A finite subset of the reals is closed. It is bounded above by its greatest element, which is s. In this case s is in A, but is not an accumulation point of A.
Let A be a nonempty set of real numbers that is bounded from above, and let s = sup(A). Prove that either s ∈ A or s is an accumulation point of A. Hence, in either case, s is in the closure.
For this, would my conclusion be that s is an accumulation point or both s is in A and is an accumulation point. If I know which direction I am aiming for, I think I can prove it.
Oh, I've seen a few proofs of this but I was trying to work it out myself but one proof that I've seen showed that s is not the least upper bound. Then I've seen a contradiction that showed that s being an accumulation point and then ANOTHER proof i've seen is a proof by contraposition showing that s is neither a point of A or an accumulation point of A. So im not sure which way to direct my proof.
Your original post said "For this, would my conclusion be that s is an accumulation point or both s is in A and is an accumulation point."
Dr. Steve's point was, "no, prove a is an accumulation point or is in A- the "and is an accumulation point" part may not be true.