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Math Help - supremum

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    Quote Originally Posted by suliman View Post
    Suppose:

     <br />
\sup_{m\in \mathbb{N}}(b_m) > \sup_{m\in \mathbb{N}}(a_m)<br />

    This implies that there exists a k\in \mathbb{N} such that:

     <br />
b_k > \sup_{m\in \mathbb{N}}(a_m)<br />

    which in turn implies that there exists a j \in \mathbb{N} such that:

     <br />
X_{k,j} > \sup_{m\in \mathbb{N}}(a_m)<br />

    But:

     <br />
\sup_{m\in \mathbb{N}}(a_m) \ge X_{u,v}<br />

    for all u,v \in \mathbb{N} .

    So we have a contradiction; so the premiss that:

     <br />
\sup_{m\in \mathbb{N}}(b_m) > \sup_{m\in \mathbb{N}}(a_m)<br />

    is false.

    In a similar manner we can show that:

     <br />
\sup_{m\in \mathbb{N}}(b_m) < \sup_{m\in \mathbb{N}}(a_m)<br />

    is false, so we are forcesd to conclude that:

     <br />
\sup_{m\in \mathbb{N}}(b_m) = \sup_{m\in \mathbb{N}}(a_m)<br />

    RonL
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