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Math Help - supremum

  1. #1
    Senior Member Sampras's Avatar
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    supremum

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

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

    Now  -x \leq \sup(-A) \leq \inf A and  \sup(-A) \leq \inf A \leq x . I took  A to have a lower bound greater than  0 . You can apply the same argument if you set  A to have a lower bound less than  0 (e.g. break it into cases)? From these inequalities, we can conclude that  \inf A = -\sup(-A) ? Adding them we get  \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.

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

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

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

    But that means that \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.

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

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

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

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