Proof that $(1/n)_{n\in\mathbb N}$ does not converge in $(0, 1)$? Consider the metric space $((0, 1), |\cdot|)$. I'd like to show the sequence $(1/n)_{n\in\mathbb N}$ does not converge in $(0, 1)$. 
I wrote the following proof: Suppose $\lim_n 1/n =a\in (0, 1)$. By the archimedean property  there exists $N\in\mathbb N$ such that $$Na>1.$$
Now, corresponding to $\varepsilon:=a-1/N$ there exists $n_0\in\mathbb N$ such that $$\forall n\geq n_0\Rightarrow \left|\frac{1}{n}-a\right|<\varepsilon=a-1/N,$$
and therefore $$\forall n\geq n_0 \Rightarrow \frac{1}{n}\in \left( \frac{1}{N}, 2a-\frac{1}{N}\right).$$
But this is an absurd, for instance, if $n\geq \max\{N, n_0\}$ it follows $n\geq N$ and therefore $1/n\leq 1/N$ so that $$ \frac{1}{n}\not\in \left(\frac{1}{N}, 2a+\frac{1}{2N} \right)$$ although $n\geq n_0$.
I have two questions concerning this proof. 
(1) Is it correct?
(2) Is there a direct proof for that fact using the definition of convergence of sequences?
Thanks.
 A: (1) Your proof is correct, although one line where you wrote $$\forall n\geq n_0\Rightarrow\frac1n\in\left(\frac1N,2a+\frac1N\right)$$
should probably be
$$\forall n\geq n_0\Rightarrow\frac1n\in\left(\frac1N,2a-\frac1N\right).$$
Or at least, that's what is directly deduced from the previous line: the interval is of center $a$ and radius $a-\frac1N$.
(2) Regarding your second question: you are trying to prove that a sequence does not converge. In formal logic, if $P$ is some property, proving $\neg P$ ("$P$ is not the case") is equivalent to proving $P\rightarrow\bot$ ("$P$ leads to a contradiction"), where $\bot$ represents False, or contradiction. In fact, the symbol $\neg$ is superfluous, and you could replace it everywhere with $\rightarrow\bot$. So in your case, you are proving the result directly.
However, if you want to make the relationship with the original definition clearer, you can state the contraposition of the definition: a sequence does not converge if
$$\forall a\in(0,1),\;\exists\varepsilon>0,\;\forall N\geq0,\;\exists n\geq N,\;\left|\frac1n-a\right|\geq\varepsilon.$$
If you reread your proof, you'll find that you have proven exactly this!
A: Yes, your proof is correct.
Regarding your second question, your proof is good if you don't want to leave your metric space (otherwise it is a one-line proof).
Just to make it look simpler, you could simply say that let $a \in (0,1)$ be the limit of the sequence. Then take $\,\varepsilon = a/2.\;$ So, $\;\exists\, n_0 \ni \frac{1}{n_0} \lt a-\varepsilon \,= a/2$ (due to archimedean property). Then $\,\forall\, n \geq n_0,\; |a - \frac{1}{n}| \gt \varepsilon$.
Hence the sequence does not converge.
