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 satisfies conditions (1) and (2) in the definition of supremum. By the definition of the given set, we see that for all x ∈[0,3], x ≤3. Thus 3 is an upper bound. This verifies 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 satisfies conditions (1) and (2) in the definition 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 Mis an upper bound of [0,3) and assume, by way of contradiction, that 3>M. Since Mis an upper bound of [0,3), we have that M>0. Set x=M+3 2 . Then 0<M= M+M 2 < M+3 2 < 3+3 2 =3 or M<x <3. This is a contradiction since Mis an upper bound of [0,3) and x ∈[0,3). We conclude that 3≤Mand, 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 boundMmust satisfy 10≤Mas 10 is an element of the set. Thus 10 is the supremum. (c) sup (−1)n n : n∈N = 1 2 . Note that if n∈Nis even, then n≥2 and (−1)n n = 1 n ≤ 1 2 . If n∈Nis odd, then (−1)n n =− 1 n <0< 1 2 . This shows that (−1)n/n≤1 2 for all n∈N. Hence 1/2 an upper bound of the set. Also 1/2 is 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 Mis an upper bound of Abut M<4. Choose a number y ∈Rsuch that M<y <4 and 0<y. Set x =− √y. Then−2<x <0<1 and, so, y =x2 ∈A. However, y >Mwhich contradicts the fact that Mis an upper bound. Thus 4≤M. This proves that 4=supA. Remark 1.5.1 Let Abe a nonempty subset of R. If there is an element aM∈Asuch that aM≥x for all x ∈A, we say that aM is the maximum of Aand 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 Aand write am=minA. It is clear that if Ahas a maximum element then it is bounded above since maxAis an upper bound. Also, if an upper bound belongs to the set then it is the maximum of the set.
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