Introduction to Mathematical Analysis I - Second Edition

23 Similarly, a number L is a lower bound of A if L ≤ x for all x ∈ A , and A is said to be bounded below if it has a lower bound. We also say that A is bounded if it is both bounded above and bounded below. It follows that a set A is bounded if and only if there exist M ∈ R such that | x | ≤ M for all x ∈ A (see Exercise 1.5.1 ) . Definition 1.5.2 Let A be a nonempty set that is bounded above. We call a number α a least upper bound or supremum of A , if (1) x ≤ α for all x ∈ A (that is, α is an upper bound of A ); (2) If M is an upper bound of A , then α ≤ M (this means α is smallest among all upper bounds). It is easy to see that if A has a supremum, then it has only one (see Exercise 1.5.2 ) . In this case, we denote such a number by sup A . Example 1.5.1 (a) sup [ 0 , 3 ) = sup [ 0 , 3 ] = 3. First consider the set [ 0 , 3 ] = { x ∈ R : 0 ≤ x ≤ 3 } . By its very definition we see that for all x ∈ [ 0 , 3 ] , x ≤ 3. Thus 3 is an upper bound. This verifies condition (1) in the definition of supremum. Next suppose M is an upper bound of [ 0 , 3 ] . Since 3 ∈ [ 0 , 3 ] , we get 3 ≤ M . This verifies condition (2) in the definition of supremum. It follows that 3 is indeed the supremum of [ 0 , 3 ] . Consider next the set [ 0 , 3 ) = { x ∈ R : 0 ≤ x < 3 } . It follows as before that 3 is an upper bound of [ 0 , 3 ) . Now suppose that M is an upper bound of [ 0 , 3 ) and assume by way of contradiction that 3 > M . If 0 > M , then M is not an upper bound of [ 0 , 3 ) as 0 is an element of [ 0 , 3 ) . If 0 ≤ M , set x = M + 3 2 . Then 0 ≤ x < 3 and x > M , which shows M is not an upper bound of [ 0 , 3 ) . Since we get a contradiction in both cases, 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. (c) sup ( − 1 ) n n : n ∈ N = 1 2 . Note that if n ∈ N is even, then n ≥ 2 and ( − 1 ) n n = 1 n ≤ 1 2 . If n ∈ N is odd, then ( − 1 ) n n = − 1 n < 0 < 1 2 . This shows that 1 2 is an upper bound of the set. Since 1 2 is an element of the set, it follows as in the previous example that 1 2 is the supremum. (d) sup { x 2 : − 2 < x < 1 , x ∈ R } = 4. Set A = { x 2 : − 2 < x < 1 , x ∈ R } . If y ∈ A , then y = x 2 for some x satisfying − 2 < x < 1 and, hence, | x | < 2. Therefore, y = x 2 = | x | 2 < 4. Thus, 4 is an upper bound of A . Suppose M is