# Jumping to conclusions here?

1. Feb 16, 2005

### semidevil

let A be an infinite subset of R and R is bounded above, and u:= sup A. show that there exist a sequence (Xn) with X(n) belongs to A, such that u = lim(Xn).

ok, so suppose that there does exist a sequence X(n) in A. We know that SupA = u. by the subsequence theorem, if A converges to u, then so will any sequence that belongs to it right? and by another theorem, the limit is the supremum....correct?

I dont know...maybe too easy? I feel I didn't cover everything

2. Feb 16, 2005

### learningphysics

Are you sure this is what the subsequence theorem says?

3. Feb 16, 2005

### matt grime

Let e>0, then there exists a(1) in A such that supA - e <= a(1) <=Sup(A)

Now let e be one half SupA -a(1)

Pick and a(2) in the range sup(A) - e to Sup A.

repeat and get a strictly increasing sequence tending to sup(A).