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Math Help - infinium and supremum question

  1. #1
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    infinium and supremum question

    I'm a little stuck on this question:

    Let S be a nonempty bounded subset of \mathbb{R} and let b<0. Show that inf(bS) = bsup(S)

    so far I have:

    \beta = inf(S) \ \therefore \ b\beta = inf(bS) \ \forall \ s \in S \  , \ \beta \leq s

    b\beta \leq bS \  \forall \ s \in S \ \rightarrow -n\beta \geq -nS

    therefore b\beta is a upper bound of bS.

    That's what I have so far.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by lllll View Post
    I'm a little stuck on this question:

    Let S be a nonempty bounded subset of \mathbb{R} and let b<0. Show that inf(bS) = bsup(S)

    so far I have:

    \beta = inf(S) \ \therefore \ b\beta = inf(bS) \ \forall \ s \in S \  , \ \beta \leq s

    b\beta \leq bS \  \forall \ s \in S \ \rightarrow -n\beta \geq -nS

    therefore b\beta is a upper bound of bS.

    That's what I have so far.
    Let S\subset\mathbb{R} be bounded, and set \beta :=\inf\{S\}.
    Then \beta is finite, therefore, we know that
    t\geq\beta \text{ for all }t\in S.\qquad(*)
    To complete the proof, we have to show that \beta b=\sup\{bS\} holds, where b<0.
    Note that bS=\{bt:t\in S\}.
    Multiplying both sides of (1) with b, we get
    bt\leq b\beta \text{ for all }t\in S,
    which is equivalent to
    bt\leq b\beta \text{ for all }bt\in bS.
    This proves that b\beta=\sup\{bS\}, which is the desired identity.

    You may wish to check the following book:
    Introduction to Calculus and ... - Google Book Search
    Last edited by bkarpuz; September 20th 2008 at 09:00 AM. Reason: Reference is added.
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