If $A\subseteq(0,+\infty)$ is nonempty and closed under addition then it is not bounded above. Let $A\subseteq(0,+\infty)$ be a nonempty set and closed under addition. Using the fact that the set of natural numbers is not bounded above, I can prove that $A$ is also not bounded above. I want to know that if is it possible to prove this without use of this fact? Thanks.
 A: Assume that $A$ is bounded above.
Thus $A$ has a supremum $s=\sup A>0$
From the fundamental property of supremum we have that $\forall \epsilon>0$ exists $x \in A$ such that $x>s-\epsilon$
For $\epsilon=\frac{s}{2}>0$ exists $x \in A$ such that $x>\frac{s}{2}$
Also for $\epsilon=\frac{s}{5}>0$ exists $y \in A$ such that $y>\frac{4s}{5}$
Thus $x+y \in A$ and $x+y>s$ which is a contradiction
A: In order for  set to be bounded above you need to have an upper bound for the set. 
Suppose A is bounded above and the upper bound is M.
Note that M is a positive number and  the sequence $(a_n)$ defined by $a_n=M/n$ approaches $0$
A is not empty, thus we can pick an element $a\in A$ and a large enough $n$ such that $M/n < a$ 
That implies $na>M$.
This is not possible due to the fact that $na = a+a+a+..+a$ is an element of A and it is larger than $M$ which is an upper bound  of  M.
Thus A is not bounded above.       
A: Pick any $a \in A$. Note that $a>0$. Since $A$ is closed under addition
an argument by induction shows that $na \in A$ for all $n=1,2,3,...$.
Hence for any $L>0$, if we pick $n > {L \over a}$ we have $na >L$, hence $A$ is not bounded above.
