Prove that the sequence $\frac{1}{n}$ does not converge in $\Bbb R\setminus \lbrace 0\rbrace$

Question

The definition of convergence is:

$(\exists x\in \Bbb R\setminus \lbrace 0\rbrace)(\forall \epsilon>0)(\exists N\in \Bbb N)(n\ge N\Rightarrow |x_n-x|<\epsilon)$.

I need to prove its negation:

$(\forall x\in \Bbb R\setminus \lbrace 0\rbrace)(\exists \epsilon>0)(\forall N\in \Bbb N)(n\ge N\land |x_n-x|\ge \epsilon)$.

Now I need an explicit construction of $\epsilon$ for each $x\in \Bbb R\setminus \lbrace 0\rbrace$. Could someone please help? Thanks in advance!

Answer 1

Fix $x\in\mathbb{R}\backslash\{0\}$. By Archimedes' axiom, there exists a natural number $N_x$ such that $\frac{1}{N_x}<|x|$. Put $\varepsilon=|x|-\frac{1}{N_x}$. Subsequently, for every $n>N$, one has: $$|\frac{1}{n}-x|\geq |x|-\frac{1}{n}>|x|-\frac{1}{N_x}=\varepsilon$$ so $\{\frac{1}{n}\}$ does not converge to $x$.

Answer 2

Hint: Look up https://en.wikipedia.org/wiki/Archimedean_property. This might help.