Let xninfinityn 1ba a bounded sequence Prove that it contai

Let {x_n}^infinity_n = 1ba a bounded sequence. Prove that it contains a monotone subsequence that converges to x = lim sup_n right arrow x_n.

Solution

Let (xn) be a bounded sequence. For each n N, let yn = sup mn xm = sup{xn, xn+1, . . .}. Then {yn} is decreasing, and it is bounded since (xn) is bounded. By Proposition 5.12, (yn) converges. The limit is called upper limit of (xn), and is denoted by lim sup xn. Similarly, let zn = infmn xm = inf{xn, xn+1, . . .}. Then (zn) is increasing, and it is bounded since (xn) is bounded. The limit of (zn) is called lower limit of (xn), and is denoted by lim inf xn. If (xn) is not bounded from above, then its upper limit is equal to and if (xn) is not bounded from below, then its lower limit is equal to . Summarizing lim sup xn = lim xn = inf nm sup kn xk = limn sup kn xk lim inf xn = lim xn = sup nm inf kn xk = limn inf kn xk.