# Thread: Prove s is in A or is an accumulation point of A..

1. ## Prove s is in A or is 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.

2. 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.

3. If $\displaystyle s\in A$ we are done. Right? So suppose that $\displaystyle s\notin A$.
Because $\displaystyle s-1<s$ we know that $\displaystyle s-1$ is not an upper bound for $\displaystyle A$.
So that implies $\displaystyle \left( {\exists a_1 \in A} \right)\left[ {s - 1 < a_1 < s} \right]$. WHY!

Next let $\displaystyle \delta_2=\max\left\{a_1,s-\frac{1}{2}\right\}.$
Can you say $\displaystyle \delta_2<s~?$ How do we select $\displaystyle a_2~?$

Carry on.

4. Originally Posted by DrSteve
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.
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.