Results 1 to 3 of 3

Thread: analysis : addition of supremums

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    2

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,774
    Thanks
    2823
    Awards
    1
    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 $\displaystyle \alpha + \beta$.
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Supremums: sup AC = supA supC
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Feb 25th 2010, 08:36 PM
  2. infimums and supremums
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 20th 2009, 03:25 AM
  3. inequality of supremums
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 27th 2009, 12:32 AM
  4. infimum and supremums
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 16th 2008, 03:07 AM
  5. Replies: 3
    Last Post: Jan 15th 2008, 10:46 PM

Search Tags


/mathhelpforum @mathhelpforum