Let A be non-empty & bounded above. Show that there exist a monotone sequence (a_n) such that a_n goes to SupA

Answer 1

let s = sup A. For every eps >0, s - eps is not an upper bound of A, so that there exists a in A such that s - eps < a <= s. Since eps >0 is arbitrary, it follows s belongs to the closure of A.

If s is not a limit point of A, then S belongs to A, so that the constant sequence a_n = s satisfies the required condition.

If s is a limit point of A, then s may or may not belong to A. In any case, since s = Sup A, there exists a_1 in A such that s -1 < a_1 < s.

Suppose that, for some positive integer n, there are a_1 <, a_2 <....a_n in A such that, for each i=1,2...n, we have

s - 1/i < a_i < s

Then, d = min {1/(n+1), a - a_n}/ > 0, so that there exists a_(n+1) in a such that

s - d < a_(n+1) < s. In virtue of the definition of d, it follows that

a_(n +1) > a_n and s - 1/(n+1) < a_(n+1) < s. This completes the induction and show there exists a sequence a_n in A such that, for each n,

a_n < a_(n+1) and s -1/n < a_n < s. Since 1/n -> 0 it follows a_n is increasing and converges to s.

This completes the proof.