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Math Help - Suprema question

  1. #1
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    Suprema question

    Let S,T be non-empty bounded subsets of the real numbers. Prove that S \cup T is bounded above.
    How would you start this question?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by worc3247 View Post
    Let S,T be non-empty bounded subsets of the real numbers. Prove that S \cup T is bounded above.
    How would you start this question?

    If s\geq 0,\;t\geq 0 are upper bounds for S and T respectively, prove that x\leq s+t for all x\in S\cup T .

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  3. #3
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    That`s not right, you should take max{s,t} as upper bound. For example think of S=T=[-2,-1] and s=t=-1. In this case s+t=-2 is no upper bound of the union.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Kevin123 View Post
    That`s not right. mFor example think of S=T=[-2,-1] and s=t=-1. In this case s+t=-2 is no upper bound of the union.[/
    It is right, I said

    s\geq 0,\;t\geq 0

    upper bounds, not minimum upper bounds.

    you should take max{s,t}
    Or

    \max \{s,t\}+\pi



    Fernando Revilla
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  5. #5
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    Oh sorry, I did not see that you assume s,t >= 0. Of course then your answer is right... :-)
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  6. #6
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    I think I can see the general idea, but am having trouble writing a formal proof. Here is what I have thus far:
    Let \alpha \geq 0and \beta \geq 0 such that \alpha , \beta are upper bounds for S and T respectively.
    If we set l = max{ \alpha , \beta}, then:
    l \geq x, \forall x \in S \cup T. Therefore S \cup T is bounded above.

    Is this along the right lines or am I way off?
    Last edited by worc3247; January 2nd 2011 at 12:36 PM.
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  7. #7
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    You cannot assume that the upper bound is in S. Think of S=(0,1). Then any upper bound is greater or equal 1.
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  8. #8
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    Ok, so I need to set \alpha and \beta so that they are greater than the supremum of the sets?
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