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.

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

Re: Cauchy sequence in Q not converging to zero.

Out of the definition of a zero sequence it follows there exists a number wherefore:

(this is the negation of the definition of a zero sequence because it's supposed to be no zero sequence)

Let and because is a cauchy sequence there exists a natural number wherefore:

either

Now choose a term wherefore at the same time and .

You can consider two cases:

(1) so ... ?

(2) so ... ?

Finish it ...

Re: Cauchy sequence in Q not converging to zero.