24 1.5 The Completeness Axiom for the Real Numbers ■ Example 1.5.1 (a) sup[0,3)= sup[0,3]= 3. First consider the set [0,3]= {x ∈ R : 0 ≤ x ≤ 3}. We will show that 3 satisfes conditions (1) and (2) in the defnition of supremum. By the defnition of the given set, we see that for all x ∈ [0,3], x ≤ 3. Thus 3 is an upper bound. This verifes condition (1). To verify condition (2) suppose that M is an upper bound of [0,3]. Since 3 ∈ [0,3], we get 3 ≤ M. Therefore condition (2) holds. It follows that 3 is indeed the supremum of [0,3]. Consider next the set [0,3)= {x ∈ R :0 ≤ x < 3}. We again verify that 3 satisfes conditions (1) and (2) in the defnition of supremum. Condition (1) follows as before since 3 is an upper bound of [0,3). For condition (2), because 3 is not in the set we cannot proceed as before. Suppose that M is an upper bound of [0,3) and assume, by way of contradiction, that 3 > M. M+3 M+M Since M is an upper bound of [0,3), we have that M > 0. Set x = 2 . Then 0 < M = 2 < M+3 3+3 2 < 2 = 3 or M < x < 3. This is a contradiction since M is an upper bound of [0,3) and x ∈ [0,3). We conclude that 3 ≤ M and, hence, 3 is the supremum of [0,3). (b) sup{3,5,7,8,10} = 10. Clearly 10 is an upper bound of the set. Moreover, any upper bound M must satisfy 10 ≤ M as 10 is an element of the set. Thus 10 is the supremum. (−1)n 1 (c) sup : n ∈ N = . n 2 Note that if n ∈ N is even, then n ≥ 2 and (−1)n 1 1 = ≤ . n n 2 If n ∈ N is odd, then (−1)n −1 1 = < 0 < . n n 2 This shows that (−1)n/n ≤ 1 for all n ∈ N. Hence 1/2 an upper bound of the set. Also 1/2 is 2 an element of the set, it follows as in the previous example that 1/2 is the supremum. (d) sup{x2 : −2 < x < 1, x ∈ R} = 4. Set A = {x2 : −2 < x < 1, x ∈ R}. If y ∈ A, then y = x2 for some x ∈ (−2,1) and, hence, |x| < 2. Therefore, y = x2 = |x|2 < 4. Thus, 4 is an upper bound of A. Suppose M is an upper bound of A but M < 4. Choose a number y ∈ R such that M < y < 4 and 0 < y. Set x = − √ y. Then −2 < x < 0 < 1 and, so, y = x2 ∈ A. However, y > M which contradicts the fact that M is an upper bound. Thus 4 ≤ M. This proves that 4 = supA. Remark 1.5.1 Let A be a nonempty subset of R. If there is an element aM ∈ A such that aM ≥ x for all x ∈ A, we say that aM is the maximum of A and write aM = maxA. If there is an element am ∈ A such that am ≤ x for all x ∈ A, we say that am is the minimum of A and write am = minA. It is clear that if A has a maximum element then it is bounded above since maxA is an upper bound. Also, if an upper bound belongs to the set then it is the maximum of the set.
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