Advanced Calculus fi..

Chapter 2 Differential Calculus of Functions of Several Variablesthen E is bounded above, and B = 1 or any number larger than 1 is an upper boundof E. Similarly, E is bounded below if x 2 B for all x in E, for some number B; Bis then a lower bound of E.THEOREM B Let E be a set of real numbers that is nonempty and boundedabove. Then E has a least upper bound; that is, there is a number Bo such that Bo isan upper bound of E and Bo 5 B for every upper bound of E.Proo$ We can imitate the proof of Theorem A. There are integers that are not upperbounds of E-for we can choose an integer less than some x in E (E is nonempty).There is a largest such integer, sl. For E has an upper bound B, and every integerthat is not an upper bound must be less than B. Similarly, there is a largest integerkl, 0 5 kl 5 9, such that s2 = s1 + (kl/lO) is not an upper bound of E. Continuing,we obtain a monotone increasing sequence s, such thatRand s, is not an upper bound of E but s, + (11 lo"-') is an upper bound. It followsthat for every x in E, for all n,Now the sequence s, is bounded above (by B, as earlier). Therefore by Theorem A,s, converges to some Bo. By (2.17 I),so that Bo is an upper bound of E. On the other hand, s, < B for all n implies thatBo 5 B. Therefore Bo is the least upper bound (and is clearly unique).CORO1,LARY TO THEOREM B If a set E of real numbers is bounded below,then E has a greatest lower bound Ao.byThe proof is left as an exercise (Problem 1 following).One abbreviates the least upper bound of E and the greatest lower bound of Elub E, glb E.A sequence t, is said to be a subsequence of s, if tl = s,, , t2 =s,,, . . . , tk =s,, , . . . , where n < n2 < . . < nk c nk+~ < . . . . We remark that if the sequence s,converges, then the subsequence t, also converges, with the same limit (Problem 2).THEOREM C (Weierstrass-Bolzano Theorem) Lets, be a bounded sequenceof real numbers. Then s, has a convergent subsequence.CrProof. Let E be the set of all real numbers x such that s, >_ x for infinitely manyvalues of n. Since the sequences, satisfies A 5 s, 5 B for all n, for some A and B, E