Prove that ${x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$












0












$begingroup$


Let $(c_0,|cdot|_{infty})$ be the normed space of real sequences convergent to $0$, with the maximum norm. I need to prove that the subspace
$M = {x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$.



I am trying to find an $xin M$ such that, for a given $epsilon>0$ and a given $yin c_0$; $|x-y|_{infty}<epsilon$ , but I'm struggling with it. I have tried to construct this $x in M$ but since $y in c_0$ is arbitrary, I don't know how to deal with it. I suspect I may need to try it by contradiction, but so far I'm stuck.



Thanks in advance!






functional-analysis banach-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
    $endgroup$
    – tommy xu3
    Apr 13 '18 at 16:23
















0












$begingroup$


Let $(c_0,|cdot|_{infty})$ be the normed space of real sequences convergent to $0$, with the maximum norm. I need to prove that the subspace
$M = {x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$.



I am trying to find an $xin M$ such that, for a given $epsilon>0$ and a given $yin c_0$; $|x-y|_{infty}<epsilon$ , but I'm struggling with it. I have tried to construct this $x in M$ but since $y in c_0$ is arbitrary, I don't know how to deal with it. I suspect I may need to try it by contradiction, but so far I'm stuck.



Thanks in advance!






functional-analysis banach-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
    $endgroup$
    – tommy xu3
    Apr 13 '18 at 16:23














0












0








0


1



$begingroup$


Let $(c_0,|cdot|_{infty})$ be the normed space of real sequences convergent to $0$, with the maximum norm. I need to prove that the subspace
$M = {x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$.



I am trying to find an $xin M$ such that, for a given $epsilon>0$ and a given $yin c_0$; $|x-y|_{infty}<epsilon$ , but I'm struggling with it. I have tried to construct this $x in M$ but since $y in c_0$ is arbitrary, I don't know how to deal with it. I suspect I may need to try it by contradiction, but so far I'm stuck.



Thanks in advance!






functional-analysis banach-spaces







share|cite|improve this question











$endgroup$




Let $(c_0,|cdot|_{infty})$ be the normed space of real sequences convergent to $0$, with the maximum norm. I need to prove that the subspace
$M = {x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$.



I am trying to find an $xin M$ such that, for a given $epsilon>0$ and a given $yin c_0$; $|x-y|_{infty}<epsilon$ , but I'm struggling with it. I have tried to construct this $x in M$ but since $y in c_0$ is arbitrary, I don't know how to deal with it. I suspect I may need to try it by contradiction, but so far I'm stuck.



Thanks in advance!






functional-analysis banach-spaces




functional-analysis banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 8:30









davyjones

393213




393213










asked Apr 13 '18 at 16:05









KerguelenKerguelen

32




32












  • $begingroup$
    Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
    $endgroup$
    – tommy xu3
    Apr 13 '18 at 16:23


















  • $begingroup$
    Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
    $endgroup$
    – tommy xu3
    Apr 13 '18 at 16:23
















$begingroup$
Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
$endgroup$
– tommy xu3
Apr 13 '18 at 16:23




$begingroup$
Since $y_n$ converges to 0, for $n$ large enough $|y_n|$ is small. Then wiggle $y_n$ for small $n$ with error less than $epsilon$ to make the sum of the new one $0.$
$endgroup$
– tommy xu3
Apr 13 '18 at 16:23










2 Answers
2






active

oldest

votes


















3












$begingroup$

Since $lim_{ntoinfty}y_n=0$, there is an $N$ such that $|y_n|<epsilon/2$ for all $nge N$. Let $S=sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<epsilon/2$. Let
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very nice construction !
    $endgroup$
    – Gabriel Romon
    Apr 13 '18 at 16:32



















0












$begingroup$

For each $i=1,2ldots$, let $e_{i}=(0,ldots,0,1,0,ldots)$ (i.e.,
the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}in c_{0}$.
Denote $A={xin c_{0}midsum_{i=1}^{infty}x_{i}=0}$. We show
that for each $iinmathbb{N}$, $e_{i}inbar{A}$, where $bar{A}$
denotes the closure of $A$ with respect to the $||cdot||_{infty}$-topology.
Without loss of generality, we consider the case that $i=1$. Let
$epsilonin(0,1)$ be arbitrary. Let $a=1-epsilon/2$. Choose $n$
be sufficiently large such that $a/n<frac{epsilon}{2}$. Define
$x=(a,-frac{a}{n},ldots,-frac{a}{n},0,0,ldots)$, where $-frac{a}{n}$
appears $n$ times. Clearly $xin A$ and $||x-e_{1}||_infty<epsilon$.
This shows that $e_{i}inbar{A}$. Note that $A$ is a vector subspace
of $c_{0}$ and hence $bar{A}$ is also a vector subspace of $c_{0}$.
Now $c_{c}=spanleft{ e_{i}mid iinmathbb{N}right} $ and hence
$c_{c}subseteqbar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$,
so $c_{0}=overline{c_{c}}subseteqbar{A}$.






share|cite|improve this answer











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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

    oldest

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    3












    $begingroup$

    Since $lim_{ntoinfty}y_n=0$, there is an $N$ such that $|y_n|<epsilon/2$ for all $nge N$. Let $S=sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<epsilon/2$. Let
    $$
    x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Very nice construction !
      $endgroup$
      – Gabriel Romon
      Apr 13 '18 at 16:32
















    3












    $begingroup$

    Since $lim_{ntoinfty}y_n=0$, there is an $N$ such that $|y_n|<epsilon/2$ for all $nge N$. Let $S=sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<epsilon/2$. Let
    $$
    x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Very nice construction !
      $endgroup$
      – Gabriel Romon
      Apr 13 '18 at 16:32














    3












    3








    3





    $begingroup$

    Since $lim_{ntoinfty}y_n=0$, there is an $N$ such that $|y_n|<epsilon/2$ for all $nge N$. Let $S=sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<epsilon/2$. Let
    $$
    x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
    $$






    share|cite|improve this answer









    $endgroup$



    Since $lim_{ntoinfty}y_n=0$, there is an $N$ such that $|y_n|<epsilon/2$ for all $nge N$. Let $S=sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<epsilon/2$. Let
    $$
    x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 13 '18 at 16:23









    Julián AguirreJulián Aguirre

    68.1k24094




    68.1k24094












    • $begingroup$
      Very nice construction !
      $endgroup$
      – Gabriel Romon
      Apr 13 '18 at 16:32


















    • $begingroup$
      Very nice construction !
      $endgroup$
      – Gabriel Romon
      Apr 13 '18 at 16:32
















    $begingroup$
    Very nice construction !
    $endgroup$
    – Gabriel Romon
    Apr 13 '18 at 16:32




    $begingroup$
    Very nice construction !
    $endgroup$
    – Gabriel Romon
    Apr 13 '18 at 16:32











    0












    $begingroup$

    For each $i=1,2ldots$, let $e_{i}=(0,ldots,0,1,0,ldots)$ (i.e.,
    the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}in c_{0}$.
    Denote $A={xin c_{0}midsum_{i=1}^{infty}x_{i}=0}$. We show
    that for each $iinmathbb{N}$, $e_{i}inbar{A}$, where $bar{A}$
    denotes the closure of $A$ with respect to the $||cdot||_{infty}$-topology.
    Without loss of generality, we consider the case that $i=1$. Let
    $epsilonin(0,1)$ be arbitrary. Let $a=1-epsilon/2$. Choose $n$
    be sufficiently large such that $a/n<frac{epsilon}{2}$. Define
    $x=(a,-frac{a}{n},ldots,-frac{a}{n},0,0,ldots)$, where $-frac{a}{n}$
    appears $n$ times. Clearly $xin A$ and $||x-e_{1}||_infty<epsilon$.
    This shows that $e_{i}inbar{A}$. Note that $A$ is a vector subspace
    of $c_{0}$ and hence $bar{A}$ is also a vector subspace of $c_{0}$.
    Now $c_{c}=spanleft{ e_{i}mid iinmathbb{N}right} $ and hence
    $c_{c}subseteqbar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$,
    so $c_{0}=overline{c_{c}}subseteqbar{A}$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      For each $i=1,2ldots$, let $e_{i}=(0,ldots,0,1,0,ldots)$ (i.e.,
      the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}in c_{0}$.
      Denote $A={xin c_{0}midsum_{i=1}^{infty}x_{i}=0}$. We show
      that for each $iinmathbb{N}$, $e_{i}inbar{A}$, where $bar{A}$
      denotes the closure of $A$ with respect to the $||cdot||_{infty}$-topology.
      Without loss of generality, we consider the case that $i=1$. Let
      $epsilonin(0,1)$ be arbitrary. Let $a=1-epsilon/2$. Choose $n$
      be sufficiently large such that $a/n<frac{epsilon}{2}$. Define
      $x=(a,-frac{a}{n},ldots,-frac{a}{n},0,0,ldots)$, where $-frac{a}{n}$
      appears $n$ times. Clearly $xin A$ and $||x-e_{1}||_infty<epsilon$.
      This shows that $e_{i}inbar{A}$. Note that $A$ is a vector subspace
      of $c_{0}$ and hence $bar{A}$ is also a vector subspace of $c_{0}$.
      Now $c_{c}=spanleft{ e_{i}mid iinmathbb{N}right} $ and hence
      $c_{c}subseteqbar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$,
      so $c_{0}=overline{c_{c}}subseteqbar{A}$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        For each $i=1,2ldots$, let $e_{i}=(0,ldots,0,1,0,ldots)$ (i.e.,
        the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}in c_{0}$.
        Denote $A={xin c_{0}midsum_{i=1}^{infty}x_{i}=0}$. We show
        that for each $iinmathbb{N}$, $e_{i}inbar{A}$, where $bar{A}$
        denotes the closure of $A$ with respect to the $||cdot||_{infty}$-topology.
        Without loss of generality, we consider the case that $i=1$. Let
        $epsilonin(0,1)$ be arbitrary. Let $a=1-epsilon/2$. Choose $n$
        be sufficiently large such that $a/n<frac{epsilon}{2}$. Define
        $x=(a,-frac{a}{n},ldots,-frac{a}{n},0,0,ldots)$, where $-frac{a}{n}$
        appears $n$ times. Clearly $xin A$ and $||x-e_{1}||_infty<epsilon$.
        This shows that $e_{i}inbar{A}$. Note that $A$ is a vector subspace
        of $c_{0}$ and hence $bar{A}$ is also a vector subspace of $c_{0}$.
        Now $c_{c}=spanleft{ e_{i}mid iinmathbb{N}right} $ and hence
        $c_{c}subseteqbar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$,
        so $c_{0}=overline{c_{c}}subseteqbar{A}$.






        share|cite|improve this answer











        $endgroup$



        For each $i=1,2ldots$, let $e_{i}=(0,ldots,0,1,0,ldots)$ (i.e.,
        the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}in c_{0}$.
        Denote $A={xin c_{0}midsum_{i=1}^{infty}x_{i}=0}$. We show
        that for each $iinmathbb{N}$, $e_{i}inbar{A}$, where $bar{A}$
        denotes the closure of $A$ with respect to the $||cdot||_{infty}$-topology.
        Without loss of generality, we consider the case that $i=1$. Let
        $epsilonin(0,1)$ be arbitrary. Let $a=1-epsilon/2$. Choose $n$
        be sufficiently large such that $a/n<frac{epsilon}{2}$. Define
        $x=(a,-frac{a}{n},ldots,-frac{a}{n},0,0,ldots)$, where $-frac{a}{n}$
        appears $n$ times. Clearly $xin A$ and $||x-e_{1}||_infty<epsilon$.
        This shows that $e_{i}inbar{A}$. Note that $A$ is a vector subspace
        of $c_{0}$ and hence $bar{A}$ is also a vector subspace of $c_{0}$.
        Now $c_{c}=spanleft{ e_{i}mid iinmathbb{N}right} $ and hence
        $c_{c}subseteqbar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$,
        so $c_{0}=overline{c_{c}}subseteqbar{A}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '18 at 16:34

























        answered Apr 13 '18 at 16:27









        Danny Pak-Keung ChanDanny Pak-Keung Chan

        2,34138




        2,34138






























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