$(x_n)_{n in mathbb{N}}$ is cauchy if and only if $d(x_{n_k},x_{m_k}) rightarrow 0$ for any two sequences...












1












$begingroup$


I want to know if the following assertion is correct:




If $(M,d)$ is a metric space, then $(x_n)_{n in mathbb{N}}$ is cauchy if and only if $d(x_{n_k},x_{m_k}) rightarrow 0$ for any two sequences of natural numbers $n_k, m_k rightarrow infty$




Of course the direction from left to right is true. But I am unsure of the direction from right to left. Can anyone give a proof or state a counterexample?



Thanks a lot in advance!






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
    $endgroup$
    – Ingix
    Oct 23 '18 at 23:00










  • $begingroup$
    Oh yes, that was a typo. Do you know how to check if this is true?
    $endgroup$
    – vaoy
    Oct 23 '18 at 23:05










  • $begingroup$
    Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
    $endgroup$
    – Kavi Rama Murthy
    Oct 23 '18 at 23:50
















1












$begingroup$


I want to know if the following assertion is correct:




If $(M,d)$ is a metric space, then $(x_n)_{n in mathbb{N}}$ is cauchy if and only if $d(x_{n_k},x_{m_k}) rightarrow 0$ for any two sequences of natural numbers $n_k, m_k rightarrow infty$




Of course the direction from left to right is true. But I am unsure of the direction from right to left. Can anyone give a proof or state a counterexample?



Thanks a lot in advance!






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
    $endgroup$
    – Ingix
    Oct 23 '18 at 23:00










  • $begingroup$
    Oh yes, that was a typo. Do you know how to check if this is true?
    $endgroup$
    – vaoy
    Oct 23 '18 at 23:05










  • $begingroup$
    Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
    $endgroup$
    – Kavi Rama Murthy
    Oct 23 '18 at 23:50














1












1








1





$begingroup$


I want to know if the following assertion is correct:




If $(M,d)$ is a metric space, then $(x_n)_{n in mathbb{N}}$ is cauchy if and only if $d(x_{n_k},x_{m_k}) rightarrow 0$ for any two sequences of natural numbers $n_k, m_k rightarrow infty$




Of course the direction from left to right is true. But I am unsure of the direction from right to left. Can anyone give a proof or state a counterexample?



Thanks a lot in advance!






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$




I want to know if the following assertion is correct:




If $(M,d)$ is a metric space, then $(x_n)_{n in mathbb{N}}$ is cauchy if and only if $d(x_{n_k},x_{m_k}) rightarrow 0$ for any two sequences of natural numbers $n_k, m_k rightarrow infty$




Of course the direction from left to right is true. But I am unsure of the direction from right to left. Can anyone give a proof or state a counterexample?



Thanks a lot in advance!






real-analysis sequences-and-series analysis




real-analysis sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 23 '18 at 23:03







vaoy

















asked Oct 23 '18 at 22:52









vaoyvaoy

537210




537210








  • 1




    $begingroup$
    I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
    $endgroup$
    – Ingix
    Oct 23 '18 at 23:00










  • $begingroup$
    Oh yes, that was a typo. Do you know how to check if this is true?
    $endgroup$
    – vaoy
    Oct 23 '18 at 23:05










  • $begingroup$
    Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
    $endgroup$
    – Kavi Rama Murthy
    Oct 23 '18 at 23:50














  • 1




    $begingroup$
    I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
    $endgroup$
    – Ingix
    Oct 23 '18 at 23:00










  • $begingroup$
    Oh yes, that was a typo. Do you know how to check if this is true?
    $endgroup$
    – vaoy
    Oct 23 '18 at 23:05










  • $begingroup$
    Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
    $endgroup$
    – Kavi Rama Murthy
    Oct 23 '18 at 23:50








1




1




$begingroup$
I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
$endgroup$
– Ingix
Oct 23 '18 at 23:00




$begingroup$
I'd assume the sequence of the distances to tend to $0$ instead of $infty$ for a Cauchy series!
$endgroup$
– Ingix
Oct 23 '18 at 23:00












$begingroup$
Oh yes, that was a typo. Do you know how to check if this is true?
$endgroup$
– vaoy
Oct 23 '18 at 23:05




$begingroup$
Oh yes, that was a typo. Do you know how to check if this is true?
$endgroup$
– vaoy
Oct 23 '18 at 23:05












$begingroup$
Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
$endgroup$
– Kavi Rama Murthy
Oct 23 '18 at 23:50




$begingroup$
Just the negation of $(x_n)$ being Cauchy. You will get the implication from right to left easily.
$endgroup$
– Kavi Rama Murthy
Oct 23 '18 at 23:50










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$begingroup$

Since the right side is true for any sequences $n_k$ and $m_k$, simply take $n_k=k$ and $m_k=k+1$ and conclude.






share|cite|improve this answer









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    0












    $begingroup$

    Since the right side is true for any sequences $n_k$ and $m_k$, simply take $n_k=k$ and $m_k=k+1$ and conclude.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since the right side is true for any sequences $n_k$ and $m_k$, simply take $n_k=k$ and $m_k=k+1$ and conclude.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since the right side is true for any sequences $n_k$ and $m_k$, simply take $n_k=k$ and $m_k=k+1$ and conclude.






        share|cite|improve this answer









        $endgroup$



        Since the right side is true for any sequences $n_k$ and $m_k$, simply take $n_k=k$ and $m_k=k+1$ and conclude.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 18:47









        Mostafa AyazMostafa Ayaz

        14.8k3938




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