# On the greatest lower bound property

### 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?

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