12 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS Theorem 1.5.16. R is a field. Proof. We need only show that multiplicative inverses exist for nonzero elements. So assume that [ak] = I. Then, as we saw above, ak is eventually bounded below in absolute value. Hence, we can pick M N and c 0 such that |ak| c for all k M. Define a sequence (bk)k∈N as follows: bk = 1 for k M, and bk = 1/ak for k M. Observe that for n, m large enough 1 an 1 am = |an am| |anam| |an am| c2 . So (bk)k∈N is a Cauchy sequence and [bk] is the multiplicative inverse of [ak]. The next step is to define order on R. Let [ak] and [bk] represent distinct elements of R. Then [ck] = [ak bk] is not equal to I. Hence there exists N N such that all the terms of ck have the same sign for k N. Thus, either ak bk for all k N or bk ak for k N. We use this fact to define an order on R. Definition 1.5.17. Let a = [ak],b = [bk] be distinct elements of R. We define a b if ak bk eventually and b a if bk ak eventually. Exercise 1.5.18. Show that the order relation on R defined above is well-defined and makes R an ordered field. To finish this, we must show that R is an Archimedean ordered field that satisfies the least upper bound property. We will then have reached the mountain top so we can dismount the funicular and ski happily down the slope. Define a map i : Q −→ R by sending r Q to the equivalence class of (r,r,... ). It is evident that this map is injective and order-preserving, so we may consider Q R as ordered fields. Theorem 1.5.19. The field R is an Archimedean ordered field. Proof. Suppose a R and a 0. Let (ak)k∈N represent a. As noted above, the Cauchy sequence (ak)k∈N is bounded above by some integer N that is, ak N for all sufficiently large k. It follows that a is less than the integer (N,N,... ) in R (under the inclusion Q R). Theorem 1.5.20. The least upper bound property holds in R. Proof. Let A be a nonempty subset of R that is bounded above by, say, m. Then, by the Archimedean property, we can find M Z with m M. Let a be in A and let n be an integer with n a. For p N set Sp = {k2−p | k Z and n k2−p M} {m}. Note that Sp = and is finite. Now let ap = min{x | x Sp and x is an upper bound for A}. Note that if p q, then ap 2−p aq ap,
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