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 $n > n_0$

  • 1
    $\begingroup$ 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 $\delta=0$ $\endgroup$
    – Alessandro
    Commented Apr 18, 2021 at 8:15
  • $\begingroup$ Thank you. I have edited, so it is $- \delta$ in the second case $\endgroup$
    – bestmate21
    Commented Apr 18, 2021 at 8:20
  • $\begingroup$ Think about the sequence $\{1+1/n\}$ and take $\delta$ to be $1+\epsilon$ and $1-\epsilon$ $\endgroup$
    – rostader
    Commented Apr 18, 2021 at 8:22

1 Answer 1


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 $x\geqslant \mu$)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .