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&gt;N_1 such that |a_m-a_n|&lt;\epsilon/8 for m,n&gt;N_2, right? Any suggestions for picking j_2?

 

[autre]

Yes, now there is an N_2&gt;N_1 such that |a_m-a_n|&lt;\epsilon/8 for m,n&gt;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&lt;N_2&lt;N_3&lt;.... And you forgot to define k_1.