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Math Help - Supremum and infimum

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    Supremum and infimum

    Let S and T be non-empty bounded subsets of \mathbb{R}, and suppose that for all s \in S and t \in T, we have s \leq t.

    Prove that supremum of S \leq infimum T.
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  2. #2
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    Quote Originally Posted by cgiulz View Post
    Let S and T be non-empty bounded subsets of \mathbb{R}, and suppose that for all s \in S and t \in T, we have s \leq t.
    Prove that supremum of S \leq infimum T.
    There is no problem knowing that the two exist. Why?
    So let \sigma  = \sup (S)\;\& \;\tau  = \inf (T).
    Working for a contradiction, suppose that \tau < \sigma.
    So \tau is not an upper bound of S. Why?
    What does that imply about some element of S?
    Where is the contradiction?
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  3. #3
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    Since S and T are bounded there must be an infinimum and supremum.

    We can see that \tau is not an upper bound of S since it is smaller than the least upper bound.

    So clearly there is some element t \in T and s \in S such that s > t. So we have our contradiction.

    Thus, Sup S \leq Inf T.

    Beautiful guidelines.
    Last edited by cgiulz; September 29th 2009 at 04:49 PM.
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