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Real Analysis Least Upper Bound Question

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data

    If S1, S2 are nonempty subsets of ℝ that are bounded from above, prove that

    l.u.b. {x+y : x \in S1, y \in S2 } = l.u.b. S1 + l.u.b. S2
    2. Relevant equations

    Least Upper Bound Property

    3. The attempt at a solution

    Using the least upper bound property, let us suppose that a is an upper bound for S1 and b is an upper bound for S2, as they are both a set of real numbers. Then there exists an x in S1 s.t. x≤a and there exists a y in S2 s.t. y≤b. By adding them together, x+y≤a+b where a and b are the upper bounds for their respected set of real numbers. Thus, it is equal to l.u.b. S1 + l.u.b. S2

    I think this is how it goes. I'm sorry if my proof is horrible; it has been a significant period of time since I've done proofs, as statistics and probability classes don't use proofs! Also, I can't get \in to work. Sorry about that!
     
  2. jcsd
  3. Sep 12, 2012 #2

    jbunniii

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    All that you have shown thus far is that if [itex]x \in S_1[/itex] and [itex]y \in S_2[/itex], then [itex]x + y \leq lub(S_1) + lub(S_2)[/itex].

    This implies that [itex]lub(x + y : x\in S_1, y\in S_2) \leq lub(S_1) + lub(S_2)[/itex].

    However, you still have to prove the opposite inequality.
     
  4. Sep 12, 2012 #3
    Wait, sorry if I don't understand. Are you saying I have to prove the inequality when instead of it being ≤, I prove ≥?
     
  5. Sep 12, 2012 #4

    jbunniii

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    Yes. If your goal is to prove that A = B, and you have only proved A <= B, then you also need to show that A >= B before you can conclude that A = B.
     
  6. Sep 12, 2012 #5
    ...oh. WELL, excuse me while I think about this some more and get back to you. Haha.
     
  7. Sep 12, 2012 #6
    So let me get this straight. I now have to prove that x ≥ a in S1 and y ≥ b in S2. BUT, isn't that a contradiction? If you add them, you get x+y≥a+b, but if the elements of x, y in their respective sets are greater than the l.u.b., then a and b are not the l.u.b.

    I don't quite understand. Perhaps you can guide me in the right direction. Much thanks.

    -J
     
  8. Sep 13, 2012 #7

    jbunniii

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    No, you have to show that [itex]lub(x + y : x \in S_1, y \in S_2) \geq lub(S_1) + lub(S_2)[/itex]

    I'm going to introduce some shorthand so it's easier to type:

    a = lub(S_1)
    b = lub(S_2)
    c = lub(x+y : x in S_1, y in S_2)

    If there were specific elements x in S_1 and y in S_2 such that x + y >= a + b, then you'd be done, because x + y <= c. But there might not be such elements. In fact, if there were, then you would have x = a and y = b. Thus each of the two sets S_1 and S_2 would contain its supremum. This is not guaranteed.

    However, if you let [itex]\epsilon > 0[/itex], can you find x in S_1 and y in S_2 such that [itex]x > a - \epsilon[/itex] and [itex]y > b - \epsilon[/itex]? i.e. you might not be able to find elements that achieve the upper bound, but can you get them arbitrarily close? If so, what can you do with that?
     
    Last edited: Sep 13, 2012
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