So I'm just not sure if the proof I have given for this claim is acceptable, and I was wondering if anyone could help me smooth it out. Thanks:
Let E be a nonempty subset of real numbers which is bounded below and let F be the set of all numbers $ -x $ such that $ x \in E $. Let $ \alpha_o $ be the greatest lower bound of E and let $ \beta_o $ be the least upper bound of F. Then: $ \alpha_o = -\beta_o$.
Proof of claim. If $ \alpha_o $ is the greatest lower bound of E then for every $ x \in E $, $ \alpha_o\leq x $. Then, by the properties of the ordered field $ \mathbb R $, for every $ -x \in F $, $ -\alpha_o\geq -x $. So $ -\alpha_o $ is an upper bound of F and $ \beta_o \leq -\alpha_o $, or rather, $ -\beta_o \geq \alpha_o $.
Similarly, if $ \beta_o $ is the least upper bound of F then for every $ -x \in F $, $ \beta_o \geq -x $. Then, by the properties of the ordered field $ \mathbb R $, for every $ x \in E $, $ -\beta_o\leq x $. So $ -\beta_o $ is a lower bound of $ E $ and $ -\beta_o \leq \alpha_o $.
This proves the claim.