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Supremum / Infimum

  1. Aug 11, 2008 #1

    danago

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    Gold Member

    Suppose that [tex]A,B \subseteq \Re^+[/tex] are non empty and bounded sets. Define + and . as the following set opetations:

    [tex]
    \begin{array}{l}
    A + B = \{ a + b|a \in A,b \in B\} \\
    A.B = \{ ab|a \in A,b \in B\} \\
    \end{array}
    [/tex]

    Prove that [tex]\sup (A + B) \le \sup A + \sup B[/tex]


    I started by letting [tex]a \in A,b \in B[/tex]. From the definition of supremum:

    [tex]
    \begin{array}{l}
    a \le \sup A \\
    b \le \sup B \\
    \end{array}
    [/tex]

    I then added the two inequalities to give:

    [tex]
    a + b \le \sup A + \sup B
    [/tex]

    Since this holds true for any a and b, sup A+sup B is an upper bound on the set A+B. This is where im a bit stuck. I cant really see why sup(A+B) isnt strictly equal to sup A + sup B. I tried coming up with a few example sets A and B but that didnt really help anything.

    Any suggestions?

    Thanks,
    Dan,
     
  2. jcsd
  3. Aug 11, 2008 #2
    You've already established that sup A + sub B is an upper bound for the A + B. All that remains is to show that it's the least upper bound.
     
  4. Aug 11, 2008 #3

    danago

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    I guess thats what im having trouble with. I tried assuming that there exists some other upper bound 'r' which is less than sup A + sup B, and then showing that this leads to a contradiction. No luck though :(
     
  5. Aug 11, 2008 #4

    Defennder

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    Homework Helper

    He isn't required to show that it is the least upper bound, only that it is an upper bound.
     
  6. Aug 11, 2008 #5
    Yes, but he already showed this in the first post, my understanding was that he was curious to see why it's the least upper bound.

    danago, if k is the supremum of A + B, then for any epsilon > 0, there's an element x in A + B satisfying |k - x| < epsilon. All you have to show is that sup A + sup B has this property.
     
  7. Aug 11, 2008 #6

    Defennder

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    If you want to prove that anyway, just note that [tex] \sup A - \frac{\epsilon}{2} < a \ \mbox{for} \ \exists a \ \mbox{and} \ \forall \epsilon > 0 [/tex]. The same applies for b. See how to continue from here?
     
  8. Aug 11, 2008 #7

    HallsofIvy

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    Staff Emeritus
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    But his question was, "Why is it not the least upper bound?"

    danago, the fact that you are asked to prove that [itex]sup A+B\le \sup A+ \sup B[/itex] does not necessarily imply they are not equal. Only that you are not asked to prove that.
     
  9. Aug 11, 2008 #8

    danago

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    Ok fair enough. So what ive done is sufficient to answer the question then?
     
  10. Aug 11, 2008 #9

    Defennder

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    Homework Helper

  11. Aug 11, 2008 #10

    danago

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    Gold Member

    Alright thanks everyone :smile:
     
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