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Math Help - analysis : addition of supremums

  1. #1
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    analysis : addition of supremums

    prove or disprove the following statement. Let A and B be sets that are bound above. Define A+B to be the following set: (A+B) = {a+b|a is an element of A and b is an element of b}. If supA=alpha and supB=beta then sup(A+B) alpha + beta.

    Thanks for your time and consideration.

    decohen
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  2. #2
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    Quote Originally Posted by decohen@purdue.edu View Post
    Define A+B to be the following set: (A+B) = {a+b|a is an element of A and b is an element of b}. If supA=alpha and supB=beta then sup(A+B) alpha + beta.
    Clearly A+B is bounded above by \alpha + \beta.
    So let \gamma  = \sup (A + B) and suppose that \gamma < \alpha + \beta.
    Then we know that \gamma  < \frac{{\alpha  + \beta  + \gamma }}{2} < \alpha  + \beta \, \Rightarrow \,\left\{ \begin{gathered}  \frac{{\beta  + \gamma  - \alpha }}{2} < \beta  \hfill \\  \frac{{\alpha  + \gamma  - \beta }}{2} < \alpha  \hfill \\ \end{gathered}  \right.
    By definition of supremum, \left( {\exists a' \in A} \right)\left( {\exists b' \in B} \right)\left[ {\frac{{\beta  + \gamma  - \alpha }}{2} < b' \leqslant \beta  \wedge \frac{{\alpha  + \gamma  - \beta }}{2} < a' \leqslant \alpha } \right].
    By adding we get \gamma  < a' + b' which is a clear contradiction.
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  3. #3
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    wouldn't you have to show that the inequality isn't satisfied in the reverse direction as well?

    that if \gamma = \sup (A + B) then if we have \alpha + \beta < \gamma then we get a contradiction as well.
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