Subsequence of a cauchy sequence in R

[autre]

Homework Statement

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

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 $\frac{\epsilon}{2^{j+1}}$.

 

[Dick]

Homework Statement

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

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 $\frac{\epsilon}{2^{j+1}}$.

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 $|a_{k_{j}}-a_{k_{j+1}}|<\frac{\epsilon}{2^{j+1}}$

 

[autre]

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 $|a_{k_{j}}-a_{k_{j+1}}|<\frac{\epsilon}{2^{j+1}}$

Sorry, typo.

 

[Dick]

Sorry, typo.

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 $k_1$?

 

[autre]

How about this --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, $\exists N_{j+1}>N_{j} s.t. \forall m,n>N_{j+1},|a_{m}-a_{n}|<\frac{\epsilon}{2^{j+1}},k_{j}=m$.

 

[Dick]

How about this --


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, $\exists N_{j+1}>N_{j} s.t. \forall m,n>N_{j+1},|a_{m}-a_{n}|<\frac{\epsilon}{2^{j+1}},k_{j}=m$.

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.

 

[autre]

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.

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

 

[Dick]

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

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 $k_{2}$?

 

[autre]

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 $k_{2}$?

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?

 

[Dick]

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?

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 $j_2$?

 

[autre]

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 $j_2$?

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

 

[Dick]

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

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

 

[autre]

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

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?

 

[Dick]

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?

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 $k_1$.