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Proposition:

Let $S$ be an ordered field and $S \supset E \neq \varnothing $. $E$ is bounded below. Then $ \inf E = - \sup ( - E ) $

Try:

Write $- E = \{ -x : x \in E \} $ and let $l $ be a lower bound of $E$. (This we are given). Hence, $x \geq l $ for every $x \in E$. Clearly, $-x \leq -l $. Put $u = -l$ and we observe that $u$ is an upper bound of $-E$ and consequently $\sup ( - E)$ exists. Put $\alpha = \sup ( - E ) $. It follows that $-x \leq \alpha $ for all $x$ and hence $x \geq - \alpha $. We see that $- \alpha $ is a lower bound for $E$. We need to show that $- \alpha $ is the infimum of $E$. To see this, we show that $NO$ $\beta$ with $\beta > - \alpha $ is a lower bound of $E$. Suppose the contrary, and let $\beta $ be a lower bound of $E$ such that $\beta > - \alpha $. This implies that $ - \beta < \alpha $ and $ - \beta$ cannot be an upper bound of $E$. In particular, there is some $e \in E$ with $- \beta < e $ or $\beta > - e $. So we have found some $y = -e \in E$ with $\beta > y$. In particular, $\beta$ cannot be a lower bound of $E$. Contradiction. Hence,

$$ \inf E = - \alpha = - \sup ( - E ) $$

Is this a correct proof? Also, do I have to worry whether $E$ is unbounded?

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  • 1
    $\begingroup$ See here and here. Possible duplicate. $\endgroup$ – Aaron Maroja Nov 27 '14 at 14:17
  • $\begingroup$ Your argument is fine as it is. $\endgroup$ – Brian M. Scott Nov 27 '14 at 21:03

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