# Thread: Cauchy sequence in Q not converging to zero.

1. ## Cauchy sequence in Q not converging to zero.

I have the following exercise:

Let s_n be a cauchy sequence in Q(rationals) not converging to 0. Show that there exists an e(epsilon) >0 and a natural number N such that either for all n>N, s_n > e or for all n>N, -s_n >e.

I know that since Q is not complete, we cannot assume that there exists a point (say s) such that s_n coverges to it since this s could well be in the real numbers.

I am hessitant with the the answer I came up with since i didnt use the fact that s_n is cauchy. What i did is the following:

From the fact that s_n does not converge to zero, then we can deduce that: there's an e>0 and N such that for all n>N, d(s_n, 0) is not less than e. so this implies that d(s_n, 0) >= e. So absolute value of (s_n - 0) > e in implies that s_n>e or -s_n>e. Now i need to show that s_n=e is not possible, but as of right now i havent been able to do so.

Any advice and guidence will be greatly appreciated.
Thank you very much.

2. ## Re: Cauchy sequence in Q not converging to zero.

Originally Posted by Arturo_026

From the fact that s_n does not converge to zero, then we can deduce that: there's an e>0 and N such that for all n>N, d(s_n, 0) is not less than e. so this implies that d(s_n, 0) >= e. So absolute value of (s_n - 0) > e in implies that s_n>e or -s_n>e..

The above is wrong

3. ## Re: Cauchy sequence in Q not converging to zero.

Out of the definition of a zero sequence it follows there exists a number $\displaystyle e>0$ wherefore:
$\displaystyle \forall n \in \mathbb{N}, \exists p>n: |s_p|>e$
(this is the negation of the definition of a zero sequence because it's supposed to be no zero sequence)
Let $\displaystyle \epsilon=\frac{e}{2}$ and because $\displaystyle s_n$ is a cauchy sequence there exists a natural number $\displaystyle N$ wherefore:
$\displaystyle \forall p,q \geq N: |s_p-q_q|<\epsilon$
either
$\displaystyle \forall p,q \geq N: s_p-\epsilon<s_q<s_p+\epsilon$
Now choose a term $\displaystyle s_m$ wherefore at the same time $\displaystyle m\geq N$ and $\displaystyle |s_m|>e=2\epsilon$.
You can consider two cases:
(1)$\displaystyle s_m>0$ so ... ?
(2)$\displaystyle s_m<0$ so ... ?

Finish it ...

4. ## Re: Cauchy sequence in Q not converging to zero.

Originally Posted by Arturo_026
Let s_n be a cauchy sequence in Q(rationals) not converging to 0. Show that there exists an e(epsilon) >0 and a natural number N such that either for all n>N, s_n > e or for all n>N, -s_n >e.
I know that since Q is not complete, we cannot assume that there exists a point (say s) such that s_n coverges to it since this s could well be in the real numbers.
The real numbers are still complete.
So your sequence still converges to some $\displaystyle \sigma\in\mathbb{R}$ such that $\displaystyle \sigma\ne 0$.
Therefore there is a neighborhood of $\displaystyle \sigma$ such that $\displaystyle \exists N\in\mathbb{Z}^+$ such that all $\displaystyle s_n$ has
the same sign for $\displaystyle n\ge N$.