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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 $s$...2012-03-21
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    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.2012-03-21
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    @Arturo: I didn't say it, but I can use the basic axioms of the real numbers.2012-03-21
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    "The basic axioms" may vary depending on your source. You should check your implications and justify them by explicit invocation of the relevant axioms.2012-03-21
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    Shouldn't the fact that $\mathbb{N} \subset \mathbb{R}$ suffice?2012-03-22
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    @PeterT.off: Well, wouldn't you have to show that $\mathbb{N}$ is unbounded as well, using a similar proof?2012-03-22
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    @Javier I think the proof is very simple. Suppose $n$ is the largest element of $\mathbb{N}$. Then $n+1 \in \mathbb{N}$ and $n+1>n$.2012-03-22
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    @Pedro Tamaroff with your last comment you're only proving that $\Bbb N$ is unbounded above as a subset of $\Bbb N$. But you're not proving at all that $\Bbb N$ is unbounded above as a subset of $\Bbb R$2014-02-12
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    @Matteo Sure: if $x\in\Bbb R$ is such that $n\leqslant x$ for every $n\in\Bbb N$, take $\lfloor x\rfloor +1\in\Bbb N$. (Correcting myself after two years.)2014-02-12
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    @Pedro Tamaroff Yes, now I agree :) But also the existence of $\lfloor x\rfloor$ for a real number $x$ is not so basic. Since the property of $\Bbb N$ of being bounded above seems to be also as obvious as the existence of $\lfloor x\rfloor$, I don't think you should use the last to prove the first: you should use only the property of real numbers ($\Bbb R$ is the only complete ordered field). But maybe I'm wrong.2014-02-12
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    @Pedro Tamaroff In fact, to show the existence of $\lfloor x \rfloor$ for a real $x$ you need the completeness property of $\Bbb R$ (every non-empty bounded above set has a least upper bound); but the same property is the one you need to prove that $\Bbb N$ in unbounded above in $\Bbb R$. So it seems to me a circular reasoning to use the existence of $\lfloor x \rfloor$ to prove that $\Bbb N$ is unbounded above.2014-02-12
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    you are (obviously) seeing $\mathbb{R}_{\geq 0}$ as a subset of $\mathbb{R}$ right?2017-09-20

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