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Thread: Real Analysis Proof Help Needed

  1. #1
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    Real Analysis Proof Help Needed

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

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

    Now I need to show that for any other upper bound $\displaystyle s$ for $\displaystyle B$, $\displaystyle r \leq s$. 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 $\displaystyle A $ is bounded below, $\displaystyle B $ is not empty. Every $\displaystyle x \in A $ is an upper bound of $\displaystyle B $. So $\displaystyle B $ is bounded above which implies that $\displaystyle \sup B = \alpha $ exists in $\displaystyle S $ where $\displaystyle A \subset S $ (e.g. assume $\displaystyle S $ has lub property). If $\displaystyle \gamma < \alpha $ then $\displaystyle \gamma $ is not an upper bound of $\displaystyle B $. And so $\displaystyle \gamma \not \in A $. So $\displaystyle \alpha \in B $.

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