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Suprema proof: quick check please

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Assume that [itex] X \subseteq \mathbb{R} [/itex] and [itex] Y \subseteq \mathbb{R} [/itex] are each bounded above. Then the set
    [itex] Z = \{ x + y : x \in X, y \in Y \} [/itex]
    is bounded above, and
    [itex] \text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y). \quad \quad [/itex]

    3. The attempt at a solution

    Let [itex] X \subseteq \mathbb{R} [/itex] and [itex] Y \subseteq \mathbb{R} [/itex] be bounded above, and define the set [itex] Z [/itex] as
    [itex] Z = \{ x + y : x \in X, y \in Y \}. [/itex]
    Let [itex] e_x [/itex] represent an element of [itex] X [/itex], [itex] e_y [/itex] represent an element of [itex] Y [/itex], and [itex] e_z [/itex] represent an element of [itex] Z [/itex].
    For any [itex] e_z [/itex], there exists some [itex] e_x [/itex] and [itex] e_y [/itex] such that
    [itex] e_z = e_x + e_y. \quad \quad (1) [/itex]
    Since [itex] X [/itex] and [itex] Y [/itex] are bounded above, there exists a least upper bound for [itex] X [/itex] and [itex] Y [/itex], which can be represented as [itex] x_{sup} = \text{sup}(X) \text{ and } y_{sup} = \text{sup}(Y) [/itex].
    By the definition of the least upper bound, for all [itex] e_x \in X, \, x_{sup} \geq e_x [/itex].
    Similarly, for all [itex] e_y \in Y, \, y_{sup} \geq e_y [/itex].
    From equation (1), this means that
    [itex] e_z = e_x + e_y \leq x_{sup} + y_{sup} \quad \quad (2) [/itex]
    for all [itex] e_x [/itex] and [itex] e_y [/itex].
    From equation (2), we see that [itex] x_{sup} + y_{sup} [/itex] is an upper bound for [itex] Z [/itex].
    By definition, this proves that [itex] Z [/itex] is bounded above.
    Since [itex] Z [/itex] is bounded above, note that [itex] Z [/itex] has a least upper bound which can be represented as [itex] z_{sup} = \text{sup}(Z) [/itex].
    By the definition of a least upper bound, [itex] z_{sup} \leq x_{sup} + y_{sup} [/itex], or
    [itex] \text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y).[/itex] Q.E.D.

    It wanted to be sure that this is 100% good to go before I have to turn this in. Thank you very very much for anyone that takes a look!
     
  2. jcsd
  3. Sep 3, 2011 #2
    Looks good to me. Wait for more comments
     
  4. Sep 3, 2011 #3

    micromass

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    Very nicely done!! The proof is good, but it's also written down superbly!!
    This would exactly be the kind of proof that I would like to receive as an assignment!
     
  5. Sep 3, 2011 #4
    Thank you very much for the encouraging feedback! My only small concern was the way I brought in the fact that e_z = e_x + e_y for some e_x and e_y. Intuitively, this is pretty obvious, but the course requires near perfect style as well as great logic.
     
  6. Sep 3, 2011 #5

    micromass

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    I wouldn't worry about that, it's pretty obvious.

    Maybe you will want to explain (2) some more. It's also obvious, but your professor might complain there...
     
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