Cauchy sequence and subsequence [closed]
"Let ${x_n}subset U$ be a Cauchy sequence. Give a direct proof that if a subsequence ${x_{nk}}subset S$ has a limit $L$, then the Cauchy sequence ${x_n}subset U$ has $L$ as a limit. Do not assume that general Cauchy sequences are convergent."
Ok I know a lot of people hate helping others with homework when they feel that the person asking the question hasn't worked on the problem. I have spent about 30 minutes going through all my theorems from my textbook on Cauchy sequences and I don't know where to start. Please help me
proof-writing cauchy-sequences
edited Dec 3 '18 at 2:07
Boshu
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
closed as off-topic by Did, Abcd, amWhy, Cesareo, José Carlos Santos Dec 25 '18 at 0:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Abcd, amWhy, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
"Let ${x_n}subset U$ be a Cauchy sequence. Give a direct proof that if a subsequence ${x_{nk}}subset S$ has a limit $L$, then the Cauchy sequence ${x_n}subset U$ has $L$ as a limit. Do not assume that general Cauchy sequences are convergent."
Ok I know a lot of people hate helping others with homework when they feel that the person asking the question hasn't worked on the problem. I have spent about 30 minutes going through all my theorems from my textbook on Cauchy sequences and I don't know where to start. Please help me
proof-writing cauchy-sequences
edited Dec 3 '18 at 2:07
Boshu
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
closed as off-topic by Did, Abcd, amWhy, Cesareo, José Carlos Santos Dec 25 '18 at 0:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Abcd, amWhy, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12
add a comment |
"Let ${x_n}subset U$ be a Cauchy sequence. Give a direct proof that if a subsequence ${x_{nk}}subset S$ has a limit $L$, then the Cauchy sequence ${x_n}subset U$ has $L$ as a limit. Do not assume that general Cauchy sequences are convergent."
Ok I know a lot of people hate helping others with homework when they feel that the person asking the question hasn't worked on the problem. I have spent about 30 minutes going through all my theorems from my textbook on Cauchy sequences and I don't know where to start. Please help me
proof-writing cauchy-sequences
edited Dec 3 '18 at 2:07
Boshu
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
"Let ${x_n}subset U$ be a Cauchy sequence. Give a direct proof that if a subsequence ${x_{nk}}subset S$ has a limit $L$, then the Cauchy sequence ${x_n}subset U$ has $L$ as a limit. Do not assume that general Cauchy sequences are convergent."
Ok I know a lot of people hate helping others with homework when they feel that the person asking the question hasn't worked on the problem. I have spent about 30 minutes going through all my theorems from my textbook on Cauchy sequences and I don't know where to start. Please help me
proof-writing cauchy-sequences
proof-writing cauchy-sequences
edited Dec 3 '18 at 2:07
Boshu
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
edited Dec 3 '18 at 2:07
Boshu
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
edited Dec 3 '18 at 2:07
Boshu
705315
edited Dec 3 '18 at 2:07
Boshu
705315
edited Dec 3 '18 at 2:07
Boshu
705315
705315
asked Dec 3 '18 at 1:54
Sam Cole
72
asked Dec 3 '18 at 1:54
Sam Cole
72
asked Dec 3 '18 at 1:54
Sam Cole
72
72
closed as off-topic by Did, Abcd, amWhy, Cesareo, José Carlos Santos Dec 25 '18 at 0:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Abcd, amWhy, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Abcd, amWhy, Cesareo, José Carlos Santos Dec 25 '18 at 0:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Abcd, amWhy, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12
add a comment |
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12
add a comment |
3 Answers
3
active
oldest
votes
Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.
answered Dec 3 '18 at 2:18
AnyAD
2,098812
add a comment |
The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|leqdfrac{epsilon}{2}$, and $n,m$ such that $|x_n-x_m|leq dfrac{epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then
$$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.
answered Dec 3 '18 at 2:13
Boshu
705315
add a comment |
Fix $epsilon>0.$ $x_{n_k}rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<epsilon/2$ for $kgeq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<epsilon/2$ for $m,ngeq n_0.$ Take $N=max{n_0,n_{k_0}}.$ Then for $ngeq N,$ $d(x_n,L)leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<epsilon.$ Hence $x_nrightarrow L.$
answered Dec 3 '18 at 2:15
John_Wick
1,366111
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.
answered Dec 3 '18 at 2:18
AnyAD
2,098812
add a comment |
Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.
answered Dec 3 '18 at 2:18
AnyAD
2,098812
add a comment |
Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.
answered Dec 3 '18 at 2:18
AnyAD
2,098812
Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.
answered Dec 3 '18 at 2:18
AnyAD
2,098812
answered Dec 3 '18 at 2:18
AnyAD
2,098812
answered Dec 3 '18 at 2:18
AnyAD
2,098812
answered Dec 3 '18 at 2:18
AnyAD
2,098812
2,098812
add a comment |
add a comment |
The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|leqdfrac{epsilon}{2}$, and $n,m$ such that $|x_n-x_m|leq dfrac{epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then
$$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.
answered Dec 3 '18 at 2:13
Boshu
705315
add a comment |
The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|leqdfrac{epsilon}{2}$, and $n,m$ such that $|x_n-x_m|leq dfrac{epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then
$$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.
answered Dec 3 '18 at 2:13
Boshu
705315
add a comment |
The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|leqdfrac{epsilon}{2}$, and $n,m$ such that $|x_n-x_m|leq dfrac{epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then
$$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.
answered Dec 3 '18 at 2:13
Boshu
705315
The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|leqdfrac{epsilon}{2}$, and $n,m$ such that $|x_n-x_m|leq dfrac{epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then
$$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.
answered Dec 3 '18 at 2:13
Boshu
705315
answered Dec 3 '18 at 2:13
Boshu
705315
answered Dec 3 '18 at 2:13
Boshu
705315
answered Dec 3 '18 at 2:13
Boshu
705315
705315
add a comment |
add a comment |
Fix $epsilon>0.$ $x_{n_k}rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<epsilon/2$ for $kgeq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<epsilon/2$ for $m,ngeq n_0.$ Take $N=max{n_0,n_{k_0}}.$ Then for $ngeq N,$ $d(x_n,L)leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<epsilon.$ Hence $x_nrightarrow L.$
answered Dec 3 '18 at 2:15
John_Wick
1,366111
add a comment |
Fix $epsilon>0.$ $x_{n_k}rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<epsilon/2$ for $kgeq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<epsilon/2$ for $m,ngeq n_0.$ Take $N=max{n_0,n_{k_0}}.$ Then for $ngeq N,$ $d(x_n,L)leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<epsilon.$ Hence $x_nrightarrow L.$
answered Dec 3 '18 at 2:15
John_Wick
1,366111
add a comment |
Fix $epsilon>0.$ $x_{n_k}rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<epsilon/2$ for $kgeq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<epsilon/2$ for $m,ngeq n_0.$ Take $N=max{n_0,n_{k_0}}.$ Then for $ngeq N,$ $d(x_n,L)leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<epsilon.$ Hence $x_nrightarrow L.$
answered Dec 3 '18 at 2:15
John_Wick
1,366111
Fix $epsilon>0.$ $x_{n_k}rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<epsilon/2$ for $kgeq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<epsilon/2$ for $m,ngeq n_0.$ Take $N=max{n_0,n_{k_0}}.$ Then for $ngeq N,$ $d(x_n,L)leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<epsilon.$ Hence $x_nrightarrow L.$
answered Dec 3 '18 at 2:15
John_Wick
1,366111
answered Dec 3 '18 at 2:15
John_Wick
1,366111
answered Dec 3 '18 at 2:15
John_Wick
1,366111
answered Dec 3 '18 at 2:15
John_Wick
1,366111
1,366111
add a comment |
add a comment |
Hint: You want to show that the cauchy sequence is close to $L$ from some point onwards. How can you do that using the closeness you get from the subsequence?
– NL1992
Dec 3 '18 at 2:05
is that using tails?
– Sam Cole
Dec 3 '18 at 2:08
You have that $|x_{n_k}-L|<epsilon$ for all $k>K$ and $|x_n-x_m|<epsilon$ for all $n,m>N$. How can you translate that for $|x_n-L|<epsilon$?
– NL1992
Dec 3 '18 at 2:12