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Okay so I am having trouble understanding this proof. What is really bugging me is how the Archimedean principle is referenced. How does one know that this is what is needed to proceed and how is it even relevant? I do not see the significance of N > 1/epsilon. Can someone please explain? Thank you.

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    Because you cannot prove the statement without the Archimedean principle. – 2017-02-21
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    Are you sure you don't mean the Archimedean *property* of the reals? – 2017-02-21
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    If you rewrite it as $N \epsilon > 1$, it may look familiar. For every $\epsilon \in \mathbb R_+$ no matter how small, exists $N \in \mathbb N$ such that $N \epsilon > a$ for some $a \in \mathbb R_+$. Just pick $a=1$ – 2017-02-21

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Well, to prove that $1/n\to 0$, you need to prove that for any $\epsilon>0$ there exists $N$ such that if $n\geq N$ then $|1/n|<\epsilon$ (this is just the definition of a limit). Since $1/n$ is positive, this just means you have to prove that $1/n<\epsilon$ for all sufficiently large $n$. Taking reciprocals, this is equivalent to $n>1/\epsilon$.

So in order to prove that $1/n\to 0$, you exactly need to know that if $n$ is a sufficiently large natural number, then $n>1/\epsilon$ (for any fixed positive $\epsilon$). This is exactly what the Archimedean principle says (every real number is less than some natural number), so that's why you invoke it.

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    Okay but if I choose ϵ = 1/000 and n = 4, so x_n = f(n) = 1/4, then |1/4| < 1/1000. But this is not true. At this point N seems arbitrary. I really do not see how everything is connected. It would make more sense if the definition was |f(n) - a|< N ϵ. – 2017-02-21
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    You don't need to know that $|1/n|<\epsilon$ for _all_ $n$. You just need to know that for some $N$, $|1/n|<\epsilon$ for all $n$ such that $n\geq N$. You get to choose $N$ depending on $\epsilon$. And in this case the Archimedean principle tells you exactly that it is possible to choose an $N$ that works (since what you need is that $N>1/\epsilon$). – 2017-02-21
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So, the Archimedean property essentially uses the fact that $\mathbb N \subseteq \mathbb R$ is an unbounded subset.

If for each $r \in \mathbb R$, there is a natural number $n$ so that $n-1 \leq r

To see the statement:

First, notice that $(1/n)_{n \in \mathbb N}$ is a decreasing sequence, so it will suffice to show that there exists some $N \in \mathbb N$ large enough so that $1/N<\epsilon$ which is equivalent to finding $N>1/\epsilon$. This can be done, since the natural numbers are unbounded.

This allows us to say that $(1/n) \to 0$, since for our choice of $\epsilon>0$, we can find some $N \in \mathbb N$ so that $1/k<\epsilon$ for all $k \geq N$.