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Math Help - Real Analysis Proof Help Needed

  1. #1
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    Real Analysis Proof Help Needed

    Let A be bounded below, and define B = \{b\in\Re : b is a lower bound for A\}. Show that \sup B = \inf A.

    This is what I have:
    Let r=\inf A. Let b\in B be any lower bound for A. It follows that r\geq b. Since r\geq for all b \in B, r is an upper bound for B.

    Now I need to show that for any other upper bound s for B, r \leq s. I think this has to do with showing that any upper bound for B is an element of A, but I can't figure out how to show this.

    Thanks for the help,
    Ultros
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  2. #2
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    Since  A is bounded below,  B is not empty. Every  x \in A is an upper bound of  B . So  B is bounded above which implies that  \sup B = \alpha exists in  S where  A \subset S (e.g. assume  S has lub property). If  \gamma < \alpha then  \gamma is not an upper bound of  B . And so  \gamma \not \in A . So  \alpha \in B .

    If  \alpha < \beta then  \beta \not \in B . So  \alpha = \inf A = \sup B .
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  3. #3
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    We will stick with the above notation: \alpha  = \inf (A)\,\& \,\beta  = \sup (B).
    Suppose that \alpha  < \beta then \left( {\exists k \in B} \right)\left[ {\alpha  < k \le \beta } \right] \Rightarrow \quad \left( {\exists j \in A} \right)\left[ {\alpha  \le j < k \le \beta } \right].
    But that is a contradiction because every k \in B is a lower bound for A.
    Therefore, we have \beta  \le \alpha .
    Now suppose that \beta  < \alpha. Define c = \frac{{\alpha  + \beta }}{2} \Rightarrow \quad \beta  < c < \alpha .
    This means that c is a lower bound of A which is a contradiction.
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