# Cauchy sequence in Q not converging to zero.

• arturo_026
In summary, Dick says that there is 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 he needs to show that s_n=e is not possible, but as of right now he hasn't been able to do so.
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 haven't been able to do so.

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

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.

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

arturo_026 said:
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

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.

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.

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?

Dick said:
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?

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?

arturo_026 said:
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?

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.

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

## 1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of rational numbers that gets arbitrarily close to each other as the sequence progresses. This means that for any small distance, there exists a point in the sequence where all subsequent terms are within that distance from each other.

## 2. What does it mean for a Cauchy sequence in Q to not converge to zero?

If a Cauchy sequence in Q does not converge to zero, it means that the sequence does not approach a limit of zero. In other words, the terms of the sequence do not get arbitrarily close to zero as the sequence progresses.

## 3. Why is a Cauchy sequence in Q not converging to zero significant?

This is significant because a fundamental property of rational numbers is that they are dense, meaning that between any two rational numbers, there exists an infinite number of other rational numbers. If a Cauchy sequence in Q does not converge to zero, it shows that there are some gaps in the rational numbers, and they are not a complete set.

## 4. Can a Cauchy sequence in Q converge to a number other than zero?

Yes, a Cauchy sequence in Q can converge to a number other than zero. The key is that the sequence must approach a limit, which can be any rational number. It does not have to be zero.

## 5. Why is it important to study Cauchy sequences in Q not converging to zero?

Studying Cauchy sequences in Q not converging to zero helps us understand the limitations of rational numbers and the concept of completeness. It also has significant implications in fields such as real analysis and topology, where the completeness of a set is crucial for certain theorems and proofs.

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