Cauchy sequence and subsequence











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"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







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  • 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 at 2:05










  • is that using tails?
    – Sam Cole
    Dec 3 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 at 2:12















up vote
2
down vote

favorite












"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







share|cite|improve this 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 at 2:05










  • is that using tails?
    – Sam Cole
    Dec 3 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 at 2:12













up vote
2
down vote

favorite









up vote
2
down vote

favorite











"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







share|cite|improve this question















"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






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edited Dec 3 at 2:07









Boshu

700315




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asked Dec 3 at 1:54









Sam Cole

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162












  • 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 at 2:05










  • is that using tails?
    – Sam Cole
    Dec 3 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 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 at 2:05










  • is that using tails?
    – Sam Cole
    Dec 3 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 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 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 at 2:05












is that using tails?
– Sam Cole
Dec 3 at 2:08




is that using tails?
– Sam Cole
Dec 3 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 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 at 2:12










3 Answers
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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 $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.






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    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



    $$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.






    share|cite|improve this answer




























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      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 $x_nrightarrow L.$






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        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 $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.






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          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 $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.






          share|cite|improve this answer























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            up vote
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            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 $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.






            share|cite|improve this answer












            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 $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|leq |x_n-(x_n)_k|+|(x_n)_k-L|<epsilon$.







            share|cite|improve this answer












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            share|cite|improve this answer










            answered Dec 3 at 2:18









            AnyAD

            2,098812




            2,098812






















                up vote
                1
                down vote













                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



                $$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  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



                  $$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    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



                    $$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.






                    share|cite|improve this answer












                    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



                    $$|x_n-L|leq|x_n-x_m|+|x_{n_k}-L|$$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 at 2:13









                    Boshu

                    700315




                    700315






















                        up vote
                        1
                        down vote













                        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 $x_nrightarrow L.$






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          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 $x_nrightarrow L.$






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            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 $x_nrightarrow L.$






                            share|cite|improve this answer












                            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 $x_nrightarrow L.$







                            share|cite|improve this answer












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                            share|cite|improve this answer










                            answered Dec 3 at 2:15









                            John_Wick

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