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Math Help - Prove s is in A or is an accumulation point of A..

  1. #1
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    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.
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  2. #2
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    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.
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    If s\in A we are done. Right? So suppose that s\notin A.
    Because s-1<s 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<s~? How do we select a_2~?

    Carry on.
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    Quote Originally Posted by DrSteve View Post
    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.
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  5. #5
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    Look at reply #3.
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    Quote Originally Posted by DrSteve View Post
    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.
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    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.
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  8. #8
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    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.
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