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  1. #1
    Senior Member Sampras's Avatar
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    supremum

    Let $\displaystyle A $ be a nonempty set of real numbers which is bounded below. Let $\displaystyle -A = \{-x: x \in A \} $. Prove that $\displaystyle \inf A = -\sup(-A) $.

    So I want to show that (i) $\displaystyle \inf A \leq -\sup(-A) $ and (ii) $\displaystyle \inf A \geq -\sup(-A) $. Now $\displaystyle \inf A $ exists because $\displaystyle A \neq \emptyset $ and $\displaystyle A $ is bounded below. Also, $\displaystyle \sup(-A) $ exists since $\displaystyle -A $ is non-empty and bounded above.

    Now $\displaystyle -x \leq \sup(-A) \leq \inf A $ and $\displaystyle \sup(-A) \leq \inf A \leq x $. I took $\displaystyle A $ to have a lower bound greater than $\displaystyle 0 $. You can apply the same argument if you set $\displaystyle A $ to have a lower bound less than $\displaystyle 0 $ (e.g. break it into cases)? From these inequalities, we can conclude that $\displaystyle \inf A = -\sup(-A) $? Adding them we get $\displaystyle \sup(-A)-x \leq \sup(-A)+ \inf A \leq \inf A + x $.
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  2. #2
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    Here is an outline that may help you get some ideas on your proof.

    $\displaystyle x \in A\, \Rightarrow \,\inf (A) \leqslant x\, \Rightarrow \, - x \leqslant - \inf (A)\, \Rightarrow \,\sup ( - A) \leqslant - \inf (A)$.

    Now suppose that $\displaystyle sup ( - A) < - \inf (A)$, then $\displaystyle \inf (A) < - \sup ( - A)$.

    By definition of inf, $\displaystyle \left( {\exists a \in A} \right)\left[ {\inf (A) \leqslant a < - \sup ( - A)} \right]$.

    But that means that $\displaystyle \sup ( - A) < - a$, a contradiction.
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Plato View Post
    Here is an outline that may help you get some ideas on your proof.

    $\displaystyle x \in A\, \Rightarrow \,\inf (A) \leqslant x\, \Rightarrow \, - x \leqslant - \inf (A)\, \Rightarrow \,\sup ( - A) \leqslant - \inf (A)$.

    Now suppose that $\displaystyle sup ( - A) < - \inf (A)$, then $\displaystyle \inf (A) < - \sup ( - A)$.

    By definition of inf, $\displaystyle \left( {\exists a \in A} \right)\left[ {\inf (A) \leqslant a < - \sup ( - A)} \right]$.

    But that means that $\displaystyle \sup ( - A) < - a$, a contradiction.
    Or $\displaystyle x \in A \Rightarrow x \geq \inf A $. Then $\displaystyle -x \leq -\inf A $. Thus $\displaystyle \sup(-A) $ exists. Now suppose there is some $\displaystyle p $ such that $\displaystyle -x \leq p < -\inf A $. Then $\displaystyle \inf A \leq -p < x $. This contradicts the definition of infimum. Hence $\displaystyle \sup(-A) = - \inf A $. Or $\displaystyle \inf A = -\sup(-A) $.
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