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Thread: infimum question

  1. #1
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    infimum question

    let A be a nonempty subset of positive real numbers with
    $\displaystyle
    \inf \ A \neq\ 0 .
    \text{ prove that } \sup \{ \frac{1}{a} \ ; a \ \in \ A \} \ = \frac {1}{\inf A} $
    Last edited by Plato; Oct 28th 2009 at 12:13 PM.
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  2. #2
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    Quote Originally Posted by flower3 View Post
    let A be a nonempty subset of positive real numbers with
    $\displaystyle
    \inf \ A \neq\ 0 .
    \text{ prove that } \sup \{ \frac{1}{a} \ ; a \ \in \ A \} \ = \frac {1}{\inf A} $
    Suppose that $\displaystyle 0<\lambda =\inf(A)$. If $\displaystyle a\in A$ then $\displaystyle \frac{1}{a}\le \frac{1}{\lambda}$.
    Therefore the set $\displaystyle \left\{\frac{1}{a}:a\in A\right\}$ has a $\displaystyle \sup$ $\displaystyle \gamma\le \frac{1}{\lambda}$.

    Suppose that $\displaystyle \gamma <\frac{1}{\lambda}$. Show that leads to a contradiction.
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