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Math Help - 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 \alpha  =   \sup (A) and \beta  =  \sup (B)
    \alpha < \beta  \Rightarrow  \alpha  \not=   \beta
    \rightarrow  B\setminus A  \not=  \emptyset
    \rightarrow \exists   b \in  B :   b \notin A
    \rightarrow \alpha \le b
    \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:

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

    \rightarrow \alpha \le b

    What if A=(1,2) and B=(0,3)? Then 0.5\in B, 0.5\not\in A, but 0.5<a for all 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 \ (\alpha  ,  \beta)
    and any point in that interval satisfies that property
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