Cauchy sequence in Q not converging to zero.

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March 7th 2012, 01:27 AM
Arturo_026

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.
March 7th 2012, 05:21 PM
psolaki

Re: Cauchy sequence in Q not converging to zero.

Quote:

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
March 8th 2012, 12:17 PM
Siron

Re: Cauchy sequence in Q not converging to zero.

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

Finish it ...
March 8th 2012, 03:09 PM
Plato

Re: Cauchy sequence in Q not converging to zero.

Quote:

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 $\sigma\in\mathbb{R}$ such that $\sigma\ne 0$ .
Therefore there is a neighborhood of $\sigma$ such that $\exists N\in\mathbb{Z}^+$ such that all $s_n$ has
the same sign for $n\ge N$ .