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Math Help - prove

  1. #1
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    prove

    inf M = -(sup (-M))

    ok i tried this:

    inf M = a => sup (-M) = -a
    - sup (-M) = -(-a) = a
    inf M = -sup(-M)

    is this good enough proof or am i missing something?
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  2. #2
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    Quote Originally Posted by metlx View Post
    inf M = -(sup (-M))

    ok i tried this:

    inf M = a => sup (-M) = -a
    This statement assumes what you are asked to prove!

    [/quote]- sup (-M) = -(-a) = a
    inf M = -sup(-M)

    is this good enough proof or am i missing something?[/QUOTE]
    No, it isn't. The fact that you haven't actually used an properties of "lim" and "sup" themselves should have been a clue! Use the definitions of "inf" and "sup". The inf M is, first of all, a lower bound on M. That is, if x\in M, then x> inf M. Can you show that -(sup(-M)) is a lower bound? If [mathj]x\in M[/tex] then -x\in -M. Since sup(-M) is an upper bound, we must have -x< sup(-M). Multiply on both sides by -1: x> -sup(-M).

    But you still have to prove that -sup(-M) is the greatest lowerbound. To do that let m be any lower bound on M. You will need to show that -m is then an upper bound on -M. Because "sup(-M)" is the least upper bound, you now have -m> sup(-M). Now multiply on both sides by -1.
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  3. #3
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    I understand what you're saying but writing it with symbols might be a problem.
    We just started with axioms, lower & upper boundaries at uni.. the professor wrote on the blackboard to prove the above and none of us have any idea how.. there's also no proof in the book :/
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