A sequence $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Separability of a product metric spaceProve equivalent metric spacesConcept of Cauchy sequenceProve that all three metrics induces the same topology on $X_1times X_2$If $(X,d_1)$ and $(X,d_2)$ two connected metric spaces if only if $Xtimes Y$ is connected metric spaceCompletion of two metric spacesProve that an open ball in $(X,d)$ is the product of open balls from $X_1, X_2, ldots, X_n$.If $d_1(x,y) leq d_2(x,y)$ and all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.$x_n rightarrow x$ iff the modified sequence is CauchyLet $tau$ denote the product topology on $X=X_1 times X_2$. Then $(X,tau)$ is a separable space.

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A sequence $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Separability of a product metric spaceProve equivalent metric spacesConcept of Cauchy sequenceProve that all three metrics induces the same topology on $X_1times X_2$If $(X,d_1)$ and $(X,d_2)$ two connected metric spaces if only if $Xtimes Y$ is connected metric spaceCompletion of two metric spacesProve that an open ball in $(X,d)$ is the product of open balls from $X_1, X_2, ldots, X_n$.If $d_1(x,y) leq d_2(x,y)$ and all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.$x_n rightarrow x$ iff the modified sequence is CauchyLet $tau$ denote the product topology on $X=X_1 times X_2$. Then $(X,tau)$ is a separable space.










3












$begingroup$


Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.



Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$



But can we generalize this result?



Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?



I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?



As far as I know, being Cauchy is a property of metric, so the result may not hold in general.



Can some one please clarify? Are there some counterexamples.



Thanks a lot.










share|cite|improve this question











$endgroup$











  • $begingroup$
    @KajHansen Thanks.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 12:22















3












$begingroup$


Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.



Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$



But can we generalize this result?



Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?



I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?



As far as I know, being Cauchy is a property of metric, so the result may not hold in general.



Can some one please clarify? Are there some counterexamples.



Thanks a lot.










share|cite|improve this question











$endgroup$











  • $begingroup$
    @KajHansen Thanks.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 12:22













3












3








3


0



$begingroup$


Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.



Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$



But can we generalize this result?



Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?



I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?



As far as I know, being Cauchy is a property of metric, so the result may not hold in general.



Can some one please clarify? Are there some counterexamples.



Thanks a lot.










share|cite|improve this question











$endgroup$




Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.



Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$



But can we generalize this result?



Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?



I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?



As far as I know, being Cauchy is a property of metric, so the result may not hold in general.



Can some one please clarify? Are there some counterexamples.



Thanks a lot.







real-analysis general-topology






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share|cite|improve this question













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share|cite|improve this question








edited Apr 8 at 12:21









Kaj Hansen

27.8k43980




27.8k43980










asked Apr 8 at 11:20









StammeringMathematicianStammeringMathematician

2,7951324




2,7951324











  • $begingroup$
    @KajHansen Thanks.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 12:22
















  • $begingroup$
    @KajHansen Thanks.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 12:22















$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22




$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22










3 Answers
3






active

oldest

votes


















3












$begingroup$

If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Does the backward implication hold, under some conditions ?
    $endgroup$
    – spkakkar
    Apr 8 at 11:30










  • $begingroup$
    What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:33










  • $begingroup$
    Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
    $endgroup$
    – Mark
    Apr 8 at 11:45



















2












$begingroup$

Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Thanks. I got it. Under what conditions does the result hold true? Thanks
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:34


















1












$begingroup$

When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Does the backward implication hold, under some conditions ?
      $endgroup$
      – spkakkar
      Apr 8 at 11:30










    • $begingroup$
      What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:33










    • $begingroup$
      Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
      $endgroup$
      – Mark
      Apr 8 at 11:45
















    3












    $begingroup$

    If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Does the backward implication hold, under some conditions ?
      $endgroup$
      – spkakkar
      Apr 8 at 11:30










    • $begingroup$
      What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:33










    • $begingroup$
      Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
      $endgroup$
      – Mark
      Apr 8 at 11:45














    3












    3








    3





    $begingroup$

    If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.






    share|cite|improve this answer









    $endgroup$



    If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 8 at 11:28









    MarkMark

    10.5k1622




    10.5k1622











    • $begingroup$
      Does the backward implication hold, under some conditions ?
      $endgroup$
      – spkakkar
      Apr 8 at 11:30










    • $begingroup$
      What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:33










    • $begingroup$
      Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
      $endgroup$
      – Mark
      Apr 8 at 11:45

















    • $begingroup$
      Does the backward implication hold, under some conditions ?
      $endgroup$
      – spkakkar
      Apr 8 at 11:30










    • $begingroup$
      What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:33










    • $begingroup$
      Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
      $endgroup$
      – Mark
      Apr 8 at 11:45
















    $begingroup$
    Does the backward implication hold, under some conditions ?
    $endgroup$
    – spkakkar
    Apr 8 at 11:30




    $begingroup$
    Does the backward implication hold, under some conditions ?
    $endgroup$
    – spkakkar
    Apr 8 at 11:30












    $begingroup$
    What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:33




    $begingroup$
    What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:33












    $begingroup$
    Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
    $endgroup$
    – Mark
    Apr 8 at 11:45





    $begingroup$
    Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
    $endgroup$
    – Mark
    Apr 8 at 11:45












    2












    $begingroup$

    Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Thanks. I got it. Under what conditions does the result hold true? Thanks
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:34















    2












    $begingroup$

    Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Thanks. I got it. Under what conditions does the result hold true? Thanks
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:34













    2












    2








    2





    $begingroup$

    Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...






    share|cite|improve this answer









    $endgroup$



    Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 8 at 11:28









    5xum5xum

    92.6k394162




    92.6k394162







    • 1




      $begingroup$
      Thanks. I got it. Under what conditions does the result hold true? Thanks
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:34












    • 1




      $begingroup$
      Thanks. I got it. Under what conditions does the result hold true? Thanks
      $endgroup$
      – StammeringMathematician
      Apr 8 at 11:34







    1




    1




    $begingroup$
    Thanks. I got it. Under what conditions does the result hold true? Thanks
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:34




    $begingroup$
    Thanks. I got it. Under what conditions does the result hold true? Thanks
    $endgroup$
    – StammeringMathematician
    Apr 8 at 11:34











    1












    $begingroup$

    When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
    $d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
    $$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
    For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
      $d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
      $$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
      For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
        $d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
        $$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
        For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.






        share|cite|improve this answer









        $endgroup$



        When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
        $d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
        $$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
        For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 8 at 12:33









        uniquesolutionuniquesolution

        9,5471823




        9,5471823



























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