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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.

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Seems right, if you know that $a\gt b$ implies $a+s\gt b+s$ for any real number $s$... – Arturo Magidin Mar 21 '12 at 21:19
Since this is such a basic exercise, I guess that you need to prove it using the axioms of the real number system. It doesn't need to 'feel' right. You just have to prove everything by using only the blocks of building the real number system. – Beni Bogosel Mar 21 '12 at 21:21
@Arturo: I didn't say it, but I can use the basic axioms of the real numbers. – Javier Mar 21 '12 at 21:31
"The basic axioms" may vary depending on your source. You should check your implications and justify them by explicit invocation of the relevant axioms. – Arturo Magidin Mar 21 '12 at 21:32
Shouldn't the fact that $\mathbb{N} \subset \mathbb{R}$ suffice? – Pedro Tamaroff Mar 22 '12 at 0:44

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