1.6. CONVERGENCE IN R 13 since, for example, ap 2−p is not an upper bound for A, while aq is an upper bound. But this implies that |ap aq| 2−p for all p q, from which it follows that (ak)k∈N is a Cauchy sequence. Let L = [ak]. We claim that L is a least upper bound for A. Suppose x A and x L. Choose p such that 2−p (x L) (using the Archimedean property). Since ap 2−p aq for p q and (ap)p∈N is a decreasing Cauchy sequence, it follows that ap 2−p L ap. In particular, if we add 2−p x L and ap 2−p L, we obtain ap x, which is a contradiction. Therefore L is an upper bound for A. Suppose that H is an upper bound for A and that H L. Choose p such that 2−p L H. Take x A such that ap 2−p x. Then ap 2−p H. Adding, we get ap L. But, as noted above, L ap for all p N, so this is a contradiction. Exercise 1.5.21. Prove that R is order-isomorphic to R. (Hint: You have already done this.) 1.6. Convergence in R We define the absolute value on R in exactly the same manner as on Q. Definition 1.6.1. Suppose x R. The absolute value of x is defined by |x| = x if x 0, −x if x 0. The following are the essential properties of the absolute value. Theorem 1.6.2 (Properties of the absolute value on R). (1) For any x R, |x| 0, and |x| = 0 iff x = 0. (2) For any x, y R, |xy| = |x||y|. (3) For any x, y R, |x + y| |x| + |y| (triangle inequality). Exercise 1.6.3. Prove these properties of the absolute value. With absolute value defined, we can talk about Cauchy and convergent sequences in R. Definition 1.6.4. A sequence (ak)k∈N of real numbers is convergent if there exists an element a R such that the sequence satisfies the following property: given any ε 0, there exists N N such that k N implies that |ak a| ε. We say that (ak)k∈N converges to a, and a is called the limit of the sequence (ak)k∈N. Symbolically, we write lim k→∞ ak = a. We will often say that a sequence of real numbers is convergent without specific reference to the limit a. Note that N depends on ε.
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