# Homework Help: Subsequence of a cauchy sequence in R

1. Jan 13, 2012

### autre

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

If $\{a_{n}\}\in\mathbb{R}$ is Cauchy, $\forall\epsilon>0,\exists$ a subsequence $\{a_{k_{j}}\}$ so that $|a_{k_{j}}-a_{k_{j+1}}|<\frac{\epsilon}{2^{j+1}}$.

3. The attempt at a solution

Since $\{a_{k_{j}}\}$ is Cauchy,$\forall\epsilon>0$,$\exists N_{\epsilon}$ such that for $j,j+1\geq N_{\epsilon}$,$|a_{k_{j}}-a_{k+1}|<\epsilon$.

I can't figure out how to incorporate $\frac{\epsilon}{2^{j+1}}$.

Last edited: Jan 13, 2012
2. Jan 13, 2012

### Dick

I don't think $|a_{k_{j}}-a_{k+1}|<\frac{\epsilon}{2^{j+1}}$ makes much sense. $k_{j}$ is a sequence of integers. k itself doesn't have a value. I think you mean $|a_{k_{j}}-a_{k_{j+1}}|<\frac{\epsilon}{2^{j+1}}$

3. Jan 13, 2012

### autre

Sorry, typo.

4. Jan 13, 2012

### Dick

Start with j=1. Then there is an N such that for all m,n>=N, $|a_m-a_n|<\epsilon/4$. Any idea how to chose a value for $k_1$?

5. Jan 16, 2012

### autre

Since $\{a_{n}\}$ is Cauchy, $\exists N_{1}>N s.t. \forall m,n>N_{1}, |a_{m}-a_{n}|<\frac{\epsilon}{4},k_{1}=m$. Continuing j+1 times, $\exists N_{j+1}>N_{j} s.t. \forall m,n>N_{j+1},|a_{m}-a_{n}|<\frac{\epsilon}{2^{j+1}},k_{j}=m$.

6. Jan 16, 2012

### Dick

How can you put e.g. $k_{1}=m$?? It was $\forall m>N$. m doesn't have a definite value. Are you sure you understand that? It kind of looks like you saw a solution and copied it in a somewhat garbled form.

7. Jan 16, 2012

### autre

You've got me there, I'm still confused as to how to approach this. Any clues at to how I might arrive at $k_{1}$?

8. Jan 16, 2012

### Dick

Why not pick $k_{1}=N_1$? Then all of the elements of the sequence with indices greater than $N_1$ are within $\epsilon/4$ of $a_{N_1}$. Do you agree? Don't write a bunch of symbols down if you don't understand what they mean. Try and explain it in words. It would be great if you could actually understand this. Any idea how to pick $k_{2}$?

9. Jan 17, 2012

### autre

Intuitively, I see {${a_{k_j}}$} converging to a point in R, so the higher $N$ the closer we get to that point. Then, we can can arbitrarily divide distance $\epsilon$ by 2,2^2,...2^j+1 as long as $N_{2},...N_{j+1}$ is sufficiently larger than the previous $N_{i}$, i = 1,...j+1.

Since {${a_{k_j}}$} is Cauchy, there exists ϵ>0 and N such that for all m,n>=N, |a_m−a_n|<ϵ/4. Since ϵ/4=ϵ/2^1+1, it is implicit that you are making j=1. Looking at the subsequence, that means you have |a_k1−a_k2|<ϵ/4. Now you are trying to move further along that what you did previously, so you would set a new $N$ that would make you move closer to the limit?

10. Jan 17, 2012

### Dick

Yes, now there is an $N_2>N_1$ such that $|a_m-a_n|<\epsilon/8$ for $m,n>N_2$, right? Any suggestions for picking $j_2$?

11. Jan 17, 2012

### autre

Do you mean $k_2$? Wouldn't $N_2$ work?

12. Jan 17, 2012

### Dick

Yes and yes, now can you define $k_3$?

13. Jan 17, 2012

### autre

I think I've got it:

Start with j=1. Then there is an N such that for all m,n>=N, |am−an|<ϵ/4. Since {$a_{{k}_{j}}$} is Cauchy, there exists an $N_2$=$k_2$ s.t. for all m,n>=$N_2$, |am−an|<ϵ/2^3. Continuing, there exists an $N_j=k_j$ s.t. for all m,n>=$N_j$, |am−an|<ϵ/2^(j+1). Does that look about right?

14. Jan 17, 2012

### Dick

It's 'about' right. It's still a little hazy around the edges, but it's a big improvement. The idea behind the whole proof is to use that the sequence ${a_n}$ (not the subsequence ${a_{k_j}}$) is Cauchy to pick the N's. Then use those N's to define the $k_j$ of the subsequence. Say so. Don't just write $N_j=k_j$ and assume that they'll know that you are intending to define $k_j$ that way. And you'll want to make sure that you pick $N_1<N_2<N_3<...$. And you forgot to define $k_1$.