# analysis : addition of supremums

• Feb 16th 2009, 10:49 AM
decohen@purdue.edu
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
• Feb 16th 2009, 01:04 PM
Plato
Quote:

Originally Posted by decohen@purdue.edu
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 $\displaystyle \alpha + \beta$.
So let $\displaystyle \gamma = \sup (A + B)$ and suppose that $\displaystyle \gamma < \alpha + \beta$.
Then we know that $\displaystyle \gamma < \frac{{\alpha + \beta + \gamma }}{2} < \alpha + \beta \, \Rightarrow \,\left\{ \begin{gathered} \frac{{\beta + \gamma - \alpha }}{2} < \beta \hfill \\ \frac{{\alpha + \gamma - \beta }}{2} < \alpha \hfill \\ \end{gathered} \right.$
By definition of supremum, $\displaystyle \left( {\exists a' \in A} \right)\left( {\exists b' \in B} \right)\left[ {\frac{{\beta + \gamma - \alpha }}{2} < b' \leqslant \beta \wedge \frac{{\alpha + \gamma - \beta }}{2} < a' \leqslant \alpha } \right]$.
By adding we get $\displaystyle \gamma < a' + b'$ which is a clear contradiction.
• Oct 24th 2009, 09:01 PM
lllll
wouldn't you have to show that the inequality isn't satisfied in the reverse direction as well?

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