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Math Help - prove or disprove question

  1. #1
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    prove or disprove question

    Let S be a non-empty set of real numbers and suppose that L is the least upper bound for S. Prove that there is a sequence of points x_n in S such that x_n --> L as n --> \infty.
    Prove or disprove:
    For non-empty bounded sets S and T: lub(SUT) = max{lub(S), lub(T)}.
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  2. #2
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    I see that you have over fifty postings.
    By now you should understand that this is not a homework service
    So you need to either post some of your work on a problem or you need to explain what you do not understand about the question.
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  3. #3
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    okay, here is what i got so far

    for the first one, Let x_0 by any point of S. Since L is a least upper bound, for each n >0, there must be an element (x_n) of S in the open interval ( max(x_n-1,L-1/n),L). Use these x_n.

    For the second, i am thinking to prove by contradiction
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  4. #4
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    For the first one there are two cases: i) L\in S and ii) L\notin S.
    In case i) take a constant sequence of Lís.
    In case ii) there is a sequence of distinct terms such that x_n  \in \left( {L - \frac{1}{n},L} \right).
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  5. #5
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    Quote Originally Posted by Plato View Post
    For the first one there are two cases: i) L\in S and ii) L\notin S.
    In case i) take a constant sequence of L’s.
    In case ii) there is a sequence of distinct terms such that x_n  \in \left( {L - \frac{1}{n},L} \right).
    I think the question stated that we need to PROVE that there is a sequence of points x_n such that x_n\rightarrow L as n\rightarrow\infty.

    I've done some work on this thanks to other sources, but I'm sure it either has some errors or is completely wrong.

    For some x_n\in S < L=lub(S), \exists x_{n+1}\in S = L such that \frac{x_n+L}{2}<x_{n+1}<L where x_n\rightarrow L as n\rightarrow\infty.

    Can anyone tell me if I'm missing anything or if I did something wrong?
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  6. #6
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    If S=[0,1]\cup \{2\} then L=2.
    The only sequence of points from S converging to 2 is come variation of \left( {\forall n} \right)\left[ {a_n  = 2} \right].

    Does what you have work for this example?
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  7. #7
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    Quote Originally Posted by Plato View Post
    If S=[0,1]\cup \{2\} then L=2.
    The only sequence of points from S converging to 2 is come variation of \left( {\forall n} \right)\left[ {a_n  = 2} \right].

    Does what you have work for this example?
    I'm not exactly sure how that is meant to work into the first half of the question, or I'm mistaking it for being related to the first half when it's meant for the second.

    I came up with my current answer from this link: http://www.mathisfunforum.com/viewtopic.php?id=1645, but I'm sure there's something in it that's missing.
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  8. #8
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    The proof in the link is wrong. Use the same counterexample I gave.

    The usual problem that goes with a similar proof is:
    If L=\text{LUB}(S)~\&~L\notin S then there is a sequence of distinct points from S that converges to L.

    But the example where L\in S may not work except for almost constant sequence.
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