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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.

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That's perfect. $ $

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I was just worried that I couldn't say that: "if for every $ x \in E$, $ \alpha_o\leq x $ then for every $ -x \in F $, $ -\alpha_o\geq -x $." –  mkeachie Oct 9 '12 at 0:47
    
Of course you can. The justification depends on how you have defined/constructed the real numbers. This is how I would justify it: You have $\alpha_0\leq x$; by definition, this is $x-\alpha_0\geq0$; then $-(x-\alpha_0)\leq0$, and this is $-x\leq-\alpha_0$. –  Martin Argerami Oct 9 '12 at 0:58
    
Thank you very much Martin. –  mkeachie Oct 9 '12 at 1:11

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