Prove that ${x=(x_n) in c_0 mid sum_{n=1}^infty x_n = 0}$ is dense in $c_0$
$begingroup$
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
edited Dec 2 '18 at 8:30
davyjones
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
$endgroup$
add a comment |
$begingroup$
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
edited Dec 2 '18 at 8:30
davyjones
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 2 '18 at 8:30
davyjones
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
$endgroup$
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
edited Dec 2 '18 at 8:30
davyjones
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
edited Dec 2 '18 at 8:30
davyjones
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
edited Dec 2 '18 at 8:30
davyjones
393213
edited Dec 2 '18 at 8:30
davyjones
393213
edited Dec 2 '18 at 8:30
davyjones
393213
393213
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
asked Apr 13 '18 at 16:05
KerguelenKerguelen
32
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Very nice construction !
$endgroup$
– Gabriel Romon
Apr 13 '18 at 16:32
add a comment |
$begingroup$
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 $c_{0}=overline{c_{c}}subseteqbar{A}$.
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Very nice construction !
$endgroup$
– Gabriel Romon
Apr 13 '18 at 16:32
add a comment |
$begingroup$
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
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Very nice construction !
$endgroup$
– Gabriel Romon
Apr 13 '18 at 16:32
add a comment |
$begingroup$
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
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
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
$$
x=(y_1,dots,y_N,underbrace{-frac{S}{M},dots,-frac{S}{M}}_{Mtext{ times}},0,0,dots)
$$
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
answered Apr 13 '18 at 16:23
Julián AguirreJulián Aguirre
68.1k24094
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $c_{0}=overline{c_{c}}subseteqbar{A}$.
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
$begingroup$
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 $c_{0}=overline{c_{c}}subseteqbar{A}$.
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
add a comment |
$begingroup$
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 $c_{0}=overline{c_{c}}subseteqbar{A}$.
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
$endgroup$
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 $c_{0}=overline{c_{c}}subseteqbar{A}$.
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
edited Apr 13 '18 at 16:34
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
answered Apr 13 '18 at 16:27
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,34138
2,34138
add a comment |
add a comment |
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$begingroup$
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