I need to prove that $\mathbb{R}_{>0}$ (that is, the set positive real numbers) does not have an upper bound. I've come up with a proof that seems simple enough, but I wanted to check that I haven't assumed anything non-obvious. Here it goes.
Suppose $\mathbb{R}_{>0}$ was bounded from above. Then, by definition, there would be an $s \in \mathbb{R}$ such that $s \ge x \ \forall x \in \mathbb{R}_{>0}$. Now, since there is at least one element of $\mathbb{R}_{>0}$ greater than $0$, $s$ must be greater than $0$ too (because of transitivity of $>$). But consider $s+1$. Since $s$ is positive, $s+1$ is positive too (because $a > b \implies a+s>b+s$), and therefore an element of $\mathbb{R}_{>0}$. But $s+1>s$ (using previously mentioned property in reverse), which contradicts the original assumption. Therefore, $\mathbb{R}_{>0}$ has no upper bound.
Is this correct? It seems that the point of this exercise is to use the bare basics, otherwise you could just say it is obvious and be done with it.
Edit: I have added what I think are the axioms and properties I'm using.
Best Answer
Suppose R>0 has an upper bound, let it be s. Then for all elements in R>0, say x, we have x < s. Since x > 0, and x < s, we have s > 0, and then s + 1 > 0. So s + 1 belong to R>0. However, s + 1 > s, we get contridiction with the definition of upper bound (every element in R>0 must be smaller than s, we find an element s + 1 which is > s).