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Thread: Analysis - supA<sup B

  1. #1
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    Analysis - supA<sup B

    If sup A<sup B, then show that there exists an element bB that is an upper bound for A.
    This one is really frustrating me. I'm having trouble even beginning.
    I know sup A implies s<=b where c is an upper bound for A.
    Similarily sup B implies s<= d where d is an upper bound for B.
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  2. #2
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    Let $\displaystyle \alpha = \sup (A) $ and $\displaystyle \beta = \sup (B) $
    $\displaystyle \alpha < \beta \Rightarrow \alpha \not= \beta$
    $\displaystyle \rightarrow B\setminus A \not= \emptyset$
    $\displaystyle \rightarrow \exists b \in B : b \notin A$
    $\displaystyle \rightarrow \alpha \le b$
    $\displaystyle \rightarrow \forall a \in A ; a \le b$
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  3. #3
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    Reply to bubble86:

    I'm not sure you can just go from this line:

    $\displaystyle \rightarrow \exists b \in B : b \notin A$ to

    $\displaystyle \rightarrow \alpha \le b$

    What if $\displaystyle A=(1,2)$ and $\displaystyle B=(0,3)$? Then $\displaystyle 0.5\in B$, $\displaystyle 0.5\not\in A$, but $\displaystyle 0.5<a$ for all $\displaystyle a\in A.$ Your proof is the right outline. You just need to do a bit more to ensure the second line I've quoted.
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  4. #4
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    Yes sorry i was referring to the interval $\displaystyle \ (\alpha , \beta)$
    and any point in that interval satisfies that property
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