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Math Help - supremum if bounded subset

  1. #1
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    supremum if bounded subset

    Let A be a non-empty bounded subset of \mathbb{R} Define:

    D=\{|x-y|:x,y\in A\}

    show that sup D = sup A - inf A.

    I can intuitively understand this however I am not too sure how to go about a rigorous proof.

    Thanks for any help
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  2. #2
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    Re: supremum if bounded subset

    Quote Originally Posted by hmmmm View Post
    Let A be a non-empty bounded subset of \mathbb{R} Define:

    D=\{|x-y|:x,y\in A\}

    show that sup D = sup A - inf A.
    Let U=\sup( A)~\&~L=\inf( A) and \{x,y\}\subset A.

    Then L \leqslant x \leqslant U\;\& \, - U \leqslant  - y \leqslant  - L which implies
     - (U - L) \leqslant x - y \leqslant U - L or |x-y|\le U-L.

    Can you finish?
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  3. #3
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    Re: supremum if bounded subset

    errm no sorry, I get that you have shown that it is an upper bound, but i don't know how to show that it is the least upper bound, sorry.

    thanks for any help
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  4. #4
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    Re: supremum if bounded subset

    Quote Originally Posted by hmmmm View Post
    errm no sorry, I get that you have shown that it is an upper bound, but i don't know how to show that it is the least upper bound, sorry.
    Let \alpha=\sup(D) suppose that \alpha<U-L.
    Use \beta=\frac{U-L-\alpha}{4}>0.
    There must be y'\in A such that L\le y'<L+\beta.
    There must be x'\in A such that U-\beta< x'\le U.

    Now you show that \alpha< |x'-y'| which is a contradiction.
    So what does that prove?
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  5. #5
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    Re: supremum if bounded subset

    It shows that if there is any number less than U-L then you can find a member of the set that is bigger than it and so it is not the supermum.

    Thanks for the help
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