# Help proving a condition where the greatest lower bound of a set is equal to the least upper bound of another.

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. 
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