Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space











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20
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I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.



Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.



Thanks for helping me.






functional-analysis metric-spaces







share|cite|improve this question




















  • 5




    The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
    – Chris Eagle
    May 31 '12 at 22:54






  • 4




    Try the sequence $x_n = n$. Draw a picture.
    – copper.hat
    May 31 '12 at 23:00








  • 1




    You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
    – copper.hat
    May 31 '12 at 23:10








  • 1




    $d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
    – copper.hat
    May 31 '12 at 23:26








  • 2




    I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
    – copper.hat
    May 31 '12 at 23:37

















up vote
20
down vote

favorite
12












I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.



Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.



Thanks for helping me.






functional-analysis metric-spaces







share|cite|improve this question




















  • 5




    The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
    – Chris Eagle
    May 31 '12 at 22:54






  • 4




    Try the sequence $x_n = n$. Draw a picture.
    – copper.hat
    May 31 '12 at 23:00








  • 1




    You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
    – copper.hat
    May 31 '12 at 23:10








  • 1




    $d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
    – copper.hat
    May 31 '12 at 23:26








  • 2




    I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
    – copper.hat
    May 31 '12 at 23:37















up vote
20
down vote

favorite
12









up vote
20
down vote

favorite
12






12





I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.



Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.



Thanks for helping me.






functional-analysis metric-spaces







share|cite|improve this question















I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.



Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.



Thanks for helping me.






functional-analysis metric-spaces




functional-analysis metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '14 at 13:37









Martin Sleziak

44.6k7115269




44.6k7115269










asked May 31 '12 at 22:49









srijan

6,31563979




6,31563979








  • 5




    The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
    – Chris Eagle
    May 31 '12 at 22:54






  • 4




    Try the sequence $x_n = n$. Draw a picture.
    – copper.hat
    May 31 '12 at 23:00








  • 1




    You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
    – copper.hat
    May 31 '12 at 23:10








  • 1




    $d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
    – copper.hat
    May 31 '12 at 23:26








  • 2




    I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
    – copper.hat
    May 31 '12 at 23:37
















  • 5




    The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
    – Chris Eagle
    May 31 '12 at 22:54






  • 4




    Try the sequence $x_n = n$. Draw a picture.
    – copper.hat
    May 31 '12 at 23:00








  • 1




    You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
    – copper.hat
    May 31 '12 at 23:10








  • 1




    $d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
    – copper.hat
    May 31 '12 at 23:26








  • 2




    I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
    – copper.hat
    May 31 '12 at 23:37










5




5




The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54




The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54




4




4




Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00






Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00






1




1




You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10






You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10






1




1




$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26






$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26






2




2




I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37






I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37












3 Answers
3






active

oldest

votes

















up vote
9
down vote



accepted










Sorry for reviving such an old problem...



Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.






share|cite|improve this answer






























    up vote
    9
    down vote













    Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.






    share|cite|improve this answer



















    • 1




      I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
      – Jonathan
      Sep 29 '17 at 17:11




















    up vote
    6
    down vote













    Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.






    share|cite|improve this answer





















    • thanks i took sequence $x_n = n$ as suggested by copper.hat
      – srijan
      May 31 '12 at 23:33











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    9
    down vote



    accepted










    Sorry for reviving such an old problem...



    Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.






    share|cite|improve this answer



























      up vote
      9
      down vote



      accepted










      Sorry for reviving such an old problem...



      Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.






      share|cite|improve this answer

























        up vote
        9
        down vote



        accepted







        up vote
        9
        down vote



        accepted






        Sorry for reviving such an old problem...



        Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.






        share|cite|improve this answer














        Sorry for reviving such an old problem...



        Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 4 at 23:11









        Adam

        32




        32










        answered Dec 9 '14 at 12:50









        user149792

        4,06842467




        4,06842467






















            up vote
            9
            down vote













            Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.






            share|cite|improve this answer



















            • 1




              I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
              – Jonathan
              Sep 29 '17 at 17:11

















            up vote
            9
            down vote













            Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.






            share|cite|improve this answer



















            • 1




              I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
              – Jonathan
              Sep 29 '17 at 17:11















            up vote
            9
            down vote










            up vote
            9
            down vote









            Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.






            share|cite|improve this answer














            Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 21 at 19:05









            Yadati Kiran

            1,327418




            1,327418










            answered Feb 3 '13 at 10:48









            Robert Cardona

            5,20023398




            5,20023398








            • 1




              I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
              – Jonathan
              Sep 29 '17 at 17:11
















            • 1




              I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
              – Jonathan
              Sep 29 '17 at 17:11










            1




            1




            I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
            – Jonathan
            Sep 29 '17 at 17:11






            I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
            – Jonathan
            Sep 29 '17 at 17:11












            up vote
            6
            down vote













            Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.






            share|cite|improve this answer





















            • thanks i took sequence $x_n = n$ as suggested by copper.hat
              – srijan
              May 31 '12 at 23:33















            up vote
            6
            down vote













            Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.






            share|cite|improve this answer





















            • thanks i took sequence $x_n = n$ as suggested by copper.hat
              – srijan
              May 31 '12 at 23:33













            up vote
            6
            down vote










            up vote
            6
            down vote









            Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.






            share|cite|improve this answer












            Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 31 '12 at 22:58









            Cameron Buie

            84.6k771155




            84.6k771155












            • thanks i took sequence $x_n = n$ as suggested by copper.hat
              – srijan
              May 31 '12 at 23:33


















            • thanks i took sequence $x_n = n$ as suggested by copper.hat
              – srijan
              May 31 '12 at 23:33
















            thanks i took sequence $x_n = n$ as suggested by copper.hat
            – srijan
            May 31 '12 at 23:33




            thanks i took sequence $x_n = n$ as suggested by copper.hat
            – srijan
            May 31 '12 at 23:33


















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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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