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

• Feb 5th 2011, 08:39 AM
alice8675309
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.
• Feb 5th 2011, 08:47 AM
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.
• Feb 5th 2011, 08:54 AM
Plato
If $s\in A$ we are done. Right? So suppose that $s\notin A$.
Because $s-1 we know that $s-1$ is not an upper bound for $A$.
So that implies $\left( {\exists a_1 \in A} \right)\left[ {s - 1 < a_1 < s} \right]$. WHY!

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

Carry on.
• Feb 5th 2011, 08:55 AM
alice8675309
Quote:

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.
• Feb 5th 2011, 08:58 AM
Plato
• Feb 5th 2011, 09:01 AM
alice8675309
Quote:

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.

I completely didn't see this part on my paper. Its a reminder: A point p is in the closure of a set A if and only if every neighborhood of p contains an element of A. So I'm probably going to have to use that fact.
• Feb 5th 2011, 09:35 AM
Hartlw
A set S of real numbers bounded from above has a lub (sup) which is either an isolated point or an accumulation point of S. In either case it satisfies the definition of a boundary point: every neighborhood has a point in S and a point not in S.
• Feb 6th 2011, 04:13 AM
HallsofIvy
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.