# Homework Help: Cauchy sequence in Q not converging to zero.

1. Mar 7, 2012

### arturo_026

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. Mar 7, 2012

### Dick

Here's why you need Cauchy. Define s_n to be 1 if n is even and 1/n if n is odd. I.e. {1,1,1/3,1,1/5,1,1/7,1...}. s_n does not converge to zero, but it doesn't avoid any neighborhood of zero either. And it's not Cauchy.

3. Mar 7, 2012

### arturo_026

I see that makes sense. Thank you Dick
Now, as far as my semi-complete proof goes, is it correct? and how could I implement the fact that s_n is cauchy in the proof?
Thank you again

4. Mar 7, 2012

### Dick

Start by getting the correct statement that corresponds to "s_n does not converge to 0". Write down the definition of s_n converges to 0 and negate it. Carefully. You didn't get it right the first time.

5. Mar 7, 2012

### arturo_026

Thank you for your patience Dick.

For all ε>0, there exists a natural number N such that n≥N implies that abs. value of (s_n - 0) < ε.

Now I tried negating every part but it doesn't seem to be right.
What seems somewhat correct is that the negation will read:
There exists an ε>0, for all natural numbers N such that n≥N implies that abs. value of (s_n - 0) ≥ ε

If this is correct, will s_N=ε? so that way we take n=N out of the final statement.

6. Mar 7, 2012

### Dick

Almost correct. Change it to this:
There exists an ε>0 such that for all natural numbers N THERE EXISTS an n≥N such that abs. value of (s_n - 0) ≥ ε

In other words, there is an infinite subsequence s_k of s_n that satisfies |s_k|> ε. That's makes sense, doesn't it?

7. Mar 7, 2012

### arturo_026

ok, so now if I replace ≤ by < then I'm able to include all n greater then N and abs. value (s_n) thus becomes only > ε. And this is it?

8. Mar 7, 2012

### Dick

Well, no! You want to say that ALL of the points with n greater than some N are farther from 0 than some ε, so far you only have an infinite sequence of them. That's where being Cauchy comes in.

9. Mar 7, 2012

### arturo_026

Yes, now it's clear.
Thank you so much Dick, I think I got it!