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Math Help - Prove a sup

  1. #1
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    Prove a sup

    LET A={a: a>0}, where a is a rea No ,and A is bounded from above.

    Define \sqrt A= { \sqrt a:a\in A}

    Prove : Sup\sqrt A = \sqrt{Sup A}
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  2. #2
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    Here is an outline.
    For the set \alpha=\sup{A}.
    If a\le\alpha then \sqrt{a}\le\sqrt{\alpha}.
    If \beta=\sup\sqrt{A} and \beta<\sqrt{\alpha} then \beta^2<\alpha.
    So some element of A is between \beta^2~\&~\alpha.
    That gives a contradiction. I leave it to you to find it.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Here is an outline.
    For the set \alpha=\sup{A}.
    If a\le\alpha then \sqrt{a}\le\sqrt{\alpha}.
    If \beta=\sup\sqrt{A} and \beta<\sqrt{\alpha} then \beta^2<\alpha.
    So some element of A is between \beta^2~\&~\alpha.
    That gives a contradiction. I leave it to you to find it.
    Does your method work for proving : SupAB = SupASupB , where A={a: a>0},B={b:b>0} e.t.c e.t.c??
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  4. #4
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    Quote Originally Posted by alexandros View Post
    Does your method work for proving : SupAB = SupASupB , where A={a: a>0},B={b:b>0} e.t.c e.t.c??
    You know that I am just not inclined to answer your questions.
    You should either post some of your own work on problems or explain what you do not understand about the question. But instead you seem to demand a solution without any effort on your part.
    Is that being unfair to you?
    If so, please explain yourself.
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  5. #5
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    Quote Originally Posted by Plato View Post
    You know that I am just not inclined to answer your questions.
    You should either post some of your own work on problems or explain what you do not understand about the question. But instead you seem to demand a solution without any effort on your part.
    Is that being unfair to you?
    If so, please explain yourself.
    Check the thread :Supremum and infemum
    Last edited by alexandros; December 6th 2010 at 06:57 PM. Reason: correction
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