# Cauchy Sequence for rational numbers (proof)

Im wokring on this proof and would be very gratefulf for help. I think I have the overarching idea behind the Cauchy. What confuses me is probably the notation. In the definition of Cauchy sequence there is this use of Epsilon. Here in the lemma I have to prove (below) there is use of delta.

Problem: Prove for alle positive rational numbers that

If $$\{a_n \}$$ is a rational Cauchy sequence which does not tend to 0, then there exist a positive $$\delta \in \mathbb{Q}$$ and $$n_0 \in \mathbb{N}$$ such that either

• $$a_n > \delta$$ for all $$n \geq n_0$$ or,
• $$a_n < -\delta$$ for alle $$n \geq n_0$$

This implies that $$a_n \neq 0$$ for all $$n > n_0$$

• Actually, the two options only imply that $a_n\neq\delta$ for all $n\geqslant n_0$. So it doesn't imply $a_n\neq0$ unless $\delta=0$ Apr 18, 2021 at 8:15
• Thank you. I have edited, so it is $- \delta$ in the second case Apr 18, 2021 at 8:20
• Think about the sequence $\{1+1/n\}$ and take $\delta$ to be $1+\epsilon$ and $1-\epsilon$ Apr 18, 2021 at 8:22

Take thes sequence $$a_n$$, then $$a_n\rightarrow0$$ if and only if for all $$\varepsilon>0$$ there is $$N\in\mathbb N$$ such that $$|a_n-0|=|a_n|<\varepsilon$$ for all $$n\geqslant N$$. If $$a_n$$ doesn't converge to $$0$$, then $$a_n$$ satisfy the negation of the last condition: There is $$\varepsilon>0$$ such that for all $$N\in\mathbb N$$ there is $$n\geqslant N$$ such that $$|a_n|\geqslant \varepsilon$$Here, use that $$a_n$$ is Cauchy, then for every $$\lambda>0$$, there is $$M\in\mathbb N$$ such that $$|a_p-a_q|<\lambda$$ for all $$p,q\geqslant M$$. Finally, using the last two results, there is $$M'\in\mathbb N$$ such that $$|a_m|=|a_n-(a_n-a_m)|\geqslant||a_n|-|a_n-a_m||\geqslant|\varepsilon-\lambda|$$for all $$m\geqslant M'$$ (the disequality is the reverse triangular inequality). Can you find appropriate $$\varepsilon,\lambda$$ and $$M'$$? (notice that the conclusion $$|x|\geqslant\mu\geqslant0$$ is equivalent to $$x\leqslant -\mu$$ or $$x\geqslant \mu$$)