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Thread: upper bound

  1. #1
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    upper bound

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

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