Every Cauchy sequence of real numbers converges

[kingwinner]

Homework Statement

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 n_k≥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 n_k≥N?
Why is it true that n≥N => |a_n-a_{n_k}| < ε/2 ?
And why is it true that n≥N => |a_{n_k}-L| < ε/2 ?
Can somebody kindly explain the last two lines of the proof in more detail?

Homework Equations

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The Attempt at a Solution

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Any help is much appreciated!

 

[HallsofIvy]

Homework Statement

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 n_k≥N are satisfied, then the last line is true. But why does the condition n≥N alone implies that the last line is true? Does n≥N guarantee that k≥K

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 n_k is a sequence that increases without bound so there exist numbers in the subsequence such that n_k> N. Of all the n_k> N, choose one for which k is also > K.

, and n_k≥N?
Why is it true that n≥N => |a_n-a_{n_k}| < ε/2 ?
And why is it true that n≥N => |a_{n_k}-L| < ε/2 ?
Can somebody kindly explain the last two lines of the proof in more detail?

Homework Equations

N/A

The Attempt at a Solution

N/A

Any help is much appreciated!

 

[kingwinner]

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

Looking at the definition of a_n->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 n_k≥N which does not even appear in |a_n-L| < ε . How come? Please help...I really don't understand :(

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

 

[vela]

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.

 

[kingwinner]

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.

But what is that "N" in this case?

 

[vela]

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

 

[kingwinner]

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

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

 

[vela]

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

 

[kingwinner]

I've seen another proof of this theorem:

Given ε>0.
There exists N s.t. m,n≥N => |a_n-a_m|<ε/2
There exists M s.t. k≥M => |a_nk-L|<ε/2 and also M≥N.
Take N'=max{N,n_M}
Then if n>N',
|a_n-L|≤|a_n-a_nM|+|a_nM-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,n_M}?

Does anyone have any idea?

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