# Cauchy sequences and sequences in general

1. Nov 12, 2008

### JG89

Is every sequence that converges a Cauchy sequence (in that for every e > 0, there is an integer N such that |a_n - a_m| < e whenever n,m > N)?

I think it is because if a sequence a_n converges to L, then you can mark off an open interval of any size about L such that this interval contains all of a_n except for at most a finite amount. So as the length of this open interval decreases then the distance between each point a_n gets closer and closer together. In fact maybe this open interval would be the epsilon neighborhood, because for two points a_n and a_m that are contained in this interval, we have |a_n - a_m| < e. Then within this epsilon neighborhood, we have points, a_j and a_p, where j,p > n, and j,p > m which are even closer to the limit, L, and again we have |a_j - a_p| < e, for an even smaller value of epsilon. We can obviously continue in this pattern, taking epsilon smaller and smaller as the integer n > N for the a_n gets larger and larger. And so in general, for large enough n and m, we have |a_n - a_m| < e.

Am I correct?

Last edited: Nov 12, 2008
2. Nov 12, 2008

### nicksauce

3. Nov 12, 2008

### JG89

I have no idea what a metric space is. I've heard about it in Real Analysis, but I'm right now only learning theoretical calculus/basic analysis.

4. Nov 12, 2008

### nicksauce

Don't worry about it too much... As far as you are probably concerned, yes Cauchy sequences are the same as convergent sequences.

5. Nov 12, 2008

### Office_Shredder

Staff Emeritus
Even in other metric spaces, convergent sequences are always Cauchy. This is just a two line proof using the triangle inequality:

If an converges to a, for all e there exists N such that n>N implies |a-an|<e/2 So if n,m>N|an - am| = |an - a + a - am| <= |a-an| + |a-am| < e/2 + e/2 = e Hence you have (an) is Cauchy

6. Nov 12, 2008

### HallsofIvy

Staff Emeritus
All convergent sequence are Cauchy. In the real numbers or Rn all Cauchy sequences converge. However, in the set, Q, of all rational numbers, Cauchy sequences do not necessarily converge. For example, the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, ... (each term one more decimal place in the decimal expansion of $\pi$) is a sequence of rational numbers since every term is a terminating decimal. It is Cauchy since, given any $\epsilon> 0$, there exist n such that $10^n< \epsilon$. All terms in the sequence past the nth place are the same in the first n decimal places so their difference is less than $10^n< \epsilon$. The sequence clearly converges to $\pi$ which is not a rational number and so does not converge in the rational numbers.

7. Nov 12, 2008

### JG89

Thanks for the replies!

8. Nov 13, 2008

### mathwonk

In a convergent sequence, all the entries eventually get close to some point L. In a cauchy sequence they eventually get close to each other.

as mentioned, by the triangle inequality, things that are close to L are almost as close to each other. but things that are getting close to each other may not be converging to any limit, because the space could have a hole in it where the limit should be.

just take any sequence of non zero numbers, like 1/n, that converges to zero. then remove zero from the space. the sequence is still cauchy but no longer convergent to an element of the smaller space.

conversely, every metric space can be enlarged by adding in all potential limits of cauchy sequences, so that afterwards all cauchy sequences do converge. thats how you make the real numbers out of the rationals. you take the ring of all cauchy sequences of rationals (that is a ring since sums and products of cauchy sequences are still cauchy), and mod out by the ideal of cauchy sequences that converge to zero.

the quotient ring is the reals.