(1) Let A be a bounded above nonempty subset of R. Show that there exists an inc

Question

(1) Let A be a bounded above nonempty subset of R. Show that there exists an increasing sequence {Xn}n=1 to infinity in A which converges to sup(A). (2)Let Xn = n^(1/n) for n belong to N. show that {Xn+2}n=1 to infinity

Explanation / Answer

First consider the case when u ? A. Then, the constant sequence xn = u works. Now assume u /? A. From the de?nition of supremum, we have that for every > 0, the interval (u - , u) must contain points from A (otherwise, u - would be an upper bound for A better than u). We now construct the sequence (xn) recursively as follows. First, let x1 ? A n (u - 1, u) be arbitrary. This intersection is nonempty by the remark above applied to = 1. Next, assume we have chosen the ?rst n - 1 elements of the sequence so that x1 = x2 = . . . = xn-1 0. From this recursive construction we see that xn u - shows that u - is not an upper bound for the set {xn : n ? N}. Note that we haven’t used anywhere that A is in?nite. In fact, we proved that a subset of R which is bounded above but does not contain its supremum, must be in?nite! Namely, it must contain a strictly increasing sequence.