14 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS Exercise 1.6.5. Show that the limit a of a convergent sequence is unique. Definition 1.6.6. A sequence (ak)k∈N of real numbers is monotonic increasing if ak ak+1 for all k N. A sequence (ak)k∈N of real numbers is strictly monotonic increasing if ak ak+1 for all k N. Monotonic decreasing and strictly monotonic decreasing sequences are defined similarly. Exercise 1.6.7. Define the notion of a bounded sequence in R. The following lemma is one of the more useful lemmas in discussing convergence in R (and Rn). Lemma 1.6.8. Let (ak)k∈N be a sequence in R. Then (ak)k∈N has a monotonic subsequence. Proof. Suppose (ak)k∈N does not have a monotonic increasing subsequence. Then, there exists n1 N such that an 1 ak for all k n1. Again, since (ak)kn 1 does not have a monotonic increasing subsequence, there exists n2 n1 such that an 2 ak for all k n2. Moreover an 1 an 2 . Continuing in this way, we obtain a strictly monotonic decreasing subsequence. Lemma 1.6.9. Every bounded monotonic sequence converges in R. Proof. Suppose (ak)k∈N is monotonic increasing and bounded. Let a be the least upper bound of the set {a1,a2,...}. For all ε 0, there exists an N such that a ε aN a. Since (ak)k∈N is increasing, if k N, we have a ak aN a ε. So limk→∞ ak = a. The next lemma is basic for analysis on R. Lemma 1.6.10. Every bounded sequence in R has a convergent subse- quence. Exercise 1.6.11. Prove Lemma 1.6.10. This should not take long. The next definition should be compared with Definition 1.5.2. Definition 1.6.12. A sequence (ak)k∈N in R is a Cauchy sequence if, given any ε 0, there exists N N such that n, m N implies |am−an| ε. Exercise 1.6.13. (i) Prove that every Cauchy sequence in R is bounded. (ii) If (ak)k∈N is a Cauchy sequence in R, show that for any ε 0 there exists a subsequence (aj)j∈N so that |aj aj+1| ε/2j+1. Theorem 1.6.14 (Cauchy criterion). A sequence (ak)k∈N of real num- bers is convergent if and only if it is a Cauchy sequence. Proof. We already did half of this in Q, but we will do it again. First, we prove that if (ak)k∈N is convergent, then it is Cauchy. Suppose limk→∞ ak =
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