Proving with completeness axiom Suppose we claim that if there is a set $E := \{a \in \mathbb{R} : a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+} \}$, then it must be true that $a \leq 0$. 
I aim to prove this using only the ordered field axioms and the completeness axiom (as per the "request" by my professor).
I am going to prove it by contradiction. So suppose $\exists a>0$ so that $0 < a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+}$. Now since $a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+}$, we have $a < \frac{\epsilon}{2}$ since $\frac{\epsilon}{2} \in \mathbb{Q}^{+}$. So $2a < \epsilon$. It follows that we can always find $a' \in E$ such that $a<a'$ for any $a \in E$. 
Now here is my problem. I know that the last sentence in the previous paragraph somehow contradicts the completeness axiom. But I cannot see why. 
Do you have any suggestions?
 A: The completeness axiom states that any set that's bounded above has a least upper bound.
You use this to produce the contradiction as follows:
Let $a$ be the least upper bound of $E$. You claim $a \le 0$. By contradiction assume that it is not. Then there exists a $q \in \mathbb Q$ such that $0<q<a$ since between any two real numbers there exists a rational number (this is a consequence of the completeness axiom).
Consider any $x \in E$. Then we have $x < q$ since $q > 0$. But then $q$ is an upper bound of $E$ and furthermore smaller than $a$. This contradicts $a$ being the least upper bound. Hence $a$ must be less than or equal to $0$.  
A: The contradiction you are looking for is as follows : Let x= $\text{lub}(E)$. If $E$ has a positive member then $x>0$.Now if $x\ge q$ for any $q \in Q^+$ then $x/2 \ge q/2 >y$ for all $y \in Q^+$, so $x/2$ is an upper bound for $E$ and is less than $\text{lub}( E)$,  which is absurd, so we must have $x<q$ for all $q \in Q^+$. Therefore  $x \in E$. But as you have shown,$ 0 <x \in E \implies 2 x \in E$.  But then $x$ is not $\text{lub}( E)$ because $x$ is not an upper bound for $E$ because $E$ has a member $(2x)$ which is greater than $x$. (Equivalently we could say $ 0<x=\text{lub}( E)$ and $2x \in E$ implies $0< 2 x \le \text{lub}( E)= x$ which is absurd.
