# Homework Help: Every Cauchy sequence of real numbers converges

1. Jan 17, 2010

### kingwinner

1. The problem statement, all variables and given/known data

I understand everything except the last two lines. I am really confused about the last two lines of the proof. (actually I was never able to fully understand it since my first year calculus)
I agree that if ALL three of the conditions n≥N, k≥K, and nk≥N are satisfied, then the last line is true. But why does the condition n≥N alone imply that the last line is true? Does n≥N guarantee that k≥K, and nk≥N?
Why is it true that n≥N => |an-an_k| < ε/2 ?
And why is it true that n≥N => |an_k-L| < ε/2 ?
Can somebody kindly explain the last two lines of the proof in more detail?

2. Relevant equations
N/A

3. The attempt at a solution
N/A

Any help is much appreciated!

Last edited: Jan 17, 2010
2. Jan 17, 2010

### HallsofIvy

Knowing that $n\ge N$ doesn't tell you anything about k. That was why the proof says "Pick any $k\ge K$" k is chosen to be larger than K. That has nothing to do with n. What the proof is really saying is "K is a fixed number so certainly there exist k> K. Also N is a fixed number and nk is a sequence that increases without bound so there exist numbers in the subsequence such that nk> N. Of all the nk> N, choose one for which k is also > K.

3. Jan 17, 2010

### kingwinner

But at the end, we are supposed to prove that:
an->L
i.e. for all ε>0, there exists N such that n≥N => |an-L| < ε ?
(note that here only n appears, there is nothing about k)

Looking at the definition of an->L above, all we have to do is to construct an N that works. Just like every ε-limit proof, we have to find an N that works. And if we can find such an N, then the only restriction should be n≥N and nothing else, but in this proof we also have other restrictions k≥K and nk≥N which does not even appear in |an-L| < ε . How come? Please help...I really don't understand :(

Do we need all three conditions (n≥N, k≥K, and nk≥N) to be simultaneously satisfied in order for |an-L| < ε to hold??

4. Jan 17, 2010

### vela

Staff Emeritus
All you have to show is that given an $\epsilon$, an N exists. It doesn't really matter how you found it; you just have to show it exists.

5. Jan 17, 2010

### kingwinner

But what is that "N" in this case?

6. Jan 17, 2010

### vela

Staff Emeritus
It says what N is in the fifth sentence of the proof.

7. Jan 17, 2010

### kingwinner

So the exact same N there would work at the end?

Last edited: Jan 17, 2010
8. Jan 17, 2010

### vela

Staff Emeritus
Yes. That's the point of the rest of the proof.

9. Jan 18, 2010

### kingwinner

I've seen another proof of this theorem:

Given ε>0.
There exists N s.t. m,n≥N => |an-am|<ε/2
There exists M s.t. k≥M => |ank-L|<ε/2 and also M≥N.
Take N'=max{N,nM}
Then if n>N',
|an-L|≤|an-anM|+|anM-L|<ε/2+ε/2=ε
=====================

Is this proof right or wrong? If it's right, I have the following questions:

1) Why is M≥N?

2) Why should we take N'=max{N,nM}?

Does anyone have any idea?

(I increased the font size to make the subscripts legible)