find an s . In the case that m ≤ 0, then every element of S is greater than m , so choose s = 1, for instance. In the case that m > 0, choose s to be the midpoint of m and 4, so s = 1 2 ( m + 4) . Verify s satisfies: (i) Domain: s ∈ S (ii) Propositional Function: s > m In the case that m ≤ 0, then s = 1 ∈ S and s = 1 > m . In the case that m > 0, then m ∈ S . Since s = 1 2 ( m + 4) is the midpoint of m and 4, then s ∈ S also and s > m , as required. Therefore m is not an upper bound for S . This proves the contrapositive statement. Thus 4 is the least upper bound of S . To show that S has no maximum, note that since sup S = 4, then no number less than 4 can be an upper bound of S . Since all elements of S are less than 4, then it follows that S does not contain any upper bounds so S has no maximum....
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