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Math Help - Infinum and surpemum

  1. #1
    iva
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    Infinum and surpemum

    If i have a set S = {x| |x-2| + |3-x| = 1 }

    and i have to check what the sup and inf are for it, is this correct way of working it out:

    -1 < x-2 + 3-x < 1 or -2 < 0 < 0

    Since x cancels out can i say that S has no supremum not infinum?

    Thank you
    Last edited by iva; August 28th 2011 at 05:32 AM. Reason: correction
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    Re: Infinum and surpemum

    Quote Originally Posted by iva View Post
    If i have a set S = {x| |x-2| + |3-x| = 1 }
    and i have to check what the sup and inf are for it, is this correct way of working it out:
    Well S=[2,3]
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  3. #3
    iva
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    Re: Infinum and surpemum

    Thanks then that would make sup S = 3 and inf S = 2, is that right?
    But how did you get to S=[2,3] as with my equation above i couldn't get a sensible result for x

    Thank you
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    Re: Infinum and surpemum

    Why not consider each term separately?

    Since |3-x| is non-negative, |x-2|<=1. Solve this to get 1<=x<=3. Since 3 is in the set, it's the supremum. Now do something similar with |3-x|.
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    Re: Infinum and surpemum

    Quote Originally Posted by LoblawsLawBlog View Post
    Why not consider each term separately?
    Since |3-x| is non-negative, |x-2|<=1. Solve this to get 1<=x<=3. Since 3 is in the set, it's the supremum. Now do something similar with |3-x|.
    The solution set for |x-2|+|x-3|=1 is 2\le x\le 3

    |x-3| is the distance from x to 3.

    |x-2| is the distance from x to 2.

    So if 2\le x\le 3 we have |x-2|+|x-3|=1.
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  6. #6
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    Re: Infinum and surpemum

    I didn't mean that 1<=x<=3 is the solution set, but it's still true. |3-x|<=1 then gives 2<=x<=4, which gives the infinum. I didn't see your solution immediately so I went with the slightly longer one.
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