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Math Help - upper bound

  1. #1
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    upper bound

    Let  A \subset \mathbb{R} be bounded above and  B = \{x: x \ \text{is an upper bound of} \ A \} . Prove  \sup A = \inf B .

    Since  B is bounded below, it has a greatest lower bound  \inf B . To show that  \sup A = \inf B , we need to show that  \forall x \in A, x \leq \sup A \Longleftrightarrow \forall x \in B, x \geq \inf B ? It this all there is to it?
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  2. #2
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    Suppose that \alpha  = \sup (A)\,\& \,\beta  = \inf (B).
    If \beta < \alpha  then by definition of supremum and infimum \beta is not an upper bound of A, so \left( {\exists c \in A} \right)\left[ {\beta  < c \le \alpha } \right] but c is not a lower bound of B. That means \left( {\exists d \in B} \right)\left[ {\beta  \le d < c} \right], but that means d is a upper bound of A which is impossible because {d < c \le \alpha }. That means that \alpha  \le \beta .
    This time suppose that \alpha  < \beta which implies \alpha  < \frac{{\alpha  + \beta }}{2} < \beta .
    But this means that \frac{{\alpha  + \beta }}{2} is an upper bound for A but it cannot belong to B. That is a contradiction. Thus \alpha  = \beta .
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  3. #3
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    Quote Originally Posted by Plato View Post
    Suppose that \alpha  = \sup (A)\,\& \,\beta  = \inf (B).
    If \beta < \alpha  then by definition of supremum and infimum \beta is not an upper bound of A, so \left( {\exists c \in A} \right)\left[ {\beta  < c \le \alpha } \right] but c is not a lower bound of B. That means \left( {\exists d \in B} \right)\left[ {\beta  \le d < c} \right], but that means d is a upper bound of A which is impossible because {d < c \le \alpha }. That means that \alpha  \le \beta .
    This time suppose that \alpha  < \beta which implies \alpha  < \frac{{\alpha  + \beta }}{2} < \beta .
    But this means that \frac{{\alpha  + \beta }}{2} is an upper bound for A but it cannot belong to B. That is a contradiction. Thus \alpha  = \beta .
    Why can't  \frac{\alpha + \beta}{2} belong to  B ?
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  4. #4
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    oh ok because  \frac{\alpha + \beta}{2} < \beta .
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