Set of all sequences converging to $0$ forms a Banach Space.












1














Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.



Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
$$||u_{k}^{n}-u_{k}||<epsilon$$
for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
$$|u_k|leq epsilon$$
and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.



Is this argument correct?






real-analysis proof-verification banach-spaces







share|cite|improve this question





























    1














    Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.



    Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
    $$||u_{k}^{n}-u_{k}||<epsilon$$
    for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
    $$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
    $$|u_k|leq epsilon$$
    and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.



    Is this argument correct?






    real-analysis proof-verification banach-spaces







    share|cite|improve this question



























      1












      1








      1







      Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.



      Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
      $$||u_{k}^{n}-u_{k}||<epsilon$$
      for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
      $$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
      $$|u_k|leq epsilon$$
      and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.



      Is this argument correct?






      real-analysis proof-verification banach-spaces







      share|cite|improve this question















      Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.



      Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
      $$||u_{k}^{n}-u_{k}||<epsilon$$
      for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
      $$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
      $$|u_k|leq epsilon$$
      and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.



      Is this argument correct?






      real-analysis proof-verification banach-spaces




      real-analysis proof-verification banach-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 at 15:15









      José Carlos Santos

      148k22117218




      148k22117218










      asked Nov 25 at 15:01









      Hello_World

      3,83521630




      3,83521630






















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          Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:





          1. $(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.


          2. $(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.






          share|cite|improve this answer





















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            1 Answer
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            3














            Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:





            1. $(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.


            2. $(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.






            share|cite|improve this answer


























              3














              Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:





              1. $(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.


              2. $(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.






              share|cite|improve this answer
























                3












                3








                3






                Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:





                1. $(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.


                2. $(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.






                share|cite|improve this answer












                Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:





                1. $(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.


                2. $(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 15:12









                José Carlos Santos

                148k22117218




                148k22117218






























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