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Math Help - One more fun proof1

  1. #1
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    One more fun proof1

    Thank you in advance for any more help!

    Prove that S = { n−1/n |n ∈ N} is bounded above and that its supremumn
    is equal to 1.
    (Hint: First show that 1 is an upper bound for S. Then suppose, for a
    contradiction, that sup S = α < 1. Using a theorem from the lecture
    we can find a rational number alpha < p < 1. Deduce that we can find an
    n ∈ N with αn infiniti alpha< n/n-1 . Contradiction to infiniti alpha being an upper bound.)
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  2. #2
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    Quote Originally Posted by Swamifez View Post
    Thank you in advance for any more help!

    Prove that S = { n−1/n |n ∈ N} is bounded above and that its supremumn
    is equal to 1.
    (Hint: First show that 1 is an upper bound for S. Then suppose, for a
    contradiction, that sup S = α < 1. Using a theorem from the lecture
    we can find a rational number alpha < p < 1. Deduce that we can find an
    n ∈ N with αn infiniti alpha< n/n-1 . Contradiction to infiniti alpha being an upper bound.)
    Suppose 1 weren't an upper bound.
    Then there exists some n \in \mathbb{N} such that \frac{n-1}{n} > 1.
    Thus, n - 1 > n, particularly  -1 > 0 .
    So 1 must be an upper bound of S.

    So S is a bounded set. We still do not know if there is a supremum in the rationals because S \subset \mathbb{Q}, and \mathbb{Q} does not carry the least upper bound property.

    However, we know since the reals do have the least upper bound property that \sup S at least lies in \mathbb{R}.

    Suppose 1 were not the least upper bound of S.

    Let \sup S = \alpha \in \mathbb{R}.
    Then \alpha < 1.

    By density of the rationals in the reals, there exists \beta \in \mathbb{Q} such that \alpha < \beta < 1.
    Since \beta > \alpha, then \beta is an upper bound of S.

    Since \beta \in \mathbb{Q}, write \beta = \frac{\gamma}{\delta}, for \gamma, \delta \in \mathbb{N}, \delta \ne 0. Note that \gamma < \delta, for if not then \beta \ge 1 which cannot be true.

    Then \frac{\gamma}{\delta} \le \frac{\delta - 1}{\delta}.

    However, if this were true, then \beta would not be an upper bound of S because for n \ge \delta+1, then \frac{n-1}{n} > \beta.

    So therefore \alpha cannot be an upper bound of S (and certainly not its least upper bound).

    Therefore, the least upper bound of S is 1 as desired. QED.
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